Static friction between a cylinder and a wall

In summary, the problem involves a cylinder with a weight of 100[N] and a horizontal rope at the top preventing it from rolling. The coefficient of friction is needed to determine the tension in the rope when the cylinder is on the verge of movement. The correct calculation for the coefficient of friction involves drawing a free body diagram and taking into account the tension in the rope, which affects both the normal and tangential forces.
  • #1
Karol
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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  • #2
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
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
Karol said:
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
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
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
Karol said:
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
You are right, it comes out good
 

What is static friction?

Static friction is a force that resists the relative motion between two surfaces in contact. It occurs when there is no movement between the two surfaces, but there is a tendency for them to move.

How does static friction differ from kinetic friction?

Static friction only acts to prevent motion between two surfaces, while kinetic friction acts to slow down or stop motion that is already occurring.

What factors affect the magnitude of static friction?

The magnitude of static friction depends on the nature of the surfaces in contact, the force pressing the surfaces together, and any other external forces acting on the system.

How is the coefficient of static friction determined?

The coefficient of static friction is determined experimentally by measuring the force required to move an object horizontally across a surface, and then dividing that force by the weight of the object.

What is the role of static friction in preventing slippage between a cylinder and a wall?

Static friction plays a crucial role in preventing slippage between a cylinder and a wall by exerting a force that resists the motion of the cylinder and keeps it in place.

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