Maximum force to not rotate a cylinder

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SUMMARY

The maximum force that can be applied tangentially to a vertical cylinder against a wall, without causing it to rotate, is determined by analyzing the forces and torques acting on the cylinder. Given a weight of 20N and a coefficient of friction of 0.499, the correct approach involves accounting for the normal force exerted by the wall and the frictional forces at play. The calculations yield a maximum force of approximately 3N, derived from the equilibrium of forces and torques acting on the cylinder.

PREREQUISITES
  • Understanding of torque equations (τ = Fr)
  • Knowledge of friction coefficients and their application (f = μN)
  • Ability to draw and interpret free body diagrams
  • Familiarity with equilibrium conditions in static systems
NEXT STEPS
  • Study the principles of static equilibrium in rigid bodies
  • Learn about the application of friction in rotational dynamics
  • Explore advanced torque calculations in mechanical systems
  • Investigate the effects of varying coefficients of friction on stability
USEFUL FOR

Students in physics or engineering, mechanical engineers, and anyone studying static equilibrium and rotational dynamics in mechanical systems.

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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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