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What is the magnitude of the upward acceleration of the load of bricks?

  1. Oct 6, 2007 #1

    A load of bricks with mass m_1 = 14.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m_2 = 28.4 kg is suspended from the other end of the rope, as shown in the figure. The system is released from rest.

    What is the tension in the rope while the load is moving?


    2. Relevant equations

    3. The attempt at a solution

    Ok first I drew a diagram, then I stated that
    m1 = 14.0 kg
    m2 = 28.4 kg

    w=ma w = 137.2

    w=ma w = 278.32

    Now I got the w of both.

    What do I do know? I am lost. I do believe that I must used the Sum of Fy.

    PS: I passed my first exam. :biggrin:
  2. jcsd
  3. Oct 6, 2007 #2

    Doc Al

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    Staff: Mentor

    Do this: (1) Identify the forces acting on each mass. (2) Apply Newton's 2nd law to each.
  4. Oct 6, 2007 #3
    Where is the tension,dear?You will have to consider the tension in the rope.
  5. Oct 6, 2007 #4
    The forces acting on each mass.


    Weight downward of 137.2 N.
    Force upward of 141.12N.

    Weight downward of 278.32 N.
    Force upward of 141.12N.


    for M1, upward it would be 141.12N = 14.0kg (a)
    a= 10.08 m/s^2

    does this seem right?
  6. Oct 6, 2007 #5
    Look here
    Scroll down to "Standard Pulley Problem". I think it's all right there.
  7. Oct 6, 2007 #6

    Doc Al

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    Staff: Mentor

    Set up equations and solve for the tension and acceleration. Call the tension T and the magnitude of the acceleration a. (How did you solve for the tension?)

    So, symbolically, what's the net force on m_1? Apply Newton's 2nd law. Then do the same thing for m_2.
    Last edited: Oct 6, 2007
  8. Oct 6, 2007 #7
    thank you very much Laban, that site helped a lot in the first part. I was able to get the acceleration of the object using the formula provided. :)

    Now I will attempt finding the tension of the rope.
  9. Oct 6, 2007 #8
    "We know that the sum of forces acting on m is T – mg which is equal to ma. Therefore, T = m(g – a)."

    So if T= m(g-a)

    it would be T = 28.4 (9.8-3.33) = 183.75
    and for the other object it would be T = 14.0 ( 9.8 - 3.33) = 90.58

    ? But tension is equal on both sides, so I should get the difference of 93.17?
  10. Oct 6, 2007 #9

    Doc Al

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    Staff: Mentor

    Just plugging into a formula is not what I would recommend. Instead, understand how they set up the problem and derive it on your own.

    Again, you must understand the equation, not just plug into it. Hint: Which way does m_1 and m_2 accelerate?

    (If you get different values for T on each side, you know you're doing something wrong.)
  11. Oct 6, 2007 #10
    you are right, I have taken the easy way to this. :(

    Back to part 1.

    M_1 accelerates upwards, and M_2 downwards.
  12. Oct 6, 2007 #11
    Those equations are two equations with two unknown variables. You found the acceleration by substracting one from the other. Now it's pretty easy to find the tension, since it is also part of those equations :)
  13. Oct 6, 2007 #12
    right, again. I solved for T for both equations and reached to a find answer which ended up correct.

    I also understand from the text that: "acceleration of the rope is of the same magnitude at every point in the rope, the acceleration of the two masses will also be of equal magnitude."

    Thank you Laban and Doc Al.:approve:

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