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Homework Help: Multiple Pulleys Problem

  1. Feb 9, 2015 #1
    1. The problem statement, all variables and given/known data
    http://www.sumoware.com/images/temp/xziltxnrjlckqjqs.png [Broken]

    Two blocks ( A and B ) are linked by rope through three identical pulleys C, D, and E.
    Pulley C and E are in stationary, while D is pulled down so its constant velocity 1 m/s
    If vA = 4m/s when A goes down 2 m, determine the velocity and acceleration also the direction of block B when A goes 2 m down

    Choices :
    a) 4 m/s and 1 m/s^2 downward
    b) 4 m/s and 2 m/s^2 downward
    c) 4 m/s and 2 m/s^2 upward
    d) 6 m/s and 4 m/s^2 downward
    e) 6 m/s and 4 m/s^2 upward

    2. Relevant equations
    W = ΔKE
    V(t)^2 = V0^2 + 2 a Δs

    3. The attempt at a solution

    W = ΔKE
    ma s = (1/2) m v^2
    a (2) = (1/2) (4)^2
    a = 4 m/s^2

    So, the acceleration of block A is 4 m/s^2 downward
    But, what is the block B acceleration and its velocity when block A goes 2m down ?

    I need to find the relationship between block A acceleration and block B acceleration.
    Their accelerations are same in magnitude, right ? (Since the rope is free)
    So, block B acceleration is 4 m/s^2 downward, right ?
    But, what about block B velocity ?
    Using the equation
    V(t)^2 = V0^2 + 2 a Δs
    V(t) = √(2aΔs) = √16 = 4 m/s

    But, my answer is not in the choices..
    Please help me..
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 9, 2015 #2


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    Use the condition that the length of the rope is constant.
  4. Feb 9, 2015 #3
    If A goes 2m down, the rope C->A increases 2m in length
    So, rope D->C decreases 2m in length
    Then, rope E->D decreases 2m in length
    And, rope E->B increases 2m in length
    Right ?
    Then, how to find the acceleration ?
  5. Feb 9, 2015 #4


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    It's a given that D is pulled down at 1 m/s
  6. Feb 9, 2015 #5
    Following ehild's suggestion, you have ##L_{AC}+L_{CD}+L_{DE}+L_{EB}=Const.##

    From the figure, how is ##L_{CD}## (always) related to ##L_{DE}##?

    If mass A moves down with a velocity of vA, what is ##dL_{AC}/dt## equal to?

    If pulley C moves downward with a velocity of vC, what is ##dL_{CD}/dt## equal to? What is ##dL_{DE}/dt## equal to?

  7. Feb 9, 2015 #6


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    As D moves downward with 1 m/s , the length of the piece AC+BE shortens at rate ....... ?
    A moves downward with uniform acceleration. You got aA=4 m/s2, downward. If its velocity is VA downward, the piece of rope AC lengthens at this rate. The piece EB has to shorten at rate .......?
    From the velocity of B in terms of VA and VC what is the acceleration if B?
  8. Feb 9, 2015 #7
    LCD = LDE
    If mass A moves down with a velocity of vA, what is dLAC/dt equal to? It is vA, right ?
    If pulley C moves downward with a velocity of vC, what is dLCD/dt equal to? What is dLDE/dt equal to? It is vC, right ?

    The length piece AC+BE shortens at rate 0.5 m/s
    The piece EB has to shorten at rate vA, right ?
    vB = vA - vD, right ?
    Please explain the correct answers. I'm really confused when finding the relationship of acceleration.
  9. Feb 9, 2015 #8
    Very good. Now, if you take the time derivative of this equation ##L_{AC}+L_{CD}+L_{DE}+L_{EB}=Const.## , and use the previous results, what do you get for the downward velocity of mass B, namely ##v_B=dL_{EB}/dt## (in terms of vA and vC)?

  10. Feb 9, 2015 #9


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    This is not right. If we pull the pulley D down by a certain distance, by how much does the rope in the middle get longer?
    (This extra length comes from AC+BE being shortened.)
  11. Feb 9, 2015 #10
    LAC+LCD+LDE+LEB = const
    vA + 2 vC + vB = 0
    vB = vA - 2 vC
    Right ?
  12. Feb 9, 2015 #11
    Hmm.. It must go down in the same distance as the distance the pulley D goes down.
    So, its rate is also 1 m/s, right ?
  13. Feb 9, 2015 #12
    Check the sign on vA.
  14. Feb 9, 2015 #13


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    Take a look at this picture I made (it's harder to explain with words).

    If the pulley goes down by x then the rope gets longer by ?

  15. Feb 9, 2015 #14
    It will get longer by 2x.. Thanks for your nice illustration..
    Then the piece EB decreases x in length and AB also decreases x in length, right ?
    So, the velocity of pulley D is 0.5 ( vA + vB ), right ?
    Then, how to find the relation between the acceleration ?
  16. Feb 9, 2015 #15
    -vA + 2 vC - vB = 0
    vB = 2 vC - vA

    I think vA is negative since it goes downward ( the rope lengthens ), vC is positive since the rope shortens, vB is negative since the rope lengthens

    Is it right ? Hmmm
    Or we can just ignore the positive/negative rule
    vA + 2 vC + vB = 0
    vB = - (vA + 2 vC)

    Which one is right ?
  17. Feb 9, 2015 #16


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    Well, EB and AC don't necessarily have to decrease by the same amount.

    Yes :) this is correct (ignoring directions). Also, Va does not necessarily equal Vb, but their sum must get shorter by 2 m/s

    Differentiate the last equation with respect to time.
  18. Feb 9, 2015 #17
    vD = 0.5 ( vA + vB )
    aD = 0.5 ( aA + aB )
    Right ?

    Then, using the work-kinetic energy formula, I've found that acceleration of A is 4 m/s2
    I will use negative sign for up direction and positive for down direction
    aD = 0.5 ( aA + aB )
    0 = 0.5 (4 + aB)
    aB = -4 m/s2

    vD = 0.5 ( vA + vB )
    -1 = 0.5 ( 4 + vB )
    vB = -6 m/s2

    So, the acceleration is 4m/s2 upward ? and the velocity is 6 m/s upward ?

    Can you tell me why it is negative ?
    I imagine the scene and I think that block B will go in the same direction as block A, right ? (The block A goes down, so the middle pulley goes up, then block B goes down)
    So, could you please tell me the mistake I made ?
  19. Feb 9, 2015 #18


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    If the block A started with a velocity 4 m/s downwards and an acceleration 4 m/s2 downwards, then yes your answer is correct.

    It is negative because it is in the opposite direction as block A

    This is not true. The middle pulley is always going down at 1m/s, that is unchangeable.
  20. Feb 9, 2015 #19
    Yeah.. I forgot that the middle pulley is going down..
    Thanks for your nice help :)
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