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Two masses on inclined plane

  1. Dec 17, 2008 #1
    Two masses are being pulled up a 30 degree incline by a force F parallel to the incline. The velocity is constant and up the incline. The force is applied to a 200 kg mass and a string connects the 200 kg mass to a 150 kg mass. The coefficient of kinetic friction is 0.2. The force F = ?

    So I drew a free body diagram, and it has F parallel to the inclined plane (Fx) and the normal force = Fy. There is mg, which is pointed directly down. The friction is pointed in the -Fx direction. I know F=uFn, I also know that Fx = mgcos(theta) and Fy = mgsin(theta). Do I include the 2nd mass in my calculations? I'm assuming I do, in which case the mass is 350 kg. If the velocity is constant, acceleration = 0. Do I need to find the velocity to solve this problem?
  2. jcsd
  3. Dec 17, 2008 #2
    I don't think you need to know the velocity. It is just there to justify using the kinetic friction coefficient for both masses.
  4. Dec 17, 2008 #3
    That's what I thought, but I have no idea how to do this problem. I used the Fx equation I listed above but it doesn't include the friction coefficient.
  5. Dec 17, 2008 #4
    draw a *FBD*

    [tex] F_{y}=mg\times cos \phi [/tex]

    [tex] F_{x}=F_{A}-mg\times sin \phi + F_{f} [/tex]

    and [tex] F_{f}=\mu \times F_{N} = \mu \times mg\times cos \phi [/tex]


    [tex] F_{A}=mgsin\phi + \mu \times mg\times cos \phi [/tex]
  6. Dec 17, 2008 #5
    Thanks, Bright Wang. That helped a lot. The Fa you used in that equation, is that the parallel force?
  7. Dec 17, 2008 #6
    applied force
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