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Homework Help: Warehouse ramp pulley

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data
    You have taken a summer job at a warehouse and have designed a method to help get heavy packages up a 15º ramp. In your system a package is attached to a rope which runs parallel to the ramp and over a pulley at the top of the ramp. After passing over the pulley the other end of the rope is attached to a counterweight which hangs straight down. In your design the mass of the counterweight is always adjusted to be twice the mass of the package. Your boss is worried about this pulley system. In particular, she is concerned that the package will be too difficult to handle at the top of the ramp and tells you to calculate its acceleration. To determine the influence of friction between the ramp and the package you run some tests. You find that you can push a 50 kg package with a horizontal force of 250 Newtons at a constant speed along a level floor made of the same material as the ramp.

    2. Relevant equations

    3. The attempt at a solution
    Block 1:
    (Fnet)y = n + T - Fgcos[tex]\theta[/tex]= 0
    (Fnet)y = n = mgcos[tex]\theta[/tex]

    (Fnet)x = n + T - Fgsin[tex]\theta[/tex] - fk = ma
    (Fnet)x = T - Fgsin[tex]\theta[/tex] - fk = ma

    (Fnet)x = 980 - Fgsin[tex]\theta[/tex] - mu_k(mgcos[tex]\theta[/tex]) = ma
    (Fnet)x = 980 - gsinmgcos[tex]\theta[/tex] - (mu_k)cos[tex]\theta[/tex] = a

    Block 2:
    (Fnet)x = 0
    (Fnet)y = T - 2mg = 0
    (Fnet)y = T = 2*mg
    (Fnet)y = T = 2*50*9.8, so T= 980N

    I'm confused as to what to do next, because I don't think I'm doing this right and I don't have the co-efficient of kinetic friction.
  2. jcsd
  3. Oct 14, 2009 #2


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    In the problem it given that a 50 kg mass is moved on a horizontal floor with 250 N. From this data find the coefficient of friction.
  4. Oct 14, 2009 #3
    So, I did:

    fk = mu_k*n
    250 = mu_k*(50)(9.8)
    250/490 = mu_k = .51

    So, going back to what I originally did and plugging this back in:

    a = [(980)-(50)(9.8)sin15-(.51)(50)(9.8)cos15]/50
    a = 12.24 m/s^2

    This doesn't seem right, is it?
  5. Oct 15, 2009 #4


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    In the calculation, you have the mass as 50 kg. It is not right. It is used to find μ of the surface. For calculation you have to use m and 2m.
    Identify the forces acting on m and its acceleration.
    Find the acceleration of 2m. And the equate them to find T and a.
  6. Oct 15, 2009 #5
    I'm a bit confused. I thought I had the forces identified in (Fnet)x = n + T - Fgsin[tex]\theta[/tex] - fk = ma, would it be 2ma or?

    Sorry, :(.
  7. Oct 15, 2009 #6


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    You have to write two equations for two masses.
    For m
    ma = T -gsinθ -fk. What is the value of fk?
    For 2m
    2m*a = ....?
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