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Just for fun

  1. Jun 23, 2003 #1
    Here is a basic, yet moderately challenging physics problem just for fun:

    A large cylinder lies at rest next to a wall so that it is tangent to both the wall and the floor beneath it; (oriented similar as wheel attached to a car). You are to exert a torque on the cylinder until it starts to roll against the wall (e.g. overcome the frictional force of the part of the cylinder that is pushing against the ground). Write the equation for the force that is required as a function of the weight of the cylinder; the coefficient of friction of the wall at all sides being "u."
     
  2. jcsd
  3. Jun 24, 2003 #2
    Let's treat it in 2-D, and let's call the point where the cylinder touches the wall, W, and let's call the point where it touches the ground, G.
    Now we apply a downward force of F at W. The component in the direction of WG will be F/sqrt(2), and this will act on G, giving a downward component of F/2, and a component away from the wall of F/2.
    Since actio = reactio, the cylinder wil also push against the wall with F/2.
    So, total friction will be
    uF/2 + uF/2 + umg
    (the latter coming from the cylinder's weight).
    So,
    F > uF + umg,
    or
    F > umg/(1-u)

    (Edited: I removed some errors)

    Correct?
     
    Last edited: Jun 25, 2003
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