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Angle required to keep block from sliding down plane

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Here's a picture of the set up.

    I need to find out what angle I need to have the plane raised to so that the blocks aren't moving.


    2. Relevant equations

    F = ma

    3. The attempt at a solution
    For the first block
    [tex]F_k=μ_kF_N[/tex]
    [tex]F_T+F_N-μ_kF_N+2mgcosθ+2mgsinθ=0[/tex]

    Since the [tex]F_N[/tex] and [tex]2mgcosθ[/tex] cancel each other, I eliminated them from the problem. I set the right side equal to 0 since there is no acceleration

    Then I get
    [tex]F_T = μ_kF_N + 2mgsinθ[/tex]

    For the right block [tex]mg = F_T[/tex]
    So I input that into my equation for the other block and get
    [tex]mg=μ_kF_N+2mgsinθ[/tex]

    For [tex]F_N[/tex] I plug in [tex]2mgcosθ[/tex]

    And get [tex]mg=μ_k2mgcosθ+2mgsinθ[/tex]

    And this is when I know something is wrong. That [tex]μ_k[/tex] is the coefficient for kinetic friction, not static friction, and I have [tex]sinθ[/tex] and [tex]cosθ[/tex], so that's two unknowns.

    How do I find out the coefficient for static friction? Do I even need it?

    And what went wrong with my analysis of this problem?

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Jan 27, 2013 #2

    SammyS

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    For one thing, if the blocks are stationary, why not use μs, not μk ?

    For another: [itex]\ F_N\ [/itex] and [itex]\ 2mg\cosθ\ [/itex] would not cancel each other in your expression. [itex]\ F_N\ [/itex] is multiplied by μ .

    The first thing to do is define a coordinate system for the 2m block and write equations for force for each component. It looks as though you have things all jumbled together into one (erroneous) equation.
     
  4. Jan 28, 2013 #3
    Ok thanks.
     
    Last edited: Jan 28, 2013
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