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Homework Help: Pulley Question

  1. Oct 21, 2006 #1
    Hi, I have a forces problem involving a pulley, I think I'm almost there. I was wondering if anyone may please give me some suggestions on how to solve this. Thank you.
    Code (Text):

     |     |
     | A   |______________
                         /  |
                         |  |
                         | _|_
                         ||   | B
                         |  |
                         | _|_
                         ||   | C
                         ||   |
    Question: Three boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are [tex]m_A=30.0kg[/tex], [tex]m_B=40.0kg[/tex], and [tex]m_C=10.0kg[/tex]. When the assembly is released from rest, what is the tension in the cord connecting [tex]B[/tex] and [tex]C[/tex].
    I've started the problem by treating mass B and C as one big mass and drawing free body diagrams.
    For mass A: [tex]F_{net}=F_{T_{sys}}=m_Aa_{sys}=30a_{sys}[/tex]
    [tex]a_{sys}=\frac{F_{T_{sys}}}{30}[/tex] (1)
    For the "big mass" (mass B + mass C):
    [tex]a_{sys}=\frac{50g-F_{T_{sys}}}{50}[/tex] (2)
    Solving for [tex]F_{T_{sys}}=183.75N[/tex] and [tex]a_{sys}=6.125m/s^2[/tex]

    Now how can I used this information to find the tension between mass B and C?
  2. jcsd
  3. Oct 21, 2006 #2


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    Draw the FBD for block C alone.

    Note: in your calculations, you didn't solve for the system tension, you solved for the tension in the rope wrapped around the pulley. There is no 'system' tension as such.
    Last edited: Oct 21, 2006
  4. Oct 22, 2006 #3
    FBD for all the blocks. Not just C. Just remember that tension in the string between A and B is the same in the FBD for A and B, and the tension between B and C are the same in the FBD for B and C.

    Also remember that the system moves uniformly (i.e. The block A, B anc D all have the same velocity).

    P.S. Nice drawing using normal characters. I like it :)
  5. Oct 22, 2006 #4
    Hi thanks for replying guys, so would this work?
    The net force acting on mass B and mass C is [tex]F_{net}=(40kg+10kg)(6.125N/kg)=306.25N[/tex]
    By drawing a free body diagram for mass B and mass C, the forces acting on it is, force of tension, force between B and C acting upwards and weight acting down.
    Therefore, [tex]F_{net}=F_T+F_{T_{BC}}-F_g\Rightarrow F_{net}+F_g-F_T=F_{T_{BC}}[/tex]
  6. Oct 22, 2006 #5


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    No, not quite. Start by setting up to equations, one for block A block B&C, thus;

    [tex](m_{b}+m_{c})g - T = (m_{b}+m_{c})a[/tex]

    [tex]T = (m_{a})a[/tex]

    I am assuming here that the table is frictionless. Can you see where the above to equations come from?

    Now you can solve for T (knowing that the acceleration is uniform).
  7. Oct 22, 2006 #6
    Hi Hootenanny! Thanks for answering back to my question, so the tension in the cord connecting mass B and mass C is the same as the tension pulling mass A to the right?

    Because I think I've set up the two equations that you have and found the tension force to be [tex]183.75N[/tex]
    Last edited: Oct 22, 2006
  8. Oct 22, 2006 #7

    Doc Al

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    No. In Hoot's equation, T stands for the tension in the cord connecting A and B, which is different from the tension in the cord connecting B and C.

    In your first post you correctly calculated the acceleration of all the masses. Now just apply Newton's 2nd law to mass C alone--using that acceleration--to find the tension that you need.
  9. Oct 22, 2006 #8
    After finding the acceleration of all three masses, [tex]a=6.125m/s^2[/tex]
    The net force on mass C is given by, [tex]F_{net}=(10kg)(6.125N/kg)=61.25N[/tex]
    The forces acting on mass C are: (1)Force of tension between mass B and C (2)the weight of mass C
    Thus, [tex]F_{T_{B,C}}-98N=61.25N\Rightarrow F_{T_{B,C}}=159.25N[/tex]

    Is it correct now?
  10. Oct 22, 2006 #9


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    Looks good to me :smile:
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