Rope between inclines?

A rope rests on two platforms which are both inclined at an angle θ \theta (which you are free to pick), as shown. The rope has uniform mass density, and its coefficient of friction with the platforms is 1. The system has left-right symmetry.

What is the largest possible fraction of the rope that does not touch the platforms?

What angle θ \theta allows this maximum value?

Enter your answer as the sum of the answers of these questions.

Please take θ \theta in degrees.


The answer is 22.672.

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2 solutions

Sai Ram
Jun 20, 2015

Therefore the answer is 22.5 + 0.172 , which is equal to 22.672.

Please add that theta is an angle in degrees. I used radians.

Akash Gaonkar - 5 years, 11 months ago

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Thanks I have updated the question.

Then why don't you up vote my solution?

Sai Ram - 5 years, 11 months ago

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Because It's not yours.... http://liquids.seas.harvard.edu/oleg/competition/BAUPC_sols04.pdf

Вук Радовић - 5 years, 11 months ago

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@Вук Радовић Yes,I took the solution from there

Sai Ram - 5 years, 11 months ago
Jack Ceroni
May 12, 2020

Let α \alpha be the fraction of the rope that is suspended in the air. It follows that 1 α 1 \ - \ \alpha is the fraction of the rope that is touching the side of the inclined plane. It must be true that the gravitational force acting on the fraction the rope suspended between the two surfaces must be in balance with the force acting upwards on this component of the rope, which will be the tension at the point that the rope leaves the surface on either side of the suspended rope, which we call T 0 T_0 . Thus, we have:

2 T 0 sin θ = α L λ g 2T_0 \sin \theta \ = \ \alpha L \lambda g

Where L L is the length of the rope and λ \lambda is the mass density. Now, let's turn our attention away from the suspended component of the rope, and consider the components of the rope that are touching the surfaces. These components of the rope must also be in equilibrium. Thus, we have (on either side):

1 2 ( 1 α ) L λ g [ sin θ + cos θ μ k ] = T 0 \frac{1}{2} (1 \ - \ \alpha) L \lambda g \big [ - \sin \theta \ + \ \cos \theta \mu_k \big] \ = \ T_0

Thus, we will have:

[ sin θ + cos θ μ k ] sin θ = α 1 α \big [ - \sin \theta \ + \ \cos \theta \mu_k \big] \sin \theta \ = \ \frac{\alpha}{1 \ - \ \alpha}

And we then re-arrange, getting α ( θ ) \alpha(\theta) :

α ( θ ) = [ sin θ + cos θ μ k ] sin θ [ sin θ + cos θ μ k ] sin θ + 1 = [ sin θ + cos θ μ k ] sin θ [ sin θ + cos θ μ k ] sin θ + 1 \alpha(\theta) \ = \ \frac{[ -\sin \theta \ + \ \cos \theta \mu_k \big] \sin \theta}{[ - \sin \theta \ + \ \cos \theta \mu_k \big] \sin \theta \ + \ 1} \ = \ \frac{[ - \sin \theta \ + \ \cos \theta \mu_k \big] \sin \theta}{[ - \sin \theta \ + \ \cos \theta \mu_k \big] \sin \theta \ + \ 1}

We then take the derivative with respect to θ \theta and find the value of (\theta) that corresponds to maximum α \alpha in the range 0 0 to π 2 \frac{\pi}{2} , which turns out to be π 8 \frac{\pi}{8} , which is equivalent to 22.5 22.5 degrees. Finally, we can plug this back into the original equation, getting α Max = 0.172 \alpha_{\text{Max}} \ = \ 0.172 .

Adding these results together, we get 22.5 + 0.172 = 22.672 22.5 \ + \ 0.172 \ = \ \fbox{22.672}

By the way, you could have just considered the vertical force acting on the whole system, but I think it is slightly cleaner considering the two sub-systems of the rope!

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