Maximum force to not rotate a cylinder

[nautola]

Homework Statement

A vertical cylinder is sitting against a wall. It has a weight of 20N and no given radius. A force is applied tangentially in a vertical direction on the side not against the wall. The coefficient of friction is 0.499 for all surfaces. What is the maximum force so that the cylinder will not rotate?

Homework Equations

τ= Fr-f(bottom)r-f(wall)r=0
f=μN

The Attempt at a Solution

I set up the necessary equations and I thought that since there wasn't a normal force applied horizontally, the wall didn't have any friction so I neglected it and answered 9.98. That wasn't correct.


Sorry about the sideways picture.

 

[RTW69]

The wall exerts a normal force and has a friction force. The cylinder isn't moving so sum of torques, and x and y forces equals 0. Draw a free body diagram to account for all the forces.

 

[JHamm]

I might have a solution however I'd rather not give you a possibly incorrect set of reasoning if it turns out to be wrong, care to give the correct number?

 

[Delphi51]

I wouldn't call that a "vertical" cylinder!
It is getting very late here, and I usually write nonsense late at night, but perhaps you will find something useful in this argument.
There must be a normal force pressing it against the wall.
Consider the horizontal forces on the cylinder. You have a friction force F₁ on the bottom going to the right (opposing the turning). There must be another horizontal force on the side preventing the cylinder from accelerating to the right - the wall pushes on the cylinder with F₁. (This F₁ results from a normal force F₀ acting down on the floor and up on the cylinder F₁ = μF₀ [1]). So the cylinder must push on the wall with F₁ . This normal force results in an upward friction force μF₁ . Vertically we have no acceleration so F + μF₁ - mg + F₀ = 0 [2]. The torques about the touch point on the wall add up to zero: F*2r - mgr + F₀r = 0 [3]. We have 3 equations in unknowns F,F₀ and F₁. Solve them for F. I got a little less than 3 N but I usually calculate incorrectly late at night.

 

[asteriatic]

I would first clarify the assumptions and parameters of the problem. For example, does the cylinder have a fixed base or is it able to roll? What material is the wall made of? Is the force applied at the center of the cylinder or at a specific point on the surface? These details can greatly affect the solution.

Assuming that the cylinder is a solid, fixed base cylinder and the force is applied at the center, the maximum force to prevent rotation can be calculated using the equation τ= Fr-f(bottom)r-f(wall)r=0, where τ is the torque, F is the applied force, r is the radius of the cylinder, and f(bottom) and f(wall) are the friction forces at the bottom and against the wall, respectively. Since the cylinder is not rotating, the torque must be equal to zero.

To calculate the friction forces, we can use the equation f=μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the cylinder, which is 20N. Therefore, the friction forces are f(bottom)=0.499*20=9.98N and f(wall)=0. The friction force against the wall is zero because there is no applied force in that direction.

Substituting these values into the torque equation, we get Fr-9.98r=0. Solving for F, we get F=9.98N. This is the maximum force that can be applied to the cylinder without causing it to rotate.

However, as a scientist, I would also note that this solution is based on several assumptions and may not accurately represent a real-world scenario. For example, if the cylinder is able to roll, the maximum force to prevent rotation would be different. Additionally, the coefficient of friction may vary depending on the material of the wall and the cylinder. Therefore, it is important to carefully consider all assumptions and parameters when solving a problem like this.