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Static friction between a cylinder and a wall

  • Thread starter Karol
  • Start date
  • #1
1,380
22

Homework Statement


The cylinder in the drawing weighs 100[N]. there is a horizontal rope at the top that prevents it from rolling. what is the coefficient of friction if it's on the verge of movement

Homework Equations


Friction force: f=μN

The Attempt at a Solution


The whole weight of the cylinder lies on the contact point. the coefficient of friction is the tangent of the slope, which is here 250, thus μ=tan 250=0.422
It should be 0.22
 

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Answers and Replies

  • #2
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Have you drawn a free body diagram? The normal component of the contact force is not pointing in the vertical direction. So the normal force is not equal to mg. You need to write down the force balance equations for equilibrium, and you probably also need to write down a moment balance equation. Call N the normal contact force, and call F the tangential contact force.

Chet
 
  • #3
1,380
22
yes i did that but was is a hurry so i used the other method of the tangent which gave the same result.
$$N=100\cdot \sin 65^0,\ F=100\cdot \cos 65^0$$
$$F=\mu N\rightarrow \mu=\frac{100\cdot \cos 65^0}{100\cdot \sin 65^0}=0.466$$
 
  • #4
TSny
Homework Helper
Gold Member
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yes i did that but was is a hurry so i used the other method of the tangent which gave the same result.
$$N=100\cdot \sin 65^0,\ F=100\cdot \cos 65^0$$
Shouldn't the tension in the rope contribute to these two equations?
 
  • #5
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4,092
You left out the effect of the rope tension T. That has a component in the normal direction and a component in the tangential direction. The reason that this happened is that you did not draw a free body diagram.

Chet
 
  • #6
1,380
22
I thought about it but decided that the rope doesn't influence. the rope only balances the moment (torque) and it comes out right according to the book:
$$100\cdot R\cdot\sin 25^0=T\cdot R(1+\cos 25^0)\rightarrow T=22.17$$
I will draw a free body diagram soon
 
  • #7
19,915
4,092
I thought about it but decided that the rope doesn't influence. the rope only balances the moment (torque) and it comes out right according to the book:
$$100\cdot R\cdot\sin 25^0=T\cdot R(1+\cos 25^0)\rightarrow T=22.17$$
I will draw a free body diagram soon
T influences the values of both F and N. Did you really feel that you had advanced beyond the point where you need to draw free body diagrams?

Chet
 
  • #8
1,380
22
You are right, it comes out good
 

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