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Rotation, Torque, Static Friction

  1. Apr 2, 2007 #1
    1. The problem statement, all variables and given/known data
    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
    [​IMG]

    2. Relevant equations
    torque anticlockwise = torque clockwise


    3. 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,
    unique_pavadrin.
     
  2. jcsd
  3. Apr 2, 2007 #2
    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.
     
  4. Apr 2, 2007 #3
    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
     
  5. Apr 2, 2007 #4

    Mentz114

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    Gold Member

    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.
     
    Last edited: Apr 2, 2007
  6. Apr 2, 2007 #5
    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.
     
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