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Difficult problem involving friction and torque

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A cord connected at one end to a block which can slide on an inclined plane has its other end wrapped around a cylinder resting in a depression at the top of the plane as shown in the figure:
    http://img35.imageshack.us/img35/6743/giancolich10p098.jpg [Broken]
    Determine the speed of the block after it has traveled 1.40 m along the plane, starting from rest.

    Assume the coefficient of friction between all surfaces is μ = 3.00×10^-2. [Hint: First determine the normal force on the cylinder, and make any reasonable assumptions needed.]


    2. Relevant equations
    Newton's second law in linear and angular form

    3. The attempt at a solution
    The forces on the block are fairly obvious: tension T, friction f, weight mg, and normal F.
    However, I am not so sure about the forces on the cylinder. First there is T and Mg. I know there is a normal force N, but I wasn't sure of the direction so I drew it pointing to the upper left, an angle θ above the horizontal. I also was not sure about the frictional force. Is it perpendicular to N? If so, I got the following 4 equations in 4 unknowns:

    [tex]
    \]
    \\
    -Ncos\theta+Tcos27+\mu Nsin\theta=0 \\

    \mu Ncos\theta+Nsin\theta-Tsin27-Mg=0 \\

    T-f=.5Ma \\

    -T-mg cos27+mg sin27=ma
    \[
    [/tex]

    I can't seem to get the right answer from these though.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 14, 2009 #2
    From the diagram, I get the impression that the cylinder is not actually touching the triangle and is instead attached to a pivot point (which I will assume frictionless).

    If this is the case, the forces on the block are the gravitational force, kinetic friction force, tension force, and normal force. The only force on the cylinder is the tension force.

    You can use these to find that:
    [tex]\Sigma F = ma = mg \sin \theta - T - \mu_k N[/tex]
    and
    [tex]\Sigma \tau = I\alpha = R \times T[/tex]
     
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