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Static Friction when pushing a crate on the floor

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    You try to push a crate of weight M with a force F on a horizontal floor. The coefficient of static friction is μ and you exert the force F under an angle θ below the horizontal.

    a) Determine the minimum value of F that will move the crate
    b) Assume you have not sufficient strength to move the crate. You now do this exercise in an elevator. Which direction should you the elevator accelerate to have a chance to move the box?


    2. Relevant equations

    The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force: (f= μ Fn)

    3. The attempt at a solution

    I have no idea after that....
     
    Last edited: Sep 3, 2009
  2. jcsd
  3. Sep 3, 2009 #2

    Hootenanny

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    Re: Friction

    Welcome to Physics Forums.

    Have you drawn a free-body diagram? I always find that they are extremely helpful when solving this type of problem.
     
  4. Sep 3, 2009 #3
    Re: Friction

    Yes, i did. Just can't figure out the value F as there is no known data given.
     
  5. Sep 3, 2009 #4

    Hootenanny

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    Re: Friction

    Start by summing the forces acting in the vertical and horizontal direction. Since this is a static problem, what do you know about their sums?
     
  6. Sep 3, 2009 #5
    Re: Friction

    Horizontal: F cos θ - fs = ma
    Vertical : -mg + F sin θ + n = 0

    Am i right?
     
  7. Sep 3, 2009 #6

    Hootenanny

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    Re: Friction

    Indeed you are. As I said in my previous post, we are dealing with the static case here, so a=0.

    Now, can you expression fs in terms of n?
     
  8. Sep 3, 2009 #7
    Re: Friction

    Are you referring to the a in the net horizontal force has to be 0 always in static case? Or both for vertical as well?

    Horizontal: F cos θ - μn = 0
    Vertical : -mg + F sin θ + n = 0

    Solving the y equation gives n= F cos θ / μ
     
  9. Sep 3, 2009 #8
    Re: Friction

    To get the F value, sub n= F cos θ / μ into x equation. Is this the solution?
     
  10. Sep 3, 2009 #9

    Hootenanny

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    Re: Friction

    Sounds good to me :approve:
     
  11. Sep 4, 2009 #10
    Re: Friction

    Thank you so much!! ^.^
     
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