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Tension Force with Pulleys

  1. Jun 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A light cable passes over a frictionless pulley attached to the ceiling. A 10 kg block hangs at one end of the cable and an 8 kg block connected by a light cable to a 4 kg block hangs at the other end. a) Construct a force diagram for each block. b) Write Newton's second law for the vertical motion of each block (choose the axes so that positive acceleration is in the same direction for each block). c) Solve these equations to determine the acceleration of the blocks and the tension in the cables.


    2. Relevant equations
    F = ma
    T - w = ma
    Ab + Ac = -Aa

    3. The attempt at a solution
    Assume block a is 10 kg, block b is 8 kg, and block c is 4 kg

    Force Diagrams:

    ^ ^ ^
    | Ta | Tb | Tc
    . . .
    | w | w | w
    v v v

    I am fairly confident this is right

    Newtons 2nd Law:
    Fneta = Ta - Wa = Ta - MaG = MaAa
    Fnetb = Tb - Wb = Tb - MbG = MbAb
    Fnetc = Tc - Wc = Tc - McG = McAc

    I am pretty sure these equations are right, it just seems that when applying them to part C, I continually get 0 for the acceleration.

    Acceleration and Tension:
    Ta = Tb + Tc
    -Aa = Ab + Ac

    Ta = MaG + MaAa
    Tb = MbG + MbAb
    Tc = McG + McAc
    MaG + MaAa = MbG + MbAb + McG + McAc

    And then I get stuck trying to substitute in acceleration. I know I am going about the problem wrong in someway, but after looking at countless examples, I can't figure out where my mistake is. Thanks
     
  2. jcsd
  3. Jun 7, 2009 #2

    PhanthomJay

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    a and c are ok (but watch your plus and minus signs), but b is not. Block b has, in addition to its weight, two tension forces acting on it: Tb up and Tc down.
    why do you say this? Ta must equal Tb
    why do you say this? All accelerations are equal in magnitude, since the blocks all move together at the same rate.
    you MUST draw a FBD of each block, and identify all forces acting. Note that tensions on either side of the pulley must be equal, and the accelerations must be equal.
     
  4. Jun 7, 2009 #3
    Acceleration and Tension:
    Ta = Tb + Tc

    "why do you say this? Ta must equal Tb"

    If Ta and Tb are equal, and Ta is on one side of the pulley, while Tb and Tc are on the other, how can Ta and Tb be equal? If the tension forces on opposite sides must be equal, thus meaning that Ta would have to be equal to the tension force on the other side, or Tb + Tc?
     
  5. Jun 8, 2009 #4

    PhanthomJay

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    If I understand your problem correctly, you have block A hanging from a cable on one side of the pulley; the cable wraps around the pulley, makes a 180 degree U-turn, and block B is connected at the other end; then another cable is attached below block B, and block C hangs from that lower cable. Correct?

    If so, you should first note two things: The tension in a continuous cable wrapped around a light frictionless pulley is the same on both sides of the pulley. This means Ta =Tb. This says nothing about Tc, which is a different cable. It is only the tension in the cable wrapped around the pulley that is the same.

    Secondly, since all the blocks are connected together, they must move together at the same rate. This implies that the magnitudes of Aa, Ab, and Ac, must be the same (call their magnitudes each equal to "A").

    So draw your FBD's of each block. Your first equation is correct, with the acceleration Aa upward. Your third equation is correct, but don't forget that the acceleartion Ac is a negative term (Ac = -A). Now draw the FBD of Block B, noting that there are two tensile forces and its weight acting on it. Solve the resulting equations, unless, without a figure, I have misunsderstood the problem, please clarify if I have.
     
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