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Acceleration of a block on the floor

  1. Mar 3, 2008 #1
    6-24.gif 1. The problem statement, all variables and given/known data
    The figure below shows an initially stationary block of mass m on a floor. A force of magnitude F=0.540mg is then applied at upward angle θ = 21°.

    (a) What is the magnitude of the acceleration of the block across the floor if (a) s = 0.600 and k = 0.500?

    (b) What is the magnitude of the acceleration of the block across the floor if s = 0.400 and k = 0.300?

    2. Relevant equations
    F=ma and F=s(N) or F=k(N)

    3. The attempt at a solution

    Well i drew a Free Body Diagram and found the forces in the x and y direction:

    x: F(s)-Fcos(21)=0

    y: mg-Fsin(21)-N=0

    Then i solved for F(s)=Fcos(21)=0.54mg(cos(21))=4.94m

    then i used F=s(N) to solve for N

    Then i set F=ma and F=s(N) equal and solved for a


    I repeated the same steps for B...but the answers are not correct. Any help at all will be appreciated!
  2. jcsd
  3. Mar 3, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    This assumes equilibrium, that a = 0. Not good.

    Good. Use this to solve for N and thus compute the friction force.

    The first thing to figure out is: Does it move? Compare the applied horizontal force with the maximum static friction.

    If it accelerates, then use the kinetic friction to find the acceleration.
  4. Mar 3, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Why have you set up the initial equation to be zero? For it to accelerate there has to be some sort of force there.
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