# Rotation, Torque, Static Friction

In summary, the problem asks for the maximum force P that can be applied to a uniform cylinder without causing it to rotate, given a coefficient of static friction of 0.500. The solution involves balancing forces and torques, with the suggestion to consider the direction of friction to oppose motion. The red dot shown in Figure P10.77 may be misleading as the normal force is zero at that point. The equilibrium conditions will help determine the values of friction and calculate the torques.

## Homework Statement

Figure P10.77 shows a vertical force applied tangentially to a uniform cylinder of weight F(g). The coefficient of static friction between the cylinder and all surfaces is 0.500. In terms of F(g), find the maximum force P that can be applied that does not cause the cylinder to rotate. (SUGGESTION: when the cylinder is on the verge of slipping, both friction forces are at their maximum values. Why?)

Shown below is Figure P10.77
http://img177.imageshack.us/img177/6066/figurep1077nk7.png [Broken]

## Homework Equations

torque anticlockwise = torque clockwise

## The Attempt at a Solution

I have absolutely no idea hos to solve this problem, however my physics teacher has suggested in use the equation of torque, that the torque anticlockwise is equal to the force clockwise in a state of equilibrium. Shown by the red dot on the image, that is the chosen pivot point suggested to be used.
Any suggestion/help is greatly appreciated,

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First treat surfaces as frictionless and check in which direction the cylinder (tries to) rotate. Now take direction of friction such that it opposes the motion. Now balance the forces and torque.

If there was no friction, then the force required to pull the cylinder up would be F(g) at its center, so the tangential torque would need to be rF(g) in order for the cylinder to move, correct?
thanks

If there was no friction, then the force required to pull the cylinder up would be F(g) at its center, so the tangential torque would need to be rF(g) in order for the cylinder to move, correct?
thanks

If there was no friction the cylinder will only rotate. If there is friction, there will be a torque opposite to the applied torque at the lower point of contact. The red dot could be a red-herring because the normal force is zero there.

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Thinking for equilibrium in horizontal direction (no motion in horizontal direction) we should have friction(at lower contact) towards right(to balance the normal reaction at upper contact).
At upper contact friction will be upward to oppose (clockwise) motion of cylinder. Apply force equilibrium conditions to get values of friction & try to cal the torques.

## 1. What is rotation?

Rotation is the movement of an object around a fixed point or axis. It is a type of circular motion and can be described by its angular velocity and acceleration.

## 2. What is torque?

Torque is the measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to the object by the distance from the axis of rotation.

## 3. How is torque related to rotation?

Torque and rotation are directly related. Torque is the cause of rotational motion, and the magnitude of the torque determines the speed and direction of the rotation.

## 4. What is static friction?

Static friction is a force that exists between two surfaces in contact that are not moving relative to each other. It prevents the surfaces from sliding against each other and is always equal and opposite to the applied force.

## 5. How does static friction affect rotation and torque?

Static friction plays a crucial role in rotation and torque. It allows for objects to maintain their position and not slip when a torque is applied. Without static friction, there would be no rotational motion as the object would simply slide instead of rotating.