# Forces on ropes connected to a slanted ceiling

• SweatingBear
In summary, the conversation discusses the forces on two ropes hanging from a slanted or horizontal ceiling. The claim is that the forces will equal 150 N regardless of the ceiling's slant. The concept of equilibrium and torque is brought up and it is determined that the tension in each rope is equal to half of the girl's weight. The gruesome picture is used as a visual aid to understand this concept. Ultimately, it is concluded that the force on each rope is half of the person's weight, regardless of the ceiling's slant.
SweatingBear
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:

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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.)

Doc Al said:
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?

Doc Al said:
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.

SweatingBear said:
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.

Doc Al said:
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?

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?

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.

Sorry for the gruesome picture above

figuratively, or literally ( in which case I do not see the gory details )

Great explanation - works for me.

256bits said:
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.

Spinnor said:
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.

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).

SweatingBear said:
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.

Doc Al said:
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!

## 1. What is the relationship between the angle of the ceiling and the forces on the ropes?

The angle of the ceiling affects the direction and magnitude of the forces on the ropes. As the angle increases, the horizontal component of the force decreases while the vertical component increases. This means that the ropes will experience more tension and pull towards the ceiling at a steeper angle.

## 2. What is the difference between tension and compression forces on the ropes?

Tension forces act to stretch or elongate the ropes, while compression forces act to shorten or compress them. In the case of ropes connected to a slanted ceiling, the tension forces are pulling the ropes towards the ceiling, while the compression forces are pushing the ropes away from the ceiling.

## 3. How do the forces on the ropes change if the weight or length of the ropes is altered?

The weight of the ropes will increase the tension forces, while the length of the ropes will affect the magnitude of both the tension and compression forces. If the weight or length of the ropes is increased, the forces on the ropes will also increase, and vice versa.

## 4. Can the forces on the ropes cause the ceiling to collapse?

It is possible for the forces on the ropes to exceed the weight-bearing capacity of the ceiling, leading to collapse. This is why it is important to consider the strength of the ceiling and the weight and length of the ropes when setting up this type of system.

## 5. How can I calculate the forces on the ropes connected to a slanted ceiling?

The forces on the ropes can be calculated using the principles of trigonometry and Newton's laws of motion. The angle of the ceiling, weight and length of the ropes, and the weight-bearing capacity of the ceiling must all be taken into account. It is recommended to seek the assistance of a trained engineer or physicist for accurate calculations.

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