Maximum force to not rotate a cylinder

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Homework Help Overview

The problem involves a vertical cylinder resting against a wall, with a weight of 20N and a coefficient of friction of 0.499. A force is applied tangentially in a vertical direction, and the objective is to determine the maximum force that can be applied without causing the cylinder to rotate.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to account for the normal force exerted by the wall and the friction forces involved. There are considerations about the equilibrium of forces and torques acting on the cylinder. Some participants express uncertainty about their reasoning and seek clarification on the correct approach.

Discussion Status

Several participants are actively engaging with the problem, offering different perspectives on the forces at play. There is an acknowledgment of the complexity of the setup, with some participants suggesting the need for a free body diagram to visualize the forces. No consensus has been reached, but there are multiple lines of reasoning being explored.

Contextual Notes

Participants note the absence of a specified radius for the cylinder, which may impact the calculations. There is also mention of potential errors in reasoning due to late-night discussions, indicating a need for careful review of the assumptions made.

nautola
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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.
 

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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.
 
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?
 
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.
 

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