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Ratio of two masses connected by pulley

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Figure 1) Block 1, of mass m1, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude a and that the coefficient of kinetic friction between block 2 and the plane is μ.

    Find the ratio of the masses m1/m2.
    Express your answer in terms of some or all of the variables a, μ, and θ, as well as the magnitude of the acceleration due to gravity g.
    2. Relevant equations

    3. The attempt at a solution

    + /x is direction of acceleration

    Forces on m2

    N - mgy=0

    T - m2gx-Ff=m2a

    T - m2g Cosθ - μm2Sinθ=m2a

    Forces on m1

    m1g - T = m1a


    I just put all the equations into one and got:

    m1/m2=(g(Cosθ - μSinθ))/(g/a)

    but it still says it's wrong. It says the final answer doesn't depend on Cosθ or Sinθ
  2. jcsd
  3. Oct 20, 2015 #2
    it should be m1/m2=(g(Cosθ - μSinθ))/(g-a)

    I also probably messed up the components but I tried both ways
  4. Oct 21, 2015 #3


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    To check whether I have sin and cos the right way round, I consider an extreme case, like ##\theta=\pi/2##. Do your equations look right for that case?
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