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Frictional force on a block held against a vertical wall

  1. Feb 27, 2013 #1
    1. A 5kg block is held at rest against a vertical wall by a horizontal force of 100 newtons.
    1-What is the frictional force exerted by the wall on the block
    2-What is the minimum horizontal force needed to prevent the block from falling if the static coefficient of friction between the wall and the block is μs = .4



    2. Relevant equations
    MAx = Right forces- Left Forces ( This is a homemade equation, basically the forces pushing from the wall are subtracted from the forces pushing against the wall.


    3. The attempt at a solution
    I am unsure how to proceed with this problem. I know that for part 1, the frictional force should be about 49.1n, this comes from the equation shown above. If right forces- left forces= 0, then there is no movement, and therefore they must be equal. So 5kg*9.81m/s = 49.05N
    The second part is what confuses me, I fail to see a way to incorporate the friction coefficient into the problem in a meaningful way.
     
    Last edited: Feb 27, 2013
  2. jcsd
  3. Feb 27, 2013 #2

    Doc Al

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    You don't need the coefficient of friction for the first part. Hint: The block is in equilibrium.
     
  4. Feb 27, 2013 #3

    haruspex

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    You know how large the frictional force needs to be from the first part. If the horizontal force is reduced, what does that do to the max frictional force available? At what point will it the available frictional force equal the required frictional force?
     
  5. Feb 27, 2013 #4
    I think the part you're missing is that the frictional force is determined by:

    Friction force = μs * Normal force

    In this case you start out with the normal force pushing against the wall at 100N, but how low could it go before the block fell down?
     
  6. Feb 28, 2013 #5
    So I should set μk*Fn= Frictional Force exerted by the wall.
    .4*FN = 49.1
    Fn = 122.8 (about)

    That actually makes a lot of sense, thank you all.
     
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