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Dynamics of material point

  1. Jun 27, 2009 #1
    The two blocks shown in the figure are initially at rest. Despise the masses of the pulleys and the friction between different parts of the system. Get (a) the acceleration of each block and (b) the tension in cable.

    (a) aA = 2,56 m/s^2; aB = 1,28 m/s^2
    (b) 677 N

    Attached Files:

  2. jcsd
  3. Jun 27, 2009 #2
    People can't help you unless you post your attempted solutions. It's a place where people help you on homework, not do your homework for you.
  4. Jun 28, 2009 #3
    I'm not asking you to resolve. I would like an explanation of the exercise
  5. Jun 28, 2009 #4

    Doc Al

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    Staff: Mentor

    I presume that you are to treat the pulleys as massless and the surfaces as frictionless.
    What don't you understand?

    Hints: Identify the forces acting on each mass. (Draw a free body diagram for each.) Apply Newton's laws to each mass.
  6. Jun 28, 2009 #5
    My solution (not find the answer)

    Block A

    ZFx = m.a
    T1 - Psin(30) = mA . aA

    Block B
    zFy = m.a
    -P + T1 + T2 = mB . aB

    T2 = 2T1
    2aB = aA

    Solving the system. I find:
    aA = 1,14 m/s^2
    aB = 0,57 m/s^2
    T2 = 1097,92 N
  7. Jun 28, 2009 #6

    Doc Al

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    Staff: Mentor

    OK, I presume P is the weight of block A. Better to use mAg instead of P.

    Don't use the same symbol, P, to represent the weight of block B. Also, since the pulleys are massless and frictionless, the rope has a single tension throughout. Just call it T. So what's the total upward force on block B?

    See comments above.
    Be careful with signs. If block A moves up the incline, block B must move down.
  8. Jun 28, 2009 #7
    Block A

    ZFx = mA . aA
    T - mA . g . sin(30) = mA . aA

    Block B

    ZFy = mB . aB
    -mB.g + 2T = mB . aB

    aA = - 2aB

    New is correct thank you very much
  9. Jul 2, 2009 #8
    I find:

    Block A
    T - Psin(30) = 90,72aA

    Block B
    -2T + P = 158,7aB

    How can I relate aA and aB ?
  10. Jul 2, 2009 #9

    Doc Al

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    Staff: Mentor

    See your last post:
  11. Jul 2, 2009 #10
    if I had two pulleys in B the accelerations were aA = -4aB ?
  12. Jul 2, 2009 #11

    Doc Al

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    Staff: Mentor

    It would depend on how they are arranged.
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