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Maximum force to not rotate a cylinder

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    τ= Fr-f(bottom)r-f(wall)r=0

    3. 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.

    Attached Files:

  2. jcsd
  3. Nov 13, 2011 #2
    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.
  4. Nov 14, 2011 #3
    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?
  5. Nov 14, 2011 #4


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    Homework Helper

    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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