Forums: Climbing Information: The Lab:
tension
RSS FeedRSS Feeds for The Lab

Premier Sponsor:

 


mtnrock


Oct 4, 2008, 2:19 PM
Post #1 of 123 (4977 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

tension
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

is tension on a rope doubled in a top rope system when its on a redirect belay or a direct from the anchor belay.

it seems like it should but im draawing a blank if not please say why


Lazlo


Oct 4, 2008, 2:27 PM
Post #2 of 123 (4971 views)
Shortcut

Registered: Nov 14, 2007
Posts: 5070

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

mtnrock wrote:
is tension on a rope doubled in a top rope system when its on a redirect belay or a direct from the anchor belay.

it seems like it should but im draawing a blank if not please say why

What? I don't get your question.

When you say tension, do you mean force?
When you say top rope, do you mean belaying your follower on multi-pitch?

I really don't mean to come off as a jerk...but when asking technical questions grammar is huge.


sungam


Oct 4, 2008, 2:28 PM
Post #3 of 123 (4969 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

You need to explain it better.
But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through.
Korekt me if ayem rong, but I do not b-leave aye am.


Lazlo


Oct 4, 2008, 2:30 PM
Post #4 of 123 (4965 views)
Shortcut

Registered: Nov 14, 2007
Posts: 5070

Re: [sungam] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

sungam wrote:
You need to explain it better.
But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through.
Korekt me if ayem rong, but I do not b-leave aye am.

Yer knot Rong. Yer Gamsum.


sungam


Oct 4, 2008, 2:32 PM
Post #5 of 123 (4962 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [Lazlo] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Lazlo wrote:
sungam wrote:
You need to explain it better.
But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through.
Korekt me if ayem rong, but I do not b-leave aye am.

Yer knot Rong. Yer Gamsum.
Eye can stil b rong.


Lazlo


Oct 4, 2008, 2:34 PM
Post #6 of 123 (4960 views)
Shortcut

Registered: Nov 14, 2007
Posts: 5070

Re: [sungam] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

sungam wrote:
Lazlo wrote:
sungam wrote:
You need to explain it better.
But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through.
Korekt me if ayem rong, but I do not b-leave aye am.

Yer knot Rong. Yer Gamsum.
Eye can stil b rong.

New eww kan't.


Lazlo


Oct 4, 2008, 2:35 PM
Post #7 of 123 (4957 views)
Shortcut

Registered: Nov 14, 2007
Posts: 5070

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Sorry for hi-jacking, OP. I'll play nice now.


sungam


Oct 4, 2008, 2:38 PM
Post #8 of 123 (4951 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [Lazlo] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Lazlo wrote:
Sorry for hi-jacking, OP. I'll play nice now.
I was just thinking "should I reply, or is it going too far?"
Probably the latter...
Anyways, yeah, explain the Q farther unless my answer that I answered answered the question that you questioned us with.


Lazlo


Oct 4, 2008, 3:23 PM
Post #9 of 123 (4932 views)
Shortcut

Registered: Nov 14, 2007
Posts: 5070

Re: [sungam] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

OP'z prob'ly thinkin' "Crap, the two guyz I didn' wan' respondin' responded!"


sungam


Oct 4, 2008, 3:26 PM
Post #10 of 123 (4927 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [Lazlo] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Lazlo wrote:
OP'z prob'ly thinkin' "Crap, the two guyz I didn' wan' respondin' responded!"
LaughLaugh
Heh, I bet "meh" is having an anal period over this thread.


hafilax


Oct 4, 2008, 3:32 PM
Post #11 of 123 (4917 views)
Shortcut

Registered: Dec 11, 2007
Posts: 3025

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

To start let's ignore friction.

If you are belaying off of the anchor then the tension in the rope and the force on the anchor is equal to the force exerted by the person at the end.

[edited to change:]
With a redirect belay (and no friction) the tension is equal to the force exerted by the belayer plus that of the climber which is approximately equal to the force on the anchor.

With a redirect belay the tension on the rope is still the same since the climber is not accelerated but the belayer end of the rope has the same tension as well. This puts twice the force on the anchor.
[/edit]

Geometry and friction modify this force. I think biners are like 60% efficient as pulleys so it only requires about 60% of the force of the climber to hold her.

I hope that's what you were asking. Majid brings this up all the time so I'm sure he's proud of you for asking.


(This post was edited by hafilax on Oct 7, 2008, 2:16 PM)


sungam


Oct 4, 2008, 3:37 PM
Post #12 of 123 (4911 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [hafilax] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Are you sure the tension increases?
Maybe I'm getting confused, it is late, but I'm thinking the force on the rope can't change, as then the climber would be pulled up, no?
The force on the anchor would... aw naw! now I'm as confused as the OP!


hafilax


Oct 4, 2008, 3:41 PM
Post #13 of 123 (4909 views)
Shortcut

Registered: Dec 11, 2007
Posts: 3025

Re: [sungam] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.


rightarmbad


Oct 5, 2008, 1:58 AM
Post #14 of 123 (4847 views)
Shortcut

Registered: Mar 21, 2005
Posts: 217

Re: [hafilax] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Will depend if it is a hanging belay or not.


USnavy


Oct 5, 2008, 5:33 AM
Post #15 of 123 (4824 views)
Shortcut

Registered: Nov 5, 2007
Posts: 2658

Re: [hafilax] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (1 rating)  
Can't Post

hafilax wrote:
When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.

That’s poor logic. In fact a fall straight onto an anchor produces more force on the anchor then when the rope runs through the anchor to the belayer. When you fall the belayer is lifted off the ground which increases the fall distance and decreases the peak load on the quickdraw / cam / anchor.

Belaying straight off the anchor for a second will produce more force on the anchor then doing a redirectional into your harness.


(This post was edited by USnavy on Oct 5, 2008, 5:35 AM)


joeforte


Oct 5, 2008, 7:32 AM
Post #16 of 123 (4805 views)
Shortcut

Registered: May 9, 2005
Posts: 1092

Re: [USnavy] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

USnavy wrote:
hafilax wrote:
When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.

That’s poor logic. In fact a fall straight onto an anchor produces more force on the anchor then when the rope runs through the anchor to the belayer. When you fall the belayer is lifted off the ground which increases the fall distance and decreases the peak load on the quickdraw / cam / anchor.

Belaying straight off the anchor for a second will produce more force on the anchor then doing a redirectional into your harness.

No, he's about right... What USnavy is saying is bad logic. You are assuming slack is developed in the belay, causing a fall before the rope gets taught. If a top belay is kept tight, it will experience HALF as much force as a toprope, or slingshot belay.

Add any slack, and the dynamics of the belay come into play, and the forces change. It's much harder to give a dynamic belay off the anchor, especially with an auto-locking device, so it is very important to keep any slack out of the rope.


Partner angry


Oct 5, 2008, 7:54 AM
Post #17 of 123 (4799 views)
Shortcut

Registered: Jul 21, 2003
Posts: 8405

Re: [joeforte] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.


milesenoell


Oct 5, 2008, 10:06 AM
Post #18 of 123 (4765 views)
Shortcut

Registered: Sep 19, 2006
Posts: 1156

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

I can't believe I'm going to say this, but this is where Majid's diagrams help to make it simple. Force on the anchor from the climber falling is easy to grok, so we'll just think about the other forces that can come into play.

"is tension on a rope doubled in a top rope system when its on a redirect belay?"

Sort of. Replace "tension on the rope" with "force on the anchor". Now a portion of the belayers body weight is also added to the downward force, but only to the extent required to counter-balance the falling climber. How much force that is can be tricky to nail down due to variables like the friction and angles in the redirect/slingshot belay system. 60% seems like a reasonable estimate, but is far from a reliable constant. Is the rope new and slick or all beat to shit and fuzzy like mine? Is the blayer right under the anchor or off to the side some?


"is tension on a rope doubled in a top rope system when its ... direct from the anchor belay."

No. Because the belay device attached to to the anchor catches the falling climber and holds only the force of the falling climber.


milesenoell


Oct 5, 2008, 10:08 AM
Post #19 of 123 (4761 views)
Shortcut

Registered: Sep 19, 2006
Posts: 1156

Re: [milesenoell] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Did that answer your question or just the question I turned yours into?


jt512


Oct 5, 2008, 10:11 AM
Post #20 of 123 (4762 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [angry] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight — that is, a little more than triple the climber's weight.

Jay


Lazlo


Oct 5, 2008, 10:14 AM
Post #21 of 123 (4756 views)
Shortcut

Registered: Nov 14, 2007
Posts: 5070

Re: [USnavy] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

USnavy wrote:
Belaying straight off the anchor for a second will produce more force on the anchor then doing a redirectional into your harness.

That'a not what my Bible says.

Either Freedom or Climbing Anchors says the opposite. I think it's Climbing Anchors.

(I only know what I know. Don't fault me for what I don't know...or what I read and blindly believedTongue.)


Partner rgold


Oct 5, 2008, 10:37 AM
Post #22 of 123 (4738 views)
Shortcut

Registered: Dec 3, 2002
Posts: 1800

Re: [Lazlo] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 has given a clear and correct explanation. USNavy, I'm afraid, is misreading his bibles.

For those who are puzzled by the rope tension going up to a (momentary) peak load of twice the climber's weight, remember that there is still rope stretch when a climber just hangs on the rope, the moment of maximum stretch produces the peak tension in the rope, and that peak tension turns out to be twice the weight hung on the rope.

A redirected top-rope anchor has to be able to withstand far more than triple body weight to be even remotely safe, since most belayed falls with have small but non-zero fall factors.


jt512


Oct 5, 2008, 12:14 PM
Post #23 of 123 (4710 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [rgold] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

rgold wrote:
jt512 has given a clear and correct explanation.

Well, I had good material¹ to work from.

In reply to:
A redirected top-rope anchor has to be able to withstand far more than triple body weight to be even remotely safe, since most belayed falls with have small but non-zero fall factors.

Which leads to the question: How strong should a toprope anchor be to be more than just remotely safe?

We saw, above, that when a 200-lb climber takes a toprope fall, the peak force on the anchor will be at least 10/3 times his weight, or 667 lb. Since some climbers weigh more than 200 lb, and most toprope falls will have fall factors greater than zero, let's round this number up to 1000 lb. Now multiply by a safety factor of 3, and I claim that a toprope anchor should be built to withstand a force of at least 3000 lb.

Jay

Ref:
¹Goldstone R. The Standard Equation for Impact Force.


chriss


Oct 5, 2008, 12:19 PM
Post #24 of 123 (4707 views)
Shortcut

Registered: Jul 13, 2004
Posts: 92

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

mtnrock wrote:
is tension on a rope doubled in a top rope system when its on a redirect belay or a direct from the anchor belay.

it seems like it should but im draawing a blank if not please say why

I have read through the responses listed above and am confused. I must not understand the question. What are you calling a "redirect belay"? Is that the belayer on the ground where the climber began and the rope through a single overhead anchor?

If you tie the rope at the anchor and hang from it. It has your weight as tension. If you double the rope through the anchor and it is able to slide. Then hang from 1 end you will fall. So, you tie off the other side to counter your weight. The rope has the same tension as before. You just moved the tie off place and have twice as much rope in play.

This does assume no movement. Just a static situation. Falling on the rope and movement of the rope complicates this.

But 10/3 times the climbers weight? What Bible is that?


chris


colatownkid


Oct 5, 2008, 12:22 PM
Post #25 of 123 (4703 views)
Shortcut

Registered: Nov 27, 2007
Posts: 512

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight — that is, a little more than triple the climber's weight.

Jay

where does the 2/3 come from? is this an empirical value derived from research somewhere (in which case i'd like to read the link if you remember where to find it)? or is the 2/3 a value that can be determined by math and physics? in that case would you mind posting the math or PMing me?

also, why is the climber's force double the static weight? clearly the climber must realistically undergo some acceleration and therefore an increase in force, but why is it double? is this value research based or math based?

edited to add:
if this is all explained in the above link to Rich's paper, then never mind.


(This post was edited by colatownkid on Oct 5, 2008, 12:26 PM)


sungam


Oct 5, 2008, 12:42 PM
Post #26 of 123 (2310 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [hafilax] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

hafilax wrote:
When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.
Simplest view. Cunning. The force on the anchors is higher, but has the tension increased?
Perhaps yes...


jt512


Oct 5, 2008, 12:48 PM
Post #27 of 123 (2309 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [colatownkid] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

colatownkid wrote:
jt512 wrote:
angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight — that is, a little more than triple the climber's weight.

Jay

where does the 2/3 come from? is this an empirical value derived from research somewhere (in which case i'd like to read the link if you remember where to find it)? or is the 2/3 a value that can be determined by math and physics? in that case would you mind posting the math or PMing me?

In his paper, Richard refers to it as a "convention" to account for the reduction in force on the belayer's side of the rope due to friction between the rope and the carabiner. I presume that it is based on empirical data.

In reply to:
also, why is the climber's force double the static weight? clearly the climber must realistically undergo some acceleration and therefore an increase in force, but why is it double? is this value research based or math based?

The peak force of twice the climber's weight for a fall-factor-0 fall is a consequence of using Hooke's Law to describe rope stretch. As you'll see, Richard provides two separate derivations of this result.

Jay


colatownkid


Oct 5, 2008, 12:54 PM
Post #28 of 123 (2305 views)
Shortcut

Registered: Nov 27, 2007
Posts: 512

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
colatownkid wrote:
jt512 wrote:
angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight — that is, a little more than triple the climber's weight.

Jay

where does the 2/3 come from? is this an empirical value derived from research somewhere (in which case i'd like to read the link if you remember where to find it)? or is the 2/3 a value that can be determined by math and physics? in that case would you mind posting the math or PMing me?

In his paper, Richard refers to it as a "convention" to account for the reduction in force on the belayer's side of the rope due to friction between the rope and the carabiner. I presume that it is based on empirical data.

In reply to:
also, why is the climber's force double the static weight? clearly the climber must realistically undergo some acceleration and therefore an increase in force, but why is it double? is this value research based or math based?

The peak force of twice the climber's weight for a fall-factor-0 fall is a consequence of using Hooke's Law to describe rope stretch. As you'll see, Richard provides two separate derivations of this result.

Jay

just finished reading it; thanks.


Partner rgold


Oct 5, 2008, 12:55 PM
Post #29 of 123 (2303 views)
Shortcut

Registered: Dec 3, 2002
Posts: 1800

Re: [colatownkid] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Colatown, the 2/3 value for carabiner efficiency is an empirically derived value that has been stated over and over in various places, and now has the status of "folklore." I don't know the original reference. I have seen 1/2 stated, but much less often. The fact is that the number will be different for different rope-carabiner combinations and probably, like other friction coefficients, has a sliding and a static value.

The fact that the peak tension in a rope with applied weight W is 2W is a simple conservation of energy consequence of Hooke's Law, viewed as a reasonable approximation to the behavior of a climbing rope. (Yes, there is some acceleration involved, since a climber who weights a rope will drop as the rope stretches.) The result assumes that the climber's weight is instantaneously applied to the climbing rope with no slack in the rope. Using various strategies to ease the weight onto the rope could, in theory, keep the peak rope tension no greater than the climber's weight, but this would not apply to catching a top-rope fall.

Whether the real result is more or less than 2W would be a matter for experiment. I'd guess a little less than 2W because of the internal damping effects of rope construction, but given the manifold uncertainties and variations in real-life practice, 2W is a very reasonable working estimate.

Apparently, I derived this result in the .pdf jt512 refers to, though I'd have to check back to be sure.

If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers, or an estimation that knowing the real figures will not contribute in any meaningul way to practical behavior in the field.


colatownkid


Oct 5, 2008, 1:00 PM
Post #30 of 123 (2299 views)
Shortcut

Registered: Nov 27, 2007
Posts: 512

Re: [rgold] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

rgold wrote:
Colatown, the 2/3 value for carabiner efficiency is an empirically derived value that has been stated over and over in various places, and now has the status of "folklore." I don't know the original reference. I have seen 1/2 stated, but much less often. The fact is that the number will be different for different rope-carabiner combinations and probably, like other friction coefficients, has a sliding and a static value.

The fact that the peak tension in a rope with applied weight W is 2W is a simple conservation of energy consequence of Hooke's Law, viewed as a reasonable approximation to the behavior of a climbing rope. (Yes, there is some acceleration involved, since a climber who weights a rope will drop as the rope stretches.) The result assumes that the climber's weight is instantaneously applied to the climbing rope with no slack in the rope. Using various strategies to ease the weight onto the rope could, in theory, keep the peak rope tension no greater than the climber's weight, but this would not apply to catching a top-rope fall.

Whether the real result is more or less than 2W would be a matter for experiment. I'd guess a little less than 2W because of the internal damping effects of rope construction, but given the manifold uncertainties and variations in real-life practice, 2W is a very reasonable working estimate.

Apparently, I derived this result in the .pdf jt512 refers to, though I'd have to check back to be sure.

If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers, or an estimation that knowing the real figures will not contribute in any meaningul way to practical behavior in the field.

indeed--i find many "bibles" to be quite inadequate for my tastes. on the other hand, i suppose one need not necessarily understand the underlying science in a numeric sense to appreciate the lessons to be learned. personally, i appreciate a mathematical understanding of climbing situations.

as you suspect, you did indeed justify T=2W in the linked paper.

as for the empirical 2/3, i'll guess i'll just take that one on faith. Wink


jt512


Oct 5, 2008, 1:25 PM
Post #31 of 123 (2311 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [rgold] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

rgold wrote:
If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers...

Stop it. You're kill me.

Jay


joeforte


Oct 5, 2008, 5:26 PM
Post #32 of 123 (2297 views)
Shortcut

Registered: May 9, 2005
Posts: 1092

Re: [angry] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

You are right, the anchor would see about 200 lbs... :Engineers cringe:


milesenoell


Oct 6, 2008, 10:11 AM
Post #33 of 123 (2264 views)
Shortcut

Registered: Sep 19, 2006
Posts: 1156

Post deleted by milesenoell [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  

 


lemon_boy


Oct 6, 2008, 10:24 AM
Post #34 of 123 (2258 views)
Shortcut

Registered: Mar 12, 2002
Posts: 287

Re: [milesenoell] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 and rgold are on the money, per usual. angry gave a pretty decent blue-collar interpretation (minus the initial peak load).

by far the scariest thing in this entire thread is the quality (or lack of) that the US Navy is providing students these days.


mtnrock


Oct 6, 2008, 5:14 PM
Post #35 of 123 (2231 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

Re: [hafilax] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.

This is what i was trying to say and i was just checking thanks for the help


mtnrock


Oct 6, 2008, 5:21 PM
Post #36 of 123 (2229 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

haha this post got so off topic i just had a simple question on tention and it went into all this stuff i can't really say it was simple because i can't word anything well but how can i put a diagram so i can show what im talking about more clearly. also does any one know the spring constant of a new dynamic rope?


jdefazio


Oct 6, 2008, 7:21 PM
Post #37 of 123 (2211 views)
Shortcut

Registered: Oct 29, 2007
Posts: 228

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

mtnrock wrote:
haha this post got so off topic i just had a simple question on tention and it went into all this stuff i can't really say it was simple because i can't word anything well but how can i put a diagram so i can show what im talking about more clearly.

Run, Forrest. Run.


onceahardman


Oct 7, 2008, 1:00 PM
Post #38 of 123 (2181 views)
Shortcut

Registered: Aug 3, 2007
Posts: 2468

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

mtnrock, at great risk of furthering misunderstanding, I'm going to try.

The situation is static. Nothing is moving. Climber of weight X is hanging from a normal slingshot-type belay.

Since he is not moving, the tension in the rope between the climber and the 'biners on the toprope anchor is also X. X=X, no motion.

Between the toprope anchor 'biner and the belayer, you also have tension = X, since the rope is not moving.

You now have TWO downward tension vectors, each = X. 2X going downward. The anchor is not moving, so there must be an UPWARD tension vector on the anchor of 2X, or else the anchor would have to be accelerating.

Does that help?


Partner cracklover


Oct 7, 2008, 1:50 PM
Post #39 of 123 (2163 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
I claim that a toprope anchor should be built to withstand a force of at least 3000 lb.

I agree with both your reasoning and your result.

Furthermore, to demand this does not put an undue burden on the anchor builder. It's a pretty easy task to achieve.

Two strands of military spec webbing equalized in a power-point: 18 kN each. Conservatively lowered by 1/3 for the knots = 24 kN, or 5,400 lbs.

If you substitute in 7mm cord (with half the breaking strength), you should still be fine, so long as you use loops.

As for using gear for anchor points, again, not an undue burden. Assuming your weakest point of a three-point anchor system has a failure point of 6kN, and the anchor is reasonably well equalized, it should hold the 13 kN we're aiming for.

GO


shockabuku


Oct 7, 2008, 2:14 PM
Post #40 of 123 (2152 views)
Shortcut

Registered: May 20, 2006
Posts: 4856

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
rgold wrote:
If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers...

Stop it. You're kill me.

Jay

Damn Jay, that's right up there with "all your base are belong to us."


jt512


Oct 7, 2008, 2:18 PM
Post #41 of 123 (2147 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [shockabuku] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

shockabuku wrote:
jt512 wrote:
rgold wrote:
If this fact about minimum peak top-rope loads doesn't appear in some of the "bibles," that reflects either an unsophisticated approach to bible-writing (perhaps because the author(s) do not have a science background) or a lack of faith on the part of the authors in bible-readers...

Stop it. You're kill me.

Jay

Damn Jay, that's right up there with "all your base are belong to us."

I blame all my typos on lack of caffeine.

Jay


hafilax


Oct 7, 2008, 2:23 PM
Post #42 of 123 (2143 views)
Shortcut

Registered: Dec 11, 2007
Posts: 3025

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Minee arre usuallly due too tooo muucch cafffeine.


mtnrock


Oct 7, 2008, 6:08 PM
Post #43 of 123 (2116 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

Re: [onceahardman] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system


jt512


Oct 7, 2008, 6:32 PM
Post #44 of 123 (2114 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay


sungam


Oct 7, 2008, 6:35 PM
Post #45 of 123 (2112 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay
Shhhhh, you'll confuzle them.


mtnrock


Oct 8, 2008, 1:53 PM
Post #46 of 123 (2089 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

Re: [sungam] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

thats not what im saying in both situations there is no acceleration so T doesn't increase


colatownkid


Oct 8, 2008, 2:48 PM
Post #47 of 123 (2082 views)
Shortcut

Registered: Nov 27, 2007
Posts: 512

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

mtnrock wrote:
thats not what im saying in both situations there is no acceleration so T doesn't increase

um...i'm pretty sure that is what you're saying, if i'm not mistaken.


Partner robdotcalm


Oct 8, 2008, 3:38 PM
Post #48 of 123 (2072 views)
Shortcut

Registered: Oct 31, 2002
Posts: 1023

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

rob.calm


jt512


Oct 8, 2008, 3:53 PM
Post #49 of 123 (2069 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [robdotcalm] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay


Partner robdotcalm


Oct 8, 2008, 4:09 PM
Post #50 of 123 (2064 views)
Shortcut

Registered: Oct 31, 2002
Posts: 1023

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay

Yes, especially if I (first) stood in front of the parked car and then in front of the car going 30 mi/hr. I'm not sure, though, that I'm willing to do that experiment. Would you do it for me?

r.c


sungam


Oct 8, 2008, 4:20 PM
Post #51 of 123 (1804 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [robdotcalm] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

robdotcalm wrote:
jt512 wrote:
robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay

Yes, especially if I (first) stood in front of the parked car and then in front of the car going 30 mi/hr. I'm not sure, though, that I'm willing to do that experiment. Would you do it for me?

r.c
Missing the point or winding Jay up?


jt512


Oct 8, 2008, 4:54 PM
Post #52 of 123 (1799 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [robdotcalm] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

robdotcalm wrote:
jt512 wrote:
robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay

Yes, especially if I (first) stood in front of the parked car and then in front of the car going 30 mi/hr. I'm not sure, though, that I'm willing to do that experiment. Would you do it for me?

r.c

I think you're missing the point. There is no such thing as absolute velocity. Velocity is relative. The parked car appears to have zero velocity to you, but to an observer in another frame of reference, the parked car appears to be moving at 66,600 mi/hr.

You perceive one car to have a speed of zero and the other to have a speed of 30 mi/hr. But if the car you perceive to be traveling 30 mi/hr is traveling in the opposite direction of the earth's motion around the sun, then someone in the sun's frame of reference would observe the car that you see as the faster one to be moving 30 mi/hr slower than the car you perceive to be "parked." Neither of you would be right or wrong in the absolute sense, because there is no absolute sense in which to describe motion.

Jay


(This post was edited by jt512 on Oct 8, 2008, 9:25 PM)


curt


Oct 8, 2008, 6:03 PM
Post #53 of 123 (1795 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

Curt


mtnrock


Oct 8, 2008, 6:14 PM
Post #54 of 123 (1789 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

Re: [robdotcalm] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

ok wait cause your guys are so set on how i word something. that's not what the point i was trying to get across. i was trying to show if there is no acceleration tension does not increase. and it really seems people are just on this site to piss people off


jt512


Oct 8, 2008, 6:18 PM
Post #55 of 123 (1785 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

mtnrock wrote:
ok wait cause your guys are so set on how i word something. that's not what the point i was trying to get across. i was trying to show if there is no acceleration tension does not increase.

Oh, sorry. Next time we'll ignore what you write, and just read your mind instead.

Jay


curt


Oct 8, 2008, 6:18 PM
Post #56 of 123 (1784 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

mtnrock wrote:
ok wait cause your guys are so set on how i word something. that's not what the point i was trying to get across. i was trying to show if there is no acceleration tension does not increase. and it really seems people are just on this site to piss people off

Some of us are only here to piss-off the retards. Why? Are you pissed?

Curt


mtnrock


Oct 8, 2008, 6:24 PM
Post #57 of 123 (1777 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

na just saying all you do is insult people


curt


Oct 8, 2008, 6:31 PM
Post #58 of 123 (1774 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [mtnrock] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

mtnrock wrote:
na just saying all you do is insult people

Not true--I do that and a whole lot more. You'll see.

Curt


sungam


Oct 8, 2008, 6:36 PM
Post #59 of 123 (1771 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
mtnrock wrote:
na just saying all you do is insult people

Not true--I do that and a whole lot more. You'll see.

Curt
It's true, sometimes he stabs (ask mark if you don't believe me).


jt512


Oct 8, 2008, 6:39 PM
Post #60 of 123 (1770 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay


curt


Oct 8, 2008, 7:13 PM
Post #61 of 123 (1782 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt


mtnrock


Oct 8, 2008, 7:38 PM
Post #62 of 123 (1773 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

i see what jay is saying but yea the examples are different because to say one object has velocity you have to say that another doesn't as a frame of reference


jt512


Oct 8, 2008, 8:01 PM
Post #63 of 123 (1771 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay


(This post was edited by jt512 on Oct 8, 2008, 8:04 PM)


curt


Oct 8, 2008, 8:30 PM
Post #64 of 123 (1760 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt


jt512


Oct 8, 2008, 8:40 PM
Post #65 of 123 (1755 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt

It would probably help to have a real physicist to weigh in, instead of two guys who sometimes play one on the internet; but all the real physicists I know want to get paid to answer questions like this. But, that said, I claim that since velocity is relative then so must be momentum, since they are proportional; so a momentum of 0 must also be an artifact of having chosen a special frame of reference.

Jay


curt


Oct 8, 2008, 8:46 PM
Post #66 of 123 (1752 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt

It would probably help to have a real physicist to weigh in, instead of two guys who sometimes play one on the internet; but all the real physicists I know want to get paid to answer questions like this. But, that said, I claim that since velocity is relative then so must be momentum, since they are proportional; so a momentum of 0 must also be an artifact of having chosen a special frame of reference.

Jay

What you're actually saying is that both velocity and momentum have no generally accepted meanings in newtonian physics--and hence, can not be used to differentiate between stationary and moving objects. Obviously, I think that's silly.

Curt


jt512


Oct 8, 2008, 8:49 PM
Post #67 of 123 (1746 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt

It would probably help to have a real physicist to weigh in, instead of two guys who sometimes play one on the internet; but all the real physicists I know want to get paid to answer questions like this. But, that said, I claim that since velocity is relative then so must be momentum, since they are proportional; so a momentum of 0 must also be an artifact of having chosen a special frame of reference.

Jay

What you're actually saying is that both velocity and momentum have no generally accepted meanings in newtonian physics--and hence, can not be used to differentiate between stationary and moving objects. Obviously, I think that's silly.

Curt

So, are you saying that Newtonian physics does not recognize the relativity of motion?

Jay


jt512


Oct 8, 2008, 9:00 PM
Post #68 of 123 (1741 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Let me put it this way. There is nothing special about v=0. Velocity is relative. v=0 is exactly as distinguishable from v=30 as v=30 is from v=60. That is, independent of reference frame, the difference in velocities in each case is 30. So now we have something that is universal; that is, we can define velocity relatively. Since momentum is velocity times a constant, then the exact same argument applies to momentum.

Jay


(This post was edited by jt512 on Oct 8, 2008, 9:01 PM)


curt


Oct 8, 2008, 9:08 PM
Post #69 of 123 (1737 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt

It would probably help to have a real physicist to weigh in, instead of two guys who sometimes play one on the internet; but all the real physicists I know want to get paid to answer questions like this. But, that said, I claim that since velocity is relative then so must be momentum, since they are proportional; so a momentum of 0 must also be an artifact of having chosen a special frame of reference.

Jay

What you're actually saying is that both velocity and momentum have no generally accepted meanings in newtonian physics--and hence, can not be used to differentiate between stationary and moving objects. Obviously, I think that's silly.

Curt

So, are you saying that Newtonian physics does not recognize the relativity of motion?

Jay

I am saying (in yet another way) that the first derivative of displacement, with respect to time, is velocity--and even mathematically it is clear that an object moving at a constant velocity and a stationary one can easily be differentiated. By the way, in the first post of yours that I replied to, you fixed the frame of reference by saying that one comparative case involved "zero velocity."

Curt


jt512


Oct 8, 2008, 9:12 PM
Post #70 of 123 (1734 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt

It would probably help to have a real physicist to weigh in, instead of two guys who sometimes play one on the internet; but all the real physicists I know want to get paid to answer questions like this. But, that said, I claim that since velocity is relative then so must be momentum, since they are proportional; so a momentum of 0 must also be an artifact of having chosen a special frame of reference.

Jay

What you're actually saying is that both velocity and momentum have no generally accepted meanings in newtonian physics--and hence, can not be used to differentiate between stationary and moving objects. Obviously, I think that's silly.

Curt

So, are you saying that Newtonian physics does not recognize the relativity of motion?

Jay

I am saying (in yet another way) that the first derivative of displacement, with respect to time, is velocity--and even mathematically it is clear that an object moving at a constant velocity and a stationary one can easily be differentiated. By the way, in the first post of yours that I replied to, you fixed the frame of reference by saying that one comparative case involved "zero velocity."

Curt

Well, I don't know how to explain that there is no difference between zero velocity and non-zero velocity without using the phrase "zero velocity." I suppose that if I tried hard enough, I could do it, but it would have made the post even more abstract.

Jay


curt


Oct 8, 2008, 9:24 PM
Post #71 of 123 (1729 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt

It would probably help to have a real physicist to weigh in, instead of two guys who sometimes play one on the internet; but all the real physicists I know want to get paid to answer questions like this. But, that said, I claim that since velocity is relative then so must be momentum, since they are proportional; so a momentum of 0 must also be an artifact of having chosen a special frame of reference.

Jay

What you're actually saying is that both velocity and momentum have no generally accepted meanings in newtonian physics--and hence, can not be used to differentiate between stationary and moving objects. Obviously, I think that's silly.

Curt

So, are you saying that Newtonian physics does not recognize the relativity of motion?

Jay

I am saying (in yet another way) that the first derivative of displacement, with respect to time, is velocity--and even mathematically it is clear that an object moving at a constant velocity and a stationary one can easily be differentiated. By the way, in the first post of yours that I replied to, you fixed the frame of reference by saying that one comparative case involved "zero velocity."

Curt

Well, I don't know how to explain that there is no difference between zero velocity and non-zero velocity without using the phrase "zero velocity." I suppose that if I tried hard enough, I could do it, but it would have made the post even more abstract.

Jay

I will now happily wait for rgold to arbitrate.

Curt


colatownkid


Oct 9, 2008, 4:29 AM
Post #72 of 123 (1689 views)
Shortcut

Registered: Nov 27, 2007
Posts: 512

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
curt wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

Although most of us know what you meant, your statement is incorrect. What you meant to say is "motion at a constant velocity and motion of zero velocity both involve zero acceleration."

Obviously, if you plot position versus time for an object of constant velocity it will look quite different than a similar graph showing an object with zero velocity. The former will result in a straight line and the latter will result in a point.

I meant it more broadly. Since motion is relative, there is no Universal distinction between zero velocity and non-zero velocity. Zero velocity is just constant velocity observed from a special frame of reference.

Jay

Still, the two examples you cited are certainly not "indistinguishable" from a physics perspective. In one case, position versus time is changing at a constant rate--and in the other case position versus time is not changing.

Curt

I disagree. In fact the example you give of why they are distinguishable — that they have different displacement–time graphs — is an artifact of the fact that the frame of reference you've chosen is unique; out of an infinite set of frames of reference, you've chosen the only one in which the one object appears stationary. Stated another way, for any two objects moving relative to each other, a frame of reference can be chosen in which one of the objects appears to be stationary. But physics is the same everywhere in the Universe; physics does not depend on frame of reference.

Jay

Yet another way to differentiate a stationary object from an object moving at constant velocity (from a physics standpoint) is that any object in motion has momentum--and the one that is stationary does not. (i.e., p =mv) Looks like we need rgold to weigh in.

Curt

It would probably help to have a real physicist to weigh in, instead of two guys who sometimes play one on the internet; but all the real physicists I know want to get paid to answer questions like this. But, that said, I claim that since velocity is relative then so must be momentum, since they are proportional; so a momentum of 0 must also be an artifact of having chosen a special frame of reference.

Jay

What you're actually saying is that both velocity and momentum have no generally accepted meanings in newtonian physics--and hence, can not be used to differentiate between stationary and moving objects. Obviously, I think that's silly.

Curt

So, are you saying that Newtonian physics does not recognize the relativity of motion?

Jay

I am saying (in yet another way) that the first derivative of displacement, with respect to time, is velocity--and even mathematically it is clear that an object moving at a constant velocity and a stationary one can easily be differentiated. By the way, in the first post of yours that I replied to, you fixed the frame of reference by saying that one comparative case involved "zero velocity."

Curt

Well, I don't know how to explain that there is no difference between zero velocity and non-zero velocity without using the phrase "zero velocity." I suppose that if I tried hard enough, I could do it, but it would have made the post even more abstract.

Jay

I will now happily wait for rgold to arbitrate.

Curt

Can I arbitrate?

I know it's only wikipedia, but seriously, as is said far too often on this site, do a search.

See http://en.wikipedia.org/wiki/Force:

"Newton's first law of motion states that objects continue to move in a state of constant velocity unless acted upon by an external net force or resultant force.[10] This law is an extension of Galileo's insight that constant velocity was associated with a lack of net force (see a more detailed description of this below). Newton proposed that every object with mass has an innate inertia that functions as the fundamental equilibrium "natural state" in place of the Aristotelian idea of the "natural state of rest". That is, the first law contradicts the intuitive Aristotelian belief that a net force is required to keep an object moving with constant velocity. By making rest physically indistinguishable from non-zero constant velocity, Newton's first law directly connects inertia with the concept of relative velocities. Specifically, in systems where objects are moving with different velocities, it is impossible to determine which object is "in motion" and which object is "at rest". In other words, to phrase matters more technically, the laws of physics are the same in every inertial frame of reference, that is, in all frames related by a Galilean transformation."

Further down the page it shows why Galileo was smarter than us:

"Simple experiments showed that Galileo's understanding of the equivalence of constant velocity and rest to be correct. For example, if a mariner dropped a cannonball from the crow's nest of a ship moving at a constant velocity, Aristotelian physics would have the cannonball fall straight down while the ship moved beneath it. Thus, in an Aristotelian universe, the falling cannonball would land behind the foot of the mast of a moving ship. However, when this experiment is actually conducted, the cannonball always falls at the foot of the mast, as if the cannonball knows to travel with the ship despite being separated from it. Since there is no forward horizontal force being applied on the cannonball as it falls, the only conclusion left is that the cannonball continues to move with the same velocity as the boat as it falls. Thus, no force is required to keep the cannonball moving at the constant forward velocity.[9]"

I'm gonna say Jay is right on this one. Jay's statement could be amended to say that "The force on an object in motion at a constant velocity and motion of zero velocity are indistinguishable." This might make it a lot easier to swallow and it directly answers the OPs question about tension while a climber is being lowered (that is, in the presence of constant velocity and no acceleration the tension does not change).

However, Galilean Relativity clearly states that Jay's original statement does in fact hold true. "Galilean invariance or Galilean relativity is a principle of relativity which states that the fundamental laws of physics are the same in all inertial frames." (see http://en.wikipedia.org/.../Galilean_relativity)

(Now, if we all really want to be completely honest about it, this is all technically wrong since time can not be considered absolute as the speed of light is approached. But let's not complicate things...)


sungam


Oct 9, 2008, 7:06 AM
Post #73 of 123 (1672 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [colatownkid] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

colatownkid wrote:
But let's not complicate things...
Indeed, we wouldn't want THAT now, would we?


jdefazio


Oct 9, 2008, 7:14 AM
Post #74 of 123 (1672 views)
Shortcut

Registered: Oct 29, 2007
Posts: 228

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

JT512 wrote:
A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.
Jay
curt wrote:
Although most of us know what you meant, your statement is incorrect.

JT512 wrote:
..."your rong"...

curt wrote:
..."no yor rong"...

^^^repeat ad-infinitum.

Curt is correct here regarding the semantics of Jay's statement. The rest is like arguing over who is better looking, since the answer depends on who you ask (i.e. what reference frame they are in).

In classical physics, when comparing measurements obtained in different reference frames, the forces/accelerations are identical if the relative motion betwen these frames is of constant velocity. This is the Galilean relativity CI mentions, essentially saying the classical laws of physics (i.e. Newton's F=ma) are invariant between inertial (non-accelerating) reference frames.


curt


Oct 9, 2008, 8:18 AM
Post #75 of 123 (1661 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jdefazio] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jdefazio wrote:
JT512 wrote:
A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.
Jay
curt wrote:
Although most of us know what you meant, your statement is incorrect.

JT512 wrote:
..."your rong"...

curt wrote:
..."no yor rong"...

^^^repeat ad-infinitum.

Curt is correct here regarding the semantics of Jay's statement. The rest is like arguing over who is better looking, since the answer depends on who you ask (i.e. what reference frame they are in)...

Fine. Then I'm better looking too. Cool

Curt


Partner cracklover


Oct 9, 2008, 8:25 AM
Post #76 of 123 (1473 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO


colatownkid


Oct 9, 2008, 8:29 AM
Post #77 of 123 (1470 views)
Shortcut

Registered: Nov 27, 2007
Posts: 512

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jdefazio wrote:
JT512 wrote:
A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.
Jay
curt wrote:
Although most of us know what you meant, your statement is incorrect.

JT512 wrote:
..."your rong"...

curt wrote:
..."no yor rong"...

^^^repeat ad-infinitum.

Curt is correct here regarding the semantics of Jay's statement. The rest is like arguing over who is better looking, since the answer depends on who you ask (i.e. what reference frame they are in)...

Fine. Then I'm better looking too. Cool

Curt

i thought that was obvious.


petsfed


Oct 9, 2008, 8:44 AM
Post #78 of 123 (1464 views)
Shortcut

Registered: Sep 24, 2002
Posts: 8587

Re: [cracklover] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.


jt512


Oct 9, 2008, 9:07 AM
Post #79 of 123 (1456 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [petsfed] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay


curt


Oct 9, 2008, 9:23 AM
Post #80 of 123 (1446 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

Jay,

The problem is that you said the two cases are "indistinguishable" and that simply isn't true. Again, if I plot position versus time for an object moving at constant velocity I will get a straight line. If I plot position versus time for a stationary object, I will get a point. If I now decide to change my frame of reference and "sit" on the object moving at constant velocity, I may indeed opt to say that the "stationary" object is the one in motion, relative to me. It doesn't matter--because if I then create the same two graphs of position versus time they are merely reversed and I can still easily differentiate one graph from the other.

Curt


petsfed


Oct 9, 2008, 9:31 AM
Post #81 of 123 (1442 views)
Shortcut

Registered: Sep 24, 2002
Posts: 8587

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Both.

It is not the case that the cannonball has both zero and non-zero velocity. Which is what you seem to be implying.

From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

Which is (I think) what you're not clarifying in your statement that the two are equivalent.


jdefazio


Oct 9, 2008, 9:53 AM
Post #82 of 123 (1436 views)
Shortcut

Registered: Oct 29, 2007
Posts: 228

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
Some of you physics guys have lost the big picture... Come on: this is high school stuff...

Agreed.
OMG TEH NUT0NZ L4WZ R TEH S4ME!!! OH NOEZZZ!!!!!!!
---

The recent debate, though, was initiated by the statement:

jt512 wrote:
A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Of course some of us know that you are really talking about Galilean invariance. You do seem like a bright fellow after all! Wink
As a reputed stickler for grammar and semantics, however, you did leave this wide open for a firm-handed bitchslapping.


jt512


Oct 9, 2008, 9:54 AM
Post #83 of 123 (1435 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jt512 wrote:
petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

Jay,

The problem is that you said the two cases are "indistinguishable" and that simply isn't true. Again, if I plot position versus time for an object moving at constant velocity I will get a straight line. If I plot position versus time for a stationary object, I will get a point. If I now decide to change my frame of reference and "sit" on the object moving at constant velocity, I may indeed opt to say that the "stationary" object is the one in motion, relative to me. It doesn't matter--because if I then create the same two graphs of position versus time they are merely reversed and I can still easily differentiate one graph from the other.

Curt

You may be able to differentiate one graph from the other, but it's just begging the question. Those two graphs will completely disagree as to which object is the stationary one. How much more indistinguishable do you want than that? And in any other frame of reference neither object would appear stationary.

I hate to appeal to authority, and hate even more to appeal to Wikipedia as an authority, but the Wikipedia article that colatownkid cited completely agrees with me: "By making rest physically indistinguishable from non-zero constant velocity, Newton's first law directly connects inertia with the concept of relative velocities. ... Simple experiments showed that Galileo's understanding of the equivalence of constant velocity and rest to be correct."

Jay


Partner cracklover


Oct 9, 2008, 10:26 AM
Post #84 of 123 (1414 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Jay, you're arguing a point we all agree with. I believe that the trouble is not with what you think, but with what you wrote: "A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable. "

The simplest interpretation of the above is that one cannot distinguish two objects at zero and non-zero velocities (in a given frame of reference) from each other. You merely have to clarify that that is not what you meant, and all is solved (for those even able to follow this).

GO


curt


Oct 9, 2008, 10:29 AM
Post #85 of 123 (1412 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
curt wrote:
jt512 wrote:
petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

Jay,

The problem is that you said the two cases are "indistinguishable" and that simply isn't true. Again, if I plot position versus time for an object moving at constant velocity I will get a straight line. If I plot position versus time for a stationary object, I will get a point. If I now decide to change my frame of reference and "sit" on the object moving at constant velocity, I may indeed opt to say that the "stationary" object is the one in motion, relative to me. It doesn't matter--because if I then create the same two graphs of position versus time they are merely reversed and I can still easily differentiate one graph from the other.

Curt

You may be able to differentiate one graph from the other, but it's just begging the question. Those two graphs will completely disagree as to which object is the stationary one. How much more indistinguishable do you want than that?

Indistinguishable would mean the graphs of position versus time for the two objects would look identical--i.e. we can't tell them apart. That is clearly not the case here. In fact (as others have commented) from any given frame of reference the two graphs of position versus time will look different for the two objects. You can go with the wiki quote from colatown kid if you like--but petsfed (who has a degree in physics) jdefazio, robdotcalm and cracklover all disagree with you. I still hope rgold will comment.

Curt


colatownkid


Oct 9, 2008, 10:30 AM
Post #86 of 123 (1410 views)
Shortcut

Registered: Nov 27, 2007
Posts: 512

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
petsfed wrote:
cracklover wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

All of the misunderstanding (between intelligent parties) on this thread stems from the way Jay phrased this.

Yes, in a way, what he's saying is true. However independent of the frame of reference you choose you can distinguish between relative velocities. This is why you can tell that one car is standing still while the other is moving.

From any vantage point, the two cars have different velocities, that's how you can tell them apart. It's not necessary to take the vantage point of the earth - which is what the poster who made the observation did. From any other vantage point, you could still tell the difference.

Cheers,

GO

IZ KOREKT

Jesus guys, any idiot can see that a "stationary" car has a different velocity from a "non-stationary" car, no matter what your reference frame (given sufficiently precise instrumentation). Just because "stationary" and "non-stationary" are relative concepts does not imply that they are equivalent or interchangeable. Stationary only becomes non-stationary when you change your reference frame (and then the initial non-stationary becomes stationary). They are at no point actually the same thing.

Some of you physics guys have lost the big picture. There is nothing special about an object that appears to be stationary. It is only stationary in the frame of reference you are observing it from. It is your frame of reference that is special: out of the infinite number of frames of reference in the Newtonian Universe, you are observing it from the only one in which it appears stationary. Come on: this is high school stuff. You drop the cannonball from the top of the mast of the ship which is moving relative to the shore. The guy on the deck of the ship observes the ball falling straight down; it has zero horizontal velocity. The guy on the shore observes it moving in a parabolic arc; it has non-zero horizontal velocity. Who's right? Both? Neither?

Jay

i see Jay has read my post about Galilean relativity.


petsfed


Oct 9, 2008, 10:35 AM
Post #87 of 123 (1403 views)
Shortcut

Registered: Sep 24, 2002
Posts: 8587

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

I have a what now?

/still hasn't finished college


(This post was edited by petsfed on Oct 9, 2008, 10:35 AM)


jdefazio


Oct 9, 2008, 10:42 AM
Post #88 of 123 (1394 views)
Shortcut

Registered: Oct 29, 2007
Posts: 228

Re: [petsfed] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

petsfed wrote:
I have a what now?

/still hasn't finished college

Pffft. Then, I guess...STFU physics N00B! WinkWinkWink


jt512


Oct 9, 2008, 10:43 AM
Post #89 of 123 (1394 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [petsfed] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

petsfed wrote:
From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay


sungam


Oct 9, 2008, 10:44 AM
Post #90 of 123 (1390 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

BTW, just to stir the pot, both your sailor and your landlubber are rong.
The cannon ball is staying still (duh) and the other things are moving around it.


curt


Oct 9, 2008, 10:47 AM
Post #91 of 123 (1407 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [petsfed] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

petsfed wrote:
I have a what now?

/still hasn't finished college

Well, you're close, right? And it is a degree in physics--isn't it?

Curt


colatownkid


Oct 9, 2008, 10:49 AM
Post #92 of 123 (1401 views)
Shortcut

Registered: Nov 27, 2007
Posts: 512

Re: [sungam] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

sungam wrote:
BTW, just to stir the pot, both your sailor and your landlubber are rong.
The cannon ball is staying still (duh) and the other things are moving around it.

oh snap.


curt


Oct 9, 2008, 10:49 AM
Post #93 of 123 (1399 views)
Shortcut

Registered: Aug 26, 2002
Posts: 18227

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
petsfed wrote:
From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay

But, they will always be different--and hence not "indistinguishable."

Curt


petsfed


Oct 9, 2008, 11:00 AM
Post #94 of 123 (1395 views)
Shortcut

Registered: Sep 24, 2002
Posts: 8587

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
petsfed wrote:
I have a what now?

/still hasn't finished college

Well, you're close, right? And it is a degree in physics--isn't it?

Curt

Yeah, I have to take circuits in the spring, and then I'll have a BS in Astrophysics. As opposed to a strong desire to bs about physics.


jt512


Oct 9, 2008, 11:24 AM
Post #95 of 123 (1383 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [curt] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

curt wrote:
jt512 wrote:
petsfed wrote:
From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay

But, they will always be different--and hence not "indistinguishable."

Curt

I think you may be forgetting what I said was indistinguishable. What I said was that rest and constant velocity are indistinguishable. That the two objects have different velocities is clear. But if I were to hand you both graphs you would have no way to distinguish between which one is at rest and which one isn't. That's because the distinction is meaningless in Newtonian physics. The question implies that Newtonian velocity is an absolute quantity, which it isn't; it is a relative quantity.

Jay


Valarc


Oct 9, 2008, 11:40 AM
Post #96 of 123 (1372 views)
Shortcut

Registered: Apr 19, 2007
Posts: 1473

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
petsfed wrote:
From the same frame of reference, zero and non-zero velocities are clearly different. In fact, so long as you measure both velocities from the same frame of reference (no matter what that reference frame is), the two objects will have different velocities. This should be apparent. To say that v=0 is the same as v=30 is only true if the two measurements are taken from two very specific frames. There is no single reference frame where v=0 is the same as v=30.

True, but there is another frame of reference in which the velocities of the two objects are switched, and both frames of reference are equally valid.

Jay

I'm no Rgold, but this is korrekt.

Although, if you listen to the cosmologists, we should measure everything relative to the cosmic microwave background, which would give a preferred, or (egads!) "absolute" frame of reference to which we should compare things.

I've never liked cosmologists, personally.


jdefazio


Oct 9, 2008, 11:44 AM
Post #97 of 123 (1362 views)
Shortcut

Registered: Oct 29, 2007
Posts: 228

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post




petsfed


Oct 9, 2008, 11:45 AM
Post #98 of 123 (1362 views)
Shortcut

Registered: Sep 24, 2002
Posts: 8587

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
The question implies that Newtonian velocity is an absolute quantity, which it isn't; it is a relative quantity.

Jay

Which should be readily apparent from the definition of an inertial frame, the kind of reference frame that Newtonian mechanics is derived for. Any inertial reference frame is basically equivalent to any other in terms of how the laws of physics work.

There are non-inertial frames of reference though.



Thank you xkcd.org


petsfed


Oct 9, 2008, 11:46 AM
Post #99 of 123 (1359 views)
Shortcut

Registered: Sep 24, 2002
Posts: 8587

Re: [Valarc] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Valarc wrote:
I've never liked cosmologists, personally.

That's ok. They think you smell funny.


Partner cracklover


Oct 9, 2008, 11:49 AM
Post #100 of 123 (1356 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay

Here is the issue, in a nutshell: The phrase "at rest" or "in motion" only have meaning within a given frame of reference. So when robdot says he can tell the difference, he is correct, from his frame of reference.

While it is true that for newtonian physics to work, a frame of reference is not required, that doesn't mean that one cannot have a frame of reference and speak intelligently about the motions of objects within that frame. This is the point which JT is unwilling to acknowledge.

GO


jt512


Oct 9, 2008, 12:45 PM
Post #101 of 123 (2309 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [cracklover] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

cracklover wrote:
jt512 wrote:
robdotcalm wrote:
jt512 wrote:
mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system

A fundamental fact of physics is that motion at a constant velocity and motion of zero velocity are indistinguishable.

Jay

I must be mistaken, but I just saw two cars on the street. One was parked and one was moving at a constant velocity of 30 mi/hr. I was able to distinguish them based on the motion.

On the other hand, both cars were moving at approximately 66,600 mi/hr around the sun, as were you. Still think you can distinguish constant velocity from zero velocity?

Jay

While it is true that for newtonian physics to work, a frame of reference is not required...

Huh?

In reply to:
...that doesn't mean that one cannot have a frame of reference and speak intelligently about the motions of objects within that frame. This is the point which JT is unwilling to acknowledge.

In order to speak intelligently about Newtonian physics you have to understand that your observations are dependent on your frame of reference. My original statement was in response to a couple of people who seemed to think that the tension in their rope depended on their rope's velocity. Failure to understand frame of reference misled them into thinking rather unintelligently about the behavior of objects.

Jay


Partner cracklover


Oct 9, 2008, 1:07 PM
Post #102 of 123 (2301 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
cracklover wrote:
While it is true that for newtonian physics to work, a frame of reference is not required...

Huh?

I mean that choosing one frame of reference versus another doesn't matter. The physics works out the same.

In reply to:
In reply to:
...that doesn't mean that one cannot have a frame of reference and speak intelligently about the motions of objects within that frame. This is the point which JT is unwilling to acknowledge.

In order to speak intelligently about Newtonian physics you have to understand that your observations are dependent on your frame of reference. My original statement was in response to a couple of people who seemed to think that the tension in their rope depended on their rope's velocity.

Who the fuck knows what the OP was thinking? So far as I can tell, he hadn't actually thought about the problems he wanted answered well enough to even generate a question that made sense.

In reply to:
Failure to understand frame of reference misled them into thinking rather unintelligently about the behavior of objects.

Jay

I suppose you could be right. But when I read:
In reply to:
is tension on a rope doubled in a top rope system when its on a redirect belay or a direct from the anchor belay.

it seems like it should but im draawing a blank if not please say why

the lack of understanding about the relevance of a frame of reference certainly doesn't shine through to me.

But probably this is not the post you're referring to? Could you please direct me to the posts of the "people who seemed to think that the tension in their rope depended on their rope's velocity"?

Thanks!

GO


onceahardman


Oct 9, 2008, 3:16 PM
Post #103 of 123 (2281 views)
Shortcut

Registered: Aug 3, 2007
Posts: 2468

Re: [cracklover] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Umm, this might be my fault, when I posted this on page 2 of this thread:

In reply to:
mtnrock, at great risk of furthering misunderstanding, I'm going to try.

The situation is static. Nothing is moving. Climber of weight X is hanging from a normal slingshot-type belay.

Since he is not moving, the tension in the rope between the climber and the 'biners on the toprope anchor is also X. X=X, no motion.

Between the toprope anchor 'biner and the belayer, you also have tension = X, since the rope is not moving.

You now have TWO downward tension vectors, each = X. 2X going downward. The anchor is not moving, so there must be an UPWARD tension vector on the anchor of 2X, or else the anchor would have to be accelerating.

Does that help?

Awkwardly stated, but an effort at simplifying things for mtnrock, by showing a static situation (which is, of course, equivalent to a constant velocity situation, with frictionless ropes and biners, etc.


jt512


Oct 9, 2008, 3:59 PM
Post #104 of 123 (2270 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [cracklover] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

cracklover wrote:
Could you please direct me to the posts of the "people who seemed to think that the tension in their rope depended on their rope's velocity"?

It was the one that "onceahardman" reposted above and mtnrock's punctuation-free response:

mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system


Jay


(This post was edited by jt512 on Oct 9, 2008, 4:00 PM)


Partner cracklover


Oct 10, 2008, 7:33 AM
Post #105 of 123 (2243 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
cracklover wrote:
Could you please direct me to the posts of the "people who seemed to think that the tension in their rope depended on their rope's velocity"?

It was the one that "onceahardman" reposted above and mtnrock's punctuation-free response:

mtnrock wrote:
yea that's what i was thinking if it was moving like while letting him down to find the tenition it would be more like T=m(g-a) t is tention m is mass and g is accleration due to gravity and a is acceleration of the system


Jay

I see. Yech, what a mess. Anywey, I think we're all on the same page now.

Just for shits and giggles, mtnrock, have we answered your question? If so, do you think you could explain, in your own language, what the forces are for a slingshot belay? This means it's a toprope fall in which the belayer is at the bottom, the rope runs up through the anchor and back down to the climber.

Assume there was no slack when the climber falls. Assume that the climber weighs 150kg.

The three forces I'm looking for are: the peak force on the belayer, on the climber, and on the anchor.
For extra credit, discuss how the frame of reference informs your answer.

Okay gang, let's see how well we did in informing mtnrock.

Good luck!

GO


knudenoggin


Oct 12, 2008, 11:10 PM
Post #106 of 123 (2209 views)
Shortcut

Registered: May 6, 2004
Posts: 594

Re: tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

I'm curious if this physics is for rockclimbers only,
or if it applies to mundane things like barbell weights
and cord & 'biners and less frictive sheaves like pulleys?

For beside me are four 25# weights, in the following configuration:

two wgt.s are tied together and to a 7mm line that runs up through
a pretty low-friction (relative 'biners, anyway) pulley wheel;

which line then runs down to tie to a 'biner;

through which runs quarter-inch solid-braid nylon line tied,
at each end, to each of the other two 25# wgt.s, resp..

Now, hoisting up one of these last 25pounders, to give it a short
drop on a sling-shot-through 'biner belay by the other 25pounder,
which 'biner anchor cord itself is sling-shot through the pulley to
the 50# belay, and . . .

what has all this preceding discussion said will happen?
(aside from showing posters' excessive love the quote function!)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - -

As I read it, the dropped (a few inches) 25# generates > (2*25#),
coupled with 4/3(25#) on the flip side (or raising that 25# belay),
producing > 10/3 (25#) = 111.111111... # of tension fed into
the pulley wheel, belayed by a mere 50#, which should thus go
halfway into orbit.
(But my faith in this is such that I've not alerted NASA, or local
air-traffic control, or even a ceiling repair guy just yet.)

Is that right?

*kN*


jdefazio


Oct 13, 2008, 5:29 AM
Post #107 of 123 (2204 views)
Shortcut

Registered: Oct 29, 2007
Posts: 228

Re: [knudenoggin] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Majid-ize a schematic please.


jt512


Oct 13, 2008, 7:25 AM
Post #108 of 123 (2193 views)
Shortcut

Registered: Apr 11, 2001
Posts: 21890

Re: [knudenoggin] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

knudenoggin wrote:
I'm curious if this physics is for rockclimbers only,
or if it applies to mundane things like barbell weights
and cord & 'biners and less frictive sheaves like pulleys?

For beside me are four 25# weights, in the following configuration:

two wgt.s are tied together and to a 7mm line that runs up through
a pretty low-friction (relative 'biners, anyway) pulley wheel;

which line then runs down to tie to a 'biner;

through which runs quarter-inch solid-braid nylon line tied,
at each end, to each of the other two 25# wgt.s, resp..

Now, hoisting up one of these last 25pounders, to give it a short
drop on a sling-shot-through 'biner belay by the other 25pounder,
which 'biner anchor cord itself is sling-shot through the pulley to
the 50# belay, and . . .

what has all this preceding discussion said will happen?
(aside from showing posters' excessive love the quote function!)

- - - - - - - - - - - - - - - - - - - - - - - - - - - - -

As I read it, the dropped (a few inches) 25# generates > (2*25#),
coupled with 4/3(25#) on the flip side (or raising that 25# belay),
producing > 10/3 (25#) = 111.111111... # of tension fed into
the pulley wheel, belayed by a mere 50#, which should thus go
halfway into orbit.
(But my faith in this is such that I've not alerted NASA, or local
air-traffic control, or even a ceiling repair guy just yet.)

Is that right?

*kN*

If you want to have any chance of getting a correct answer, you're going to have provide a diagram of your setup.

And how come everybody's posts except your go all the way across the text field?

Jay


mtnrock


Oct 13, 2008, 7:43 AM
Post #109 of 123 (2191 views)
Shortcut

Registered: Aug 23, 2008
Posts: 61

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

In reply to:
My original statement was in response to a couple of people who seemed to think that the tension in their rope depended on their rope's velocity.

it doesn't depend on its velocity and i said it didn't but it does depend on its acceleration.


sungam


Oct 13, 2008, 7:51 AM
Post #110 of 123 (2190 views)
Shortcut

Registered: Jun 24, 2004
Posts: 26572

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
And how come everybody's posts except your go all the way across the text field?

Jay
He
likes
the
return
button.


knudenoggin


Oct 13, 2008, 12:29 PM
Post #111 of 123 (2154 views)
Shortcut

Registered: May 6, 2004
Posts: 594

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

jt512 wrote:
If you want to have any chance of getting a correct answer,
you're going to have provide a diagram of your setup.
C'mon, it's quite simple: a compound sling-shot, with 25# being dropped and
25# on the *belayer's* side of this, cord run through a 'biner;
this anchor 'biner is supported itself by line running up/down through
a pulley wheel, which has much less friction than a 'biner, and its *belay*
side had the 50# wgt. In crude jargon, then, its 2:1 hauling on 2:1,
theoretical MA. The drop is something >FF0, say, 3". And I guess that
the dropping side had about 1' to 'biner, and 3' down to belay wgt.; the
pulley was about 6' above its "belay" wgt., with about 1.5' down to the 'biner.
So, the idea was that the lower, 'biner system was generating the 10/3W
(W = 25#), and delivering this force into the upper and less frictive system,
which should then generate 4/3(10/3)W on the 50# anchor wg.t.

In reply to:
And how come everybody's posts except yours go all the way across the text field?
Damn good whine: how come, indeed!
Well, I don't like reading wiiiiiiiiidddddddddeeeeee lines any more than you
like a lack of punctuation. And why keep quoting texts that are already presented
above? Most folks aren't going to MajiDelete things into oblivion: what's the
point of repeating 20-50 lines just to add another line or two?

*kN*


Partner rgold


Oct 13, 2008, 12:33 PM
Post #112 of 123 (2153 views)
Shortcut

Registered: Dec 3, 2002
Posts: 1800

Re: [jt512] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Isn't it kinda silly to make up a configuration completely different from the one described and then suggest that the same analysis ought to apply?

In any case, that 111 pound result is (in the correct configuration not employed here) merely an instantaneous peak load, rather than, as implied by the orbit allusions, a constant upward force, so neither ICBM's, jets, nor ceiling repair personnel will have to be scrambled in any case.

But you knew that.

Edit: I instantaneously added a "tan" to the former "instaneous" to repair the the gap alluded to by KnudeNoggin in a post that follows.


(This post was edited by rgold on Oct 14, 2008, 12:29 AM)


jmvc


Oct 13, 2008, 1:08 PM
Post #113 of 123 (2148 views)
Shortcut

Registered: Sep 10, 2007
Posts: 647

Re: [Lazlo] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Lazlo wrote:
When you say tension, do you mean force?

I would have thought so.

http://www.thefreedictionary.com/tension


(This post was edited by jmvc on Oct 13, 2008, 1:08 PM)


Partner cracklover


Oct 13, 2008, 2:07 PM
Post #114 of 123 (2136 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [knudenoggin] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

knude, the 50# weight won't move. This diagram represents an approximation of the system at the point of peak force:



GO


Partner cracklover


Oct 13, 2008, 7:32 PM
Post #115 of 123 (2116 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [cracklover] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Wait a minute, I just re-read your post (for like the fourth time). I think this is what you're saying, right KN?



GO


knudenoggin


Oct 13, 2008, 7:33 PM
Post #116 of 123 (2115 views)
Shortcut

Registered: May 6, 2004
Posts: 594

Re: [rgold] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

rgold wrote:
Isn't it kinda silly to make up a configuration completely different from the one described and then suggest that the same analysis ought to apply?
It would be, yes; but how is what I described that? --a sling-shot system,
compounded simply to highlight/amplify the effects I questioned.
AND it is a system that folks can readily DO, here in The Lab, after all.
But, okay, your turn: how can we demonstrate the theory?
What will show this peak force?
(Because the above experiment with weights showed no hint of it
(to the untrained eye?)!)

In reply to:
that 111 pound result is ... merely an instaneous peak load,
Which, to my quick look, seemed to be missing something, like one syllable.
But age-old [ca. '95] wisdom of Ken Cline yet retrievable has it also!
In reply to:
The short answer is that fall force is not linear with free-fall
distance or rope characteristics, and top roping on static rope won't
result in equipment failure and death. A 200 pound slingshot top
roper will load the anchor with about 660 pounds of force in a fall
with zero slack, regardless of whether the rope is dynamic, static,
bungee, or steel cable. As you add free-fall, the force felt by the
anchor starts to depend on the type of rope, but the transition isn't
all that abrupt for the types of rope that climbers use.

That sure seems like a heckuva load to not show effects, as I had tried to do.
The 25# dropped even some slight distance didn't budge its belay,
nor did the anchoring line budge the 50#.

In reply to:
But you knew that.
Nope, I'd be knewdNoggin then.
Rather, I've been finding that friction is huge in such often hyped pulley
systems (such as using pure rope, nothing metal), with behaviors quite
different from what is written, which is that near-theoretical MA obtains:
it's nowhere close.

*kN*


knudenoggin


Oct 13, 2008, 8:10 PM
Post #117 of 123 (2106 views)
Shortcut

Registered: May 6, 2004
Posts: 594

Re: [cracklover] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

cracklover wrote:
knude, the 50# weight won't move. This diagram represents an approximation of the system at the point of peak force:

[IMG]http://i38.tinypic.com/11jc4du.jpg[/IMG]

GO
Thanks for a quite nice image!
But the system I described was compound: replace, in your image, the 25#wgt.
with the 'biner, and a line run though IT anchored w/25# and the other
end has the 25# wgt. that is dropped. I.e., where you show 25# dropping
to generate double its mass in force would be a 'biner fielding such generated
force amplified to the 10/3 (25#) value.
THAT is what didn't budge the 50# (or anything).

----

But I don't follow your annotations!?

How do you figure "5 lb (friction)" ? The rule-of-thumb efficiency of 60-66%
would hold that the 50# force was reduced to 30-33 lb.s.
(Or are you showing my pulley vs. generic 'biner and assuming that it
has a high, 90% efficiency? But then, still, my next question ... :-)

And what is the red, up-arrowed "5 lb.s" at the line under the 50 lb. wgt.?
I'd think that if you were taking off 5 lb. at top, you'd show 45 lb. here, maybe?!
Which would not raise the 50#.

*kN*


Partner rgold


Oct 14, 2008, 1:21 AM
Post #118 of 123 (2089 views)
Shortcut

Registered: Dec 3, 2002
Posts: 1800

Re: [knudenoggin] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

knudenoggin wrote:
rgold wrote:
Isn't it kinda silly to make up a configuration completely different from the one described and then suggest that the same analysis ought to apply?
It would be, yes; but how is what I described that? --a sling-shot system,
compounded simply to highlight/amplify the effects I questioned.
AND it is a system that folks can readily DO, here in The Lab, after all.
But, okay, your turn: how can we demonstrate the theory?

It's very late, I have much work to do and a lot of climbing coming up---it's Gunks Reunion Week, so I ain't promising any theoretical treatises.

But looky here Knude, my original comments referred to a weight applied to a rope fastened to a fixed anchor point, but you've got a moveable anchor point; the 25 lb counterweight. That anchor point will start moving up as soon as the rope tension from the drop overcomes carabiner friction, so there will be less stretch in the rope than you'd get with a fixed anchor point and so less tension to transmit to the next part of the system.

Off the top of my very sleepy head, I'd say that if the rope to the counterweight starts sliding when the tension on the dropped weight side is 4/3 X 25, then the total peak load on the lower structure will just be about 7/3 X 25 or about 67 lbs, which won't budge the 50 lb load with the same friction to overcome on the top biner. This seems to be what you are observing, which of course does not mean that what I've just said is right.


Partner cracklover


Oct 14, 2008, 8:46 AM
Post #119 of 123 (2066 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [knudenoggin] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

knudenoggin wrote:
cracklover wrote:
knude, the 50# weight won't move. This diagram represents an approximation of the system at the point of peak force:



GO
Thanks for a quite nice image!

No prob. But (as I said before) after re-reading your post again, I realized that you had two 25 pound weights, not one. Thus the second diagram (see a few posts up).

In reply to:
But the system I described was compound: replace, in your image, the 25#wgt.
with the 'biner, and a line run though IT anchored w/25# and the other
end has the 25# wgt. that is dropped. I.e., where you show 25# dropping
to generate double its mass in force would be a 'biner fielding such generated
force amplified to the 10/3 (25#) value.

Look up in the thread. I posted what I think is exactly that. I'm happy to plot out the forces for you, but only after you've verified that the image I posted jives with the question you're asking. I don't want to answer the wrong question a second time.

In reply to:
But I don't follow your annotations!?

Well, maybe there's something to be gained by looking at that first diagram - even though it's not the question you asked. So sure, let's walk through it.

In reply to:
How do you figure "5 lb (friction)" ? The rule-of-thumb efficiency of 60-66%
would hold that the 50# force was reduced to 30-33 lb.s.
(Or are you showing my pulley vs. generic 'biner and assuming that it
has a high, 90% efficiency? But then, still, my next question ... :-)

Yes, the top thing that looks kinda like a pulley is... a pulley. And yes, I'm assuming 90% efficiency, or 10% friction (5 lb).

In reply to:
And what is the red, up-arrowed "5 lb.s" at the line under the 50 lb. wgt.?

That is the force the ground is exerting to hold up the 50 lb weight, while the tension in the line from the weight to the pulley is holding the other 45 lb.

In reply to:
I'd think that if you were taking off 5 lb. at top, you'd show 45 lb. here, maybe?!
Which would not raise the 50#.

*kN*

Not exactly, but I think you get the idea.

GO


knudenoggin


Oct 14, 2008, 10:44 AM
Post #120 of 123 (2059 views)
Shortcut

Registered: May 6, 2004
Posts: 594

Re: [cracklover] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

cracklover wrote:
In reply to:
Thanks for a quite nice image!

No prob. But (as I said before) after re-reading your post again, I realized that you had two 25 pound weights, not one. Thus the second diagram (see a few posts up)
...

Look up in the thread. I posted what I think is exactly that. I'm happy to plot out the forces for you, but only after you've verified that the image I posted jives with the question you're asking.
BINGO, well done, again!
Yes, 25# dropped (or, as in OP, even just *released*) onto 'biner system,
which delivers force to pulley'd one w/50# belay wgt..

-----------

> The usual Physics 101 explanation involves a vertical spring with a mass at its bottom.

And so I guess my problem is that I substituted for this mass the lower, 2nd sling-shot (2:1 TMA) system,
a sheave with so-calculated FORCE, not mass?!
And the theory I'm slow to understand applies to the force by mass in this lower system,
but then the upper one has acting on IT not a falling mass, directly, but a force resulting
from that?! --to which there is no doubling effect? (although a suspended mass STILL exists)

-----

Now, to the first & direct point, without complication, when I dropped (just a few inches, but SOMEthing more
than mere release) the upper 25# wgt. onto the >> 'BINER-SHEAVED sling-shot with equal, 25# belay wgt. <<
there was no movement of that BELAY/anchor wgt. (or 50# one) (just a standing to attention from its leaning
at, oh, maybe 85deg angle against something).

Which to my naive eye, didn't seem to show the delivery of the predicted peak force of 4/3 25# (but which the
aspect of instantaneity might dismiss?). And one should also say that it was more than 4/3(25#) given
the few-inches drop, but maybe then the factors of rope stretch go some ways to mitigate any rise?

*kN*


Partner rgold


Oct 14, 2008, 11:44 AM
Post #121 of 123 (2046 views)
Shortcut

Registered: Dec 3, 2002
Posts: 1800

Re: [knudenoggin] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Knude, the gizmo you have at the bottom is called an Atwood Machine and is famous in elementary physics. You might google around a bit and see if you can find any treatment that describes what happens if the rope stretches; the usual case involves the use of an inelastic string.


Partner cracklover


Oct 14, 2008, 1:17 PM
Post #122 of 123 (2038 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [knudenoggin] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

To be honest, Knude, I think the amount of time of the peak force is so short that we need to think about the problem a little differently. Basically, the tiny amount of energy involved is mostly converted to heat by your cords, rather than converting it to work done in lifting objects.

Look at it this way - what if you had a rubber band attached to a five lb weight on the floor. Attach the top of the rubber band to a lever, and drop a 1000 lb weight on the other end of the lever. If the lever can only go up two inches, and the rubber band can easily stretch that much, then it doesn't matter that an enormous amount of force is involved - all of the energy is easily turned to heat by stretching the rubber band two inches.

Now your cord isn't quite that stretchy, but I think showing a simple diagram of the forces, discounting the energy your cord can convert to heat, would be meaningless.

GO


Partner cracklover


Oct 14, 2008, 1:20 PM
Post #123 of 123 (2036 views)
Shortcut

Registered: Nov 14, 2002
Posts: 9973

Re: [cracklover] tension [In reply to]
Report this Post
Average: avg_1 avg_2 avg_3 avg_4 avg_5 (0 ratings)  
Can't Post

Oops, that's what I get for doing real work and then coming back to my post. It appears that RGold explained it away in two words.

Ah well. Have fun, Rich! I'd see you at the Gunks, maybe, but I'm no longer local.

Cheers!

GO


Forums : Climbing Information : The Lab

 


Search for (options)

Log In:

Username:
Password: Remember me:

Go Register
Go Lost Password?
$10.89 (10% off)
$4.95 (10% off)
$17.95 (10% off)
$44.96 (10% off)



Follow us on Twiter Become a Fan on Facebook