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Physics: Blocks on inclines connected by pulley problem

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data
    495x250xpulley_prob_6.png.pagespeed.ic.wKnVT_8Pko.png

    The two blocks are moving rightward at a constant speed along their inclined surfaces. The coefficient of kinetic friction is the same for both blocks. a) Determine the coefficient of kinetic friction. b) Determine the tension in the cord that connects the blocks to one another.
    M=4kg; m=5kg; θ_1=28°; θ_2=36°

    2. Relevant equations
    ∑F = ma
    kinetic friction = μN

    3. The attempt at a solution
    Let N_1 = normal force of block M; N_2 = normal force of block m; f_1 = kinetic friction of block M; f_2 = kinetic friction of block m.
    For block M:
    ∑Fy = M*0 = (N_1) - M * g * cos(θ_1) ⇒ N_1 = M * g * cos(θ_1)
    ΣFx = M * a = T - (f_1) - M * g * sin(θ_1)
    For block m:
    ΣFy = m*0 = (N_2) - m * g * cos(θ_2) ⇒ N_2 = m * g * cos(θ_2)
    ΣFx = m * a = (f_2) - T + m * g * sin (θ_2)

    f_1 = μ * M * g * cos(θ_1)
    f_2 = μ * m * g * cos(θ_2)

    Since the acceleration, the kinetic friction forces, the coefficient of kinetic friction, and tension are all unknown, I couldn't come up with any way to isolate any of those values to solve for it.
     
  2. jcsd
  3. Mar 8, 2016 #2
    Try adding your two equations for the force in the x directions together - this will eliminate the tension.
    Also, I for the Fx on the block m, the friction should be opposing the motion and be negative.
    One key word in the problem you may have missed: CONSTANT speed. So this means that the acceleration is 0.
     
  4. Mar 8, 2016 #3
    Ah, I see that the friction would be negative there too, I made a mistake drawing my free body diagram of that one. And yes! Constant velocity means that the acceleration is 0, I guess this problem was a lot easier than I thought it was. Thanks!
     
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