1. Limited time only! Sign up for a free 30min personal 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!

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.

    Answer:
    (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

    User Avatar

    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

    And:
    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

    User Avatar

    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

    User Avatar

    Staff: Mentor

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

    Doc Al

    User Avatar

    Staff: Mentor

    It would depend on how they are arranged.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dynamics of material point
  1. Magnetic materials (Replies: 6)

  2. Strength of materials (Replies: 7)

Loading...