1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar

    Staff: Mentor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Looks good to me :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Pulley Question
  1. A pulley question (Replies: 7)

  2. Pulley question (Replies: 2)

  3. Pulley questions (Replies: 3)

  4. Pulley Question (Replies: 1)

  5. Pulley question (Replies: 6)