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Forces on ropes connected to a slanted ceiling

by SweatingBear
Tags: ceiling, connected, forces, ropes, slanted
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SweatingBear
#1
Dec30-12, 07:05 AM
P: 28
Hey forum.

So I was posed the question whether the forces on two ropes will vary or not depending on if the ceiling from which they are hanging is slanted or not. Here's a picture depicting the scenario:



I claim they'll equal 150 N respectively, independent of the ceiling slanting or not. But I am not sure how to motivate this (mathematically and/or conceptually). Do you guys have any idea?

Picture-source:
https://www.youtube.com/watch?v=hSQM0hoS6VE
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Doc Al
#2
Dec30-12, 07:18 AM
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Pretend the top of the picture was covered up so you had no idea about what the ropes attached to. Would it change your solution?

Do you know how to solve this problem in the first place? Assume a horizontal ceiling, if it helps. (Consider torques acting on the rod the girl hangs from.)
SweatingBear
#3
Dec30-12, 08:14 AM
P: 28
Quote Quote by Doc Al View Post
Pretend the top of the picture was covered up so you had no idea about what the ropes attached to. Would it change your solution?
No, I do not think it would. The upward-forces would still have to cancel the downward-forces. But if it was slanted, I am not sure, albeit inclined to believe that the respective forces still equal 150 N. Got any idea?

Quote Quote by Doc Al View Post
Do you know how to solve this problem in the first place? Assume a horizontal ceiling, if it helps. (Consider torques acting on the rod the girl hangs from.)
Well, in that case the upward-directed forces must equal the downward-directed forces because equilibrium in the system reigns.

Doc Al
#4
Dec30-12, 08:17 AM
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Forces on ropes connected to a slanted ceiling

Quote Quote by SweatingBear View Post
Well, in that case the upward-directed forces must equal the downward-directed forces because equilibrium in the system reigns.
That's true, but only part of the story. For equilibrium, the torques about any point must add to zero as well. You can use that fact to solve for the tension in each rope.
SweatingBear
#5
Dec30-12, 08:52 AM
P: 28
Quote Quote by Doc Al View Post
That's true, but only part of the story. For equilibrium, the torques about any point must add to zero as well. You can use that fact to solve for the tension in each rope.
Hm, I see your point. But I do not think that is necessary because in that part of the video-series, he has not even mentioned torque. So I am supposing that the question can be answered without respect to torque.

But, if do have to consider torque, I am not really sure how the calculation really goes; care to show?
256bits
#6
Dec30-12, 01:27 PM
P: 1,478
SweatingBear
Is it the slant of the ceiling or the differing lengths of the two ropes that has you wondering?

How does force transmit itself from one location on the rope to the next? Does the tension in a rope vary along its length?
If the roof was so high that you could not tell whether the roof was slanted or not by just looking at it straight up to determine an incline, could you detemine the force at each end on the bar?
Spinnor
#7
Dec30-12, 02:15 PM
P: 1,368
Cut the girl vertically in half and attach each half to each end of each rope and remove the horizontal rod that she hangs from. The tension in each rope is the same regardless of the slope of the roof assuming equal halves.

Does this help? Sorry for the gruesome picture above.
256bits
#8
Dec30-12, 02:30 PM
P: 1,478
Sorry for the gruesome picture above
figuratively, or literally ( in which case I do not see the gory details )

Great explanation - works for me.
SweatingBear
#9
Dec30-12, 03:24 PM
P: 28
Quote Quote by 256bits View Post
SweatingBear
Is it the slant of the ceiling or the differing lengths of the two ropes that has you wondering?

How does force transmit itself from one location on the rope to the next? Does the tension in a rope vary along its length?
If the roof was so high that you could not tell whether the roof was slanted or not by just looking at it straight up to determine an incline, could you detemine the force at each end on the bar?
Hm, interesting thought. I suppose they, in that case, would have to make up 150 N respectively.

Quote Quote by Spinnor View Post
Cut the girl vertically in half and attach each half to each end of each rope and remove the horizontal rod that she hangs from. The tension in each rope is the same regardless of the slope of the roof assuming equal halves.

Does this help? Sorry for the gruesome picture above.
Yes, quite helpful. :)

Concluding remark: Regardless of if the ceiling is slanted or not, the force on each rope is half of the persons weight i.e. 150 N (respectively).
Doc Al
#10
Dec30-12, 03:41 PM
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Quote Quote by SweatingBear View Post
Hm, I see your point. But I do not think that is necessary because in that part of the video-series, he has not even mentioned torque. So I am supposing that the question can be answered without respect to torque.
Sure, you can just appeal to symmetry, which is what I was leading you towards with my suggestion to pretend you couldn't see the upper part of the diagram. (It doesn't matter!) Since the bar is horizontal and the girl is even suspended (her weight acting right in the middle), you can conclude that the rope tensions must be equal.
But, if do have to consider torque, I am not really sure how the calculation really goes; care to show?
If you wanted to use torque to figure out the tension in the rope, here's how it goes. We'll use the left end of the bar as the pivot point. The torque from the girl's weight, which acts in the middle of the bar, exerts a clockwise torque of WL/2 (where L is the length of the bar). The right hand rope exerts a counterclockwise torque of TL, where T is the tension. Those torques must balance to have equilibrium, so WL/2 = TL and thus T = W/2. The tension in the right hand rope must equal half the weight. The same logic applies to either end, of course.
SweatingBear
#11
Dec30-12, 11:34 PM
P: 28
Quote Quote by Doc Al View Post
Sure, you can just appeal to symmetry, which is what I was leading you towards with my suggestion to pretend you couldn't see the upper part of the diagram. (It doesn't matter!) Since the bar is horizontal and the girl is even suspended (her weight acting right in the middle), you can conclude that the rope tensions must be equal.

If you wanted to use torque to figure out the tension in the rope, here's how it goes. We'll use the left end of the bar as the pivot point. The torque from the girl's weight, which acts in the middle of the bar, exerts a clockwise torque of WL/2 (where L is the length of the bar). The right hand rope exerts a counterclockwise torque of TL, where T is the tension. Those torques must balance to have equilibrium, so WL/2 = TL and thus T = W/2. The tension in the right hand rope must equal half the weight. The same logic applies to either end, of course.
Fantastic answer, thank you Doc Al!


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