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Homework Help: Three blocks, two pulleys, on a incline

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data
    The suspended 2.29 kg mass on the right is
    moving up, the 1.3 kg mass slides down the
    ramp, and the suspended 7.7 kg mass on the
    left is moving down. There is friction between
    the block and the ramp.
    The acceleration of gravity is 9.8 m/s2 . The
    pulleys are massless and frictionless.
    [PLAIN]http://img816.imageshack.us/img816/6190/009q.png [Broken]

    What is the acceleration of the three block
    Answer in units of m/s2

    2. Relevant equations
    The sum of Fx= N-mgcostheta=o => N=mgcostheta
    The sum of Fy= mgsintheta-usN=0=> mgsintheta-us(mgcostheta)

    3. The attempt at a solution

    are these the correct eqations for this?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 24, 2010 #2


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    Gold Member

    You'll want to look at each block in isolation. The system will have the same acceleration.

    What forces act on the two hanging blocks?
    What forces act on the block sliding down the incline?

    What direction is the net acceleration?

    Hint: there are two tension forces.
  4. Sep 24, 2010 #3


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    Science Advisor
    Homework Helper
    Gold Member

    For the mass on the incline, choose the horizontal axis as the x axis, and the axis perpendicular to the incline as the y axis. In which case you have your Fx and Fy equations reversed. Then once corected , this will give you the normal force acting on that one block on the incline. That's step one......then its free body, free body, free body diagrams! Answer jhae2.718 questions. Look at each block separately, identify alll forces acting, and apply Newton's laws.
  5. Sep 24, 2010 #4
    For block on the left... the mass I'll call m1... the forces acting on it are T1 acting up and m1g acting down

    For block on incline... the mass I'll call m2... the forces act on it in the y direction of N and m2g and in the x directuon fk and T2

    For the block on the right... the mass I'll call m3... the forces acting on it are T1 acting up and m3g acting down

    Since m1=7.7 and is bigger than 2.29 the acceleration is pulling m3 up and m1 down...

    So if I use the equations I gave before for the block on the incline and just use
    sum of Fy=T1-mg=0 => T1=mg

    but where does the acceleration come in for m1 and m3?
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