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Force required to keep a wooden block from sliding on a rough incline

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data

    This thing was a part of an entire question . I solved the other parts but I don't know why I am messing up here .

    Force required to keep a wooden block from sliding on a rough incline with coefficient of friction μ

    2. Relevant equations

    Force on a block on an inclined plane ,
    towards the ground : m*g*sin(θ)
    due to friction in opposite direction of the above force is m*g*cos(θ)
    where m is the mass of the block , g is the gravitational acceleration and θ is the angle the incline makes with the horizontal.
    3. The attempt at a solution


    So I pictured something like this
    Initially
    http://awwapp.com/s/38/18/09.png [Broken]

    Now if I apply a force to stop the block from sliding the FBD should be something like this
    http://awwapp.com/s/5a/44/08.png [Broken]

    So the force I have to apply ( the blue one ) should be Force due to gravity - Force due to friction
    F = mg(sin(θ) -μcos(θ))
    But the answer given is
    F = μ*m*g*cos(θ)
    How is this possible
    Sorry about the drawings
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 15, 2013 #2

    Doc Al

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    Staff: Mentor

    I'd say you were correct. If this is a textbook problem, what textbook is it?
     
  4. Apr 15, 2013 #3
    @Doc Al It's actually a Test prep book published by a very well known publisher in my country at least . This question was part of the actual test held in 2012 . The question goes like
    The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is 1/(2*sqrt(3)) angle of the inclined plane is
    a> 60 degrees
    b> 45 degrees
    c> 30 degrees
    d> 15 degrees

    The answer in the key is 30 degrees
     
  5. Apr 15, 2013 #4

    Doc Al

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    Staff: Mentor

    I agree with that answer.
     
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