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Acceleration on a pulley system with 3 blocks

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Three objects are connected on an inclined table. The objects have masses 8.0 kg, 4.0 kg and 2.0 kg as shown, and the pulleys are frictionless. The tabletop is rough, with a coefficient of kinetic friction of 0.47, and makes an angle of 20.0° with the horizontal. The 8.0kg block hangs on the left, the 4.0kg block in the middle on the table (20.0° t the left and above horizontal), and the 2.0kg is hanging on the right.

    Draw a free-body diagram for each object. (did this but can't display on the forum)
    Determine the acceleration and direction of motion of the system.
    Determine the tensions in the two cords.



    2. Relevant equations
    εF=ma
    trig
    T-mg=ma


    3. The attempt at a solution
    I can do this with two blocks but 3 on a angle through me off.
    mg for the 8.0kg block =78.4N and mg for the 2 kg block is 19.6kg

    I know that for the block on the table:
    Fgx = sin 20 (39.2) = 13.4N
    Fgy = cos 20 (39.2)= 36.83N - Force normal and Fgy are equal
    Ff=0.47(36.83)

    The using εFx=max
    19.6=13.4-17.3-78.4 = (14)a
    a= -4.5m/s/s or 4.5 m/s/s moving to the left.

    Not sure if I did this right and when calculating tension in a 3 block pulley system on a angle.

    Thanks
     
  2. jcsd
  3. Mar 30, 2014 #2

    Doc Al

    User Avatar

    Staff: Mentor

    It's a bit unclear what you are doing here.

    You need equations for all three masses, which you will solve together to get the acceleration.
     
  4. Mar 30, 2014 #3
    Sorry, first time using this forum. What i did was
    εFx=max and i used mg for the masses on either end of the system
    so in this case i had
    εFx = m3g+Fgx-Ff-m2g=(m1+m2+m3)a
    εFx = 8x9.8 + sin20(4x9.8) - .47xcos20(4x9.8) - 2x9.8 = (8+4+2)a
    19.6+13.4-17.3-78.4 = (14)a

    The Fgx and Ff apply to the box on a slant on the table.
     
  5. Mar 30, 2014 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I think what you did was assume that the tension in each rope equals the weight of the hanging mass. You can't do that!

    If the tension equaled the weight of the masses, then the net force on each hanging mass would equal zero. No acceleration!

    Instead, call the unknown tensions T1 and T2 (or whatever). You need to solve for the three unknowns: the acceleration and the two tensions. That's why you need three equations, one for each mass.
     
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