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Two blocks sliding!

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Two blocks, of weights 3.6 N and 7.5 N, are connected by a massless string and slide down a 30° inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10; that between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string.

    2. Relevant equations

    Fnet = ma

    3. The attempt at a solution

    Doing the free body diagram for the heavy weight,

    x-axis: Fn(Heavy) - Fg(heavy)sin(theta) = 0 so, Fn(heavy) = Fg(heavy)sin(theta)

    y-axis: Fk(heavy) - Ft - Fg(heavy)cos(theta) = -m(heavy)a

    For the light weight i got,

    x-axis: Fn(light) - Fg(light)sin(theta) = 0 so, Fn(light) = Fg(light)sin(theta)

    y-axos: Ft + Fk(light) - Fg(light)cos(theta) = -m(light)a

    I equated each block equation to equal to the Ft to get:

    M(heavy)a + fk(heavy) - Fg(hanging)cos(theta) = Fg(light)cos(theta) - fk(light) - m(light)a

    I set it equal to acceleration but I get the wrong answer.

    I also got the mass of each weight by dividing it by g (9.8 m/s^2) Can this be my mistake in the equation?
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2


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    Homework Helper

    Your last equation is wrong.
    The acceleration of the heavy block is given as
    [ Fk(heavy) - Ft - Fg(hang)cos(theta)]/M = -a
    Similarly write down the acceleration for light block.
  4. Oct 11, 2009 #3
    I setted them both equal to the tension and solved for a.
  5. Oct 11, 2009 #4


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    Homework Helper

    Will you show the calculations, please?
  6. Oct 12, 2009 #5
    M(heavy)a + fk(heavy) - Fg(hanging)cos(theta) = Fg(light)cos(theta) - fk(light) - m(light)a

    The tension force from the heavy block and the light block.

    a = [Fg(light)cos(theta) - Uk(light)Fn(light) - Uk(heavy)Fn(Heavy) + Fg(heavy)cos(theta)] / (m(hang) + m(light)

    The answer in the back should be 3.5 m/s^2
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