Solving Tension Questions: 2Ftcos(theta) = mg

[ual8658]

While doing my coursework, I ran across two things regarding tension I do not understand.
[Mentor's note: The second question has been moved into its own thread]

The first is if a rope is attached at two ends, for example attached to two canyon walls with the rope over the canyon itself, why is it that (2F_tcos(theta) = mg) if theta is the angle the rope makes below the horizontal on each side due to a weight in the exact middle of the rope? Why isn't it F_tcos(theta) = mg instead?

 

[Nugatory]

While doing my coursework, I ran across two things regarding tension I do not understand.
[Mentor's note: The second question has been moved into its own thread]

The first is if a rope is attached at two ends, for example attached to two canyon walls with the rope over the canyon itself, why is it that (2F_tcos(theta) = mg) if theta is the angle the rope makes below the horizontal on each side due to a weight in the exact middle of the rope? Why isn't it F_tcos(theta) = mg instead?

This situation is no different than if the weight were hanging from two ropes, both tied to the object at one end ans to a canyon wall at the other. What tension would you expect to find in the ropes in that case, and why?

 

[ual8658]

I still can't grasp how one rope can become two ropes. I understand what happens when there are two ropes, but how can a weight hanging on one rope be pictured as two ropes?

 

[mfig]

I think you have the wrong expression for tension. If the angle is measured from the horizontal then the tension in the rope is

T = mg/2sin(θ)

The factor of 2 comes from the fact that each half of the rope must hold up half the weight of the object. To check something like this, I often look at limiting cases where intuition is more solid. To that end, imagine we make the angle with the horizontal 90°, so that the ends of the rope are directly above the object and there is no horizontal component whatsoever. Clearly in this case each half of the rope is supporting exactly half the object's weight, and our formula agrees.

T = mg/2sin(90°) = mg/2

If the factor of 2 were not there, then we would calculate each half of the rope to be pulling vertically upwards with a force equal to the object's weight, for a total upward force of twice the weight of the object. Since the only downward force is the weight of the object itself, there would be a net acceleration without the factor of 2 in our expression.

This would violate the equilibrium condition we set at the beginning of the exercise. Thus we see the factor of 2 is correct.

 

[ual8658]

Ok so in this situation, we are imagining that each end of the rope supports half the weight because the rope is attached at two different ends which leads to the tension we see?

 

[mfig]

"Ok so in this situation, we are imagining that each end of the rope supports half the weight because the rope is attached at two different ends which leads to the tension we see?"

Yes. To see this, imagine a sequence of transformations. In the diagram, the ropes are black and the hanging object is blue. The red ring is a super-rigid steel. I think you would agree in the first image that each half of the rope is holding half the weight when the steel ring is in the middle. Correct? Now, ask how the physics would change if the ring were a semi-circle instead of a circle. I believe you can see the physics does not change: each half of the rope is supporting half the weight. Now ask yourself how the physics would change if the semi-circle of steel were replaced by a short segment of rope. The physics does not change here either: each half of the rope must still lift half of the object. As the final move, imagine the short section of rope is simply the same rope as each end so there is no separation along the rope. The physics still do not change! Each half of the rope still supports half the object's weight.

 

[ual8658]

Ok thank you. This makes sense now!

 

[mfig]

Ok thank you. This makes sense now!

You are welcome. :-)