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Dynamic Equilibrium Question (block sliding on wall)

  1. Oct 11, 2008 #1


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    Gold Member

    1. The problem statement, all variables and given/known data

    The 2.0kg wood box in the figure slides down a vertical wood wall while you push on it at a 45 degree angle.

    What magnitude of force should you apply to cause the box to slide down at a constant speed?

    2. Relevant equations

    I am having a hard time figuring out the normal force in this problem.

    I have set the problem up with four forces (f_k (friction), F_g (gravity), n (normal), F_push (push)). I got the coefficient of friction (f_k) as 0.20 from my textbook.

    Fnet(x) = F_push(x) + F_g(x) + n + f_k(x)
    Fnet(x) = F_push*cos(45) + 0 + n + 0

    Fnet(y) = F_push(y) + F_g(y) + f_k(y)
    Fnet(y) = F_push*sin(45) + (2.0)(9.8) -[0.20*2.0*(sin(45)]

    3. The attempt at a solution
    I know that the normal force acts perpendicular to the surface, so their should be no n in the y-direction. I'm stuck in finding the normal force in the x-direction and I believe this is the missing link in me figuring out this problem.
    Last edited: Oct 11, 2008
  2. jcsd
  3. Oct 11, 2008 #2


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    Homework Helper

    Welcome to PF.

    Since the angle is 45 degrees, don't you know what the normal force is in terms of the force?

    And isn't that really a part of the equation I highlighted and not the 2.0 as it relates to friction?

    As in 2.0*(9.8) needs to balance with .2 * F * cos45 as well as the vertical component F * Sin45
    Last edited: Oct 11, 2008
  4. Oct 11, 2008 #3


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    Gold Member


    Thanks for welcoming me to this site and thanks for your guidance.
    The normal force in terms of the push force is F_push*cos45. Using that to find f_k, I think I have the solution.

    Fnet(y) = f_k + F_g + F_push(y) = 0

    Fnet(y) = 0.20*F_push*cos45 - (2.0)(9.8) + F_push(sin45) = 0
    19.6 = 0.1414*F_push + 0.7071*F_push

    Therefore, F_push is roughly 23 N.
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