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Mechanic problem.

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data
    http://www.freeimagehosting.net/t/7bf5e.jpg

    Ok the initial velocity of the car is 5 m/s to the left. Calculate for each situation, the instant time when the velocity of car is :

    a)null and 5m/s to the right.



    For situation one with just one pulley:


    In the block 2 we have F = P-T (=) P = T+m(A)*a (1)
    In car, because it is going to the right F = T (=) T = m(B)*a (2)

    because he tension that block do in car is equal to the tension that car do in block so

    F = P-T (=) F = P - (m(B)*a) (=) P = m(A)*a + m(B)*a (=) P = a(mA+mB)
    a = (10*9.8)/(50+10) (=) a = 1.633 m/s^2


    So with the law of velocity v = vi + at

    v = 0 (null)
    vi = 5m/s

    a = - 1.633m/s^2 (negative because it is opposite to the car)

    t(when v=0) = -5/-1.633 = 3.06s
    t(when v = -5) = -10/-1.633 = 6.12s


    For the situation with two pulleys


    I do not know how to do it, because in the system of the two pulleys we have two tension and i know that they are T/2 each one

    but my teacher said that the acceleration of the car is 2* acceleration of the box.

    My problem is Why? and even knowing that a(A) = 2*a(B), how can i find the acceleration to the block 3 ?

    I appreciate the help.
     
  2. jcsd
  3. Mar 21, 2012 #2
    Sorry for double post. But please help.
    I really need to do this problem :/
     
  4. Mar 21, 2012 #3

    tiny-tim

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    Hi Fabio010! :smile:
    This isn't physics, it's just geometry …

    the string has fixed length,

    so if you make the distance between the pulleys increase by y, you use 2y of string, don't you?

    so when the block moves down y, the car moves along x = 2y

    differentiate once, or twice, and you get dx/dt = 2dy/dt, d2x/dt2 = 2d2y/dt2 :wink:
    when you do F1 = m1a1 and F2 = m2a2,

    you relate them by a1 = 2a2 :smile:
     
  5. Mar 21, 2012 #4
    hum..

    so i guess that for block 3 we have m3a3 = mg-T;
    and for car we have T = m2*a2

    (a2) = 2(a3)

    But solving the equations to find the acceleration, it gives a wrong a solution.
     
  6. Mar 21, 2012 #5

    tiny-tim

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    (try using the X2 and X2 buttons just above the Reply box :wink:)

    that should work :confused:

    show us your full calculations :smile:
     
  7. Mar 21, 2012 #6
    "Damn sorry i just checked you tip after post"

    ok

    m(3)a(3) = mg -T


    as we know T = m(1)*a(1)
    so:
    m(3)a(3) = m(3)g - m(1)*a(1)

    because a(1) = 2a(3)

    m(3)a(3) = m(3)g - m(2)*2a(3)
    a(3) (m(3)+2m(1) ) = m(3)g
    a(3) = m(3)g/((m(3)+2m(1) ))

    a(3) = 20*9.8/((20+100))
    a(3) = 1.633 m/s^2

    so a(1) = 3.2667 m/s^2

    law of velocity of car is v = 5 -(3.2667)t
    when v = 0 t = 1.55 s
    solution t = 2.8 s
     
    Last edited: Mar 21, 2012
  8. Mar 21, 2012 #7

    tiny-tim

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    oops!

    oops! i forgot to check the diagram! :redface:

    it isn't mg - T, is it? (draw a free body diagram) :wink:
     
  9. Mar 21, 2012 #8
    It is mg -2T because of the two pulleys :O

    Now it is correct :)

    Thanks for the help :)
     
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