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Homework Help: Inclined plane and friction

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data

    There are two wedges, each with mass m, placed on a flat floor. A cube of mass M is balanced on the wedges. Assume there is no friction between wedge and block, and the coefficient of static friction is less than 1 between the floor and wedges. What is the largest M that can be balanced without moving?

    diagram here:

    2. Relevant equations
    3. The attempt at a solution

    If we split the block in half due to symmetry, the question will get easier.

    force of friction is given by:
    [tex]F=\mu (\frac{M}{2}+m)g[/tex]
    the M/2 comes from taking only half of the block

    i want to equate the force that the block has on one of the wedge.

    Still considering half of the diagram, i want the normal force of (M/2)g to the wedge's surface. The normal force is give by the usual inclined plane equation, mgcos(theta)

    So normal force of (M/2) is:


    this normal force pushes in the opposite direction of the force of friction... but we have to consider the horizontal component, which tags on a sin(theta) term

    I finally get:

    [tex]\mu (\frac{M}{2}+m)g=\frac{Mg}{2}cos(\theta)sin(\theta)[/tex]

    And then i solve for mu. How close am I to being correct?
  2. jcsd
  3. Aug 19, 2008 #2


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    Hi learning_phys! :smile:

    No … consider the forces on the block

    there is no friction, so the only forces are Mg/2 and N …

    the horizontal components balance anyway …

    so resolve vertically, and that gives you … ? :smile:
  4. Aug 19, 2008 #3
    are you sayin N=Mg/2? but the normal force should be perpendicular to the blocks surface right? (that's why I have the cosine term)
  5. Aug 19, 2008 #4


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    Yes (because there is no friction). :smile:
    No … resolve vertically.
  6. Aug 19, 2008 #5
  7. Aug 19, 2008 #6


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    That's it! :smile:
  8. Aug 19, 2008 #7

    so I would get:
    [tex]\mu (\frac{M}{2}+m)g=\frac{Mg}{2cos(\theta)}sin(\theta )[/tex]
  9. Aug 20, 2008 #8


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    :biggrin: Woohoo! :biggrin:
    (you're easily awed! … :wink:)
    And so the largest M is … ? :smile:
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