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Laws of Motion (Pulleys, blocks)

  1. Aug 16, 2014 #1
    1. The problem statement, all variables and given/known data
    A truck shown in the figure is driven with an acceleration a =3m/s2. Find the acceleration of the
    bodies A and B of masses 10 kg and 5 kg respectively, assuming pulleys are mass less and friction is absent everywhere.



    3. The attempt at a solution

    I have written three equations i.e.

    for B: 5a1= T-5g

    for A: I dont what to write whether 10a2= 2T-10g or 10a2= 10g-2T

    and 2(a1 - a)= a2

    I have solved the equations and i have got a1=a2=2m/s2

    but the ans is a1=a2=1m/s2
     

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    Last edited: Aug 16, 2014
  2. jcsd
  3. Aug 16, 2014 #2

    Nathanael

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    Well it's arbitrary, but it has to be consistent with your first equation. So since you considered the direction of gravity to be negative in your first equation, you must do the same in this equation. So you would use [itex]10a_2=2T-10g[/itex]


    Now about the problem;
    Your first two equations combine to give the constraint [itex]a_1=a_2[/itex] That is half the battle.
    Now you just need to find a relationship between [itex]a_1[/itex] and [itex]a_2[/itex] (which will likely also relate to the known acceleration of the truck, [itex]a[/itex])



    Your math claims that this relationship is:
    Can you explain how you came up with this equation?
     
  4. Aug 17, 2014 #3
    oh sorry i have got that wrong, actually I think it should be
    a1 - a = 2a2
    is this correct???
     
  5. Aug 17, 2014 #4

    Nathanael

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    Let me ask you, after some time, does the length of the entire rope get shorter? (Or longer?)


    P.S.
    It can be difficult to help if you don't (at least slightly) explain your math. Why did that equation come to mind?
     
  6. Aug 21, 2014 #5
    Hi,
    I have also been trying to solve this problem, using the string length method:

    If the weight on the left moves up by x, then the string segment on the left decreases by x.
    If the weight on the right moves up by y, then the string segment in the middle decreases by y and the string segment on the right decreases by y.

    The sum of the differences must be zero so x + 2y = 0.

    Then I not sure what to do as this relation just gives a1 = -2a2 ?

    Thanks
     
  7. Aug 21, 2014 #6

    Nathanael

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    If the weight on the right (A) moves up by a distance y, then the string is shortened by a distance 2y, right?

    The sum of how much the rope is shortened must equal the distance that the truck moved to the right.
     
  8. Aug 22, 2014 #7
    Are you sure this is the answer given ?

    I am getting aB =19/3ms-2 downwards and aA = 14/3 ms-2 upwards .
     
  9. Aug 22, 2014 #8

    Orodruin

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    That the accelerations are the same is relatively straightforward to arrive at. Consider the string tension and the free body diagrams for B and for A+pulley.
     
  10. Aug 22, 2014 #9
    I don't think we need to consider the forces to determine the relationship between the accelerations of the two blocks .

    Could you explain just one thing and that is - how are the accelerations of the two blocks equal ?

    May be I am missing something very obvious.
     
  11. Aug 22, 2014 #10

    ehild

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    Assume both blocks move upward.

    a1 is the acceleration of A, a2 is the acceleration of B. The tension in the rope is T.

    2T-10g=10a1 divided by 2: T-5g=5a1.
    T-5g=5a2. You see that a1=a2.

    ehild
     
  12. Aug 22, 2014 #11

    Orodruin

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    We are given that the mass of B, let us call it ##m_B## is half of the mass of A ##m_A = 2 m_B##. If we call the string tension ##T##, then the free body diagram for B gives (positive direction down)
    $$
    m_B g - T = m_B a_B
    $$
    while that for A gives
    $$
    m_A g - 2T = m_A a_A \quad \Rightarrow \quad 2(m_B g -T) = 2m_B a_A.
    $$
    Inserting the first equation into the second gives
    $$
    a_A = a_B.
    $$
    In order to relate them to the acceleration of the truck you need to further assume that the string does not change its length.
     
  13. Aug 22, 2014 #12
    Sorry .I realise my mistake .

    I was wrongly using mass of A = 5kg and that of B = 10kg , when it should have been the other way round .In addition,there was a sign error . I need to be more careful .

    I think I managed to confuse myself quite well .

    Thanks Orodruin and ehild !
     
  14. Aug 22, 2014 #13
    The other part I have worked out as:

    If block B rises by x, then the string length at the truck increases by x.
    If block A rises by y, then the string length at the truck increases by 2y.
    Let the string length at the truck be z, then z=x+2y.

    Differentiating this relation twice, and sub in the known truck acceleration gives:
    aB + 2aA = 3
    which with aB = aA gives the solution, 1ms-2
     
  15. Aug 22, 2014 #14
    The problem is solved but still a few things are unclear .

    The point where I confuse myself is where acceleration of string pieces is related to the acceleration of the blocks.This is where sign error creeps in.

    Suppose ΔL1 is the change in horizontal rope length from pulley to the truck.ΔL2 is the change in vertical rope length to the right of movable pulley .ΔL3 is the change in vertical rope length to the left of movable pulley .ΔL4 is the change in vertical rope length from fixed pulley to block B .

    If I write ΔL1 = ΔL2+ΔL3+ΔL4 , I get the correct relation 2xA+xB = 3 .

    But what is the problem in writing it as ΔL2 +ΔL3= ΔL1+ΔL4 ? This gives incorrect relation 2xA=xB + 3 .

    Or, may be writing it as ΔL4 = ΔL2+ΔL3+ΔL1.This gives xB=2xA + 3 .

    What am I missing ?
     
    Last edited: Aug 22, 2014
  16. Aug 23, 2014 #15
    You are right, this is confusing. Really you don't want to consider the direction of the rope wrt the diagram, but wrt the truck. So there are only really two directions the rope moves - to the right or to the left.

    (Think about 'unwinding' the rope from the pulleys, and consider it as a straight piece of rope, then there is no distinction of vertical or horizontal rope length)
     
  17. Aug 23, 2014 #16

    ehild

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    You have to suppose some direction which is positive. For ΔL1, it is positive if the truck moves to the right. But that makes the total vertical piece of the rope shorten.

    If any of the vertical pieces gets longer and the others do not change, the horizontal rope gets shorter. If the horizontal piece gets longer, and only one vertical piece changes length, that piece gets shorter. You can decide about the signs of ΔLi. If you assume they are positive in case the rope gets shorter than you have the equation ΔL2+ΔL3+ΔL4=ΔL1.
    In case you consider ΔLi positive if the length increases, you have the equation ΔL1+ΔL2+ΔL3+ΔL4=0. But then you have to reconsider how the ΔL-s are related to the displacements of the blocks.

    ehild
     
  18. Aug 23, 2014 #17
    In this case ΔL4 would be negative as the rope from fixed pulley to block B increases .Isn't it ?

    Same issue here . I think whatever is assumption ,ΔL2+ΔL3 and ΔL4 would have opposite signs , as when the rope length to block A increases , that to block B decreases and vica versa .
     
  19. Aug 23, 2014 #18

    ehild

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    No, both blocks rise, and that makes the length of the attached pieces shorten.

    ehild
     

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    Last edited: Aug 23, 2014
  20. Aug 23, 2014 #19
    How did you conclude that ?

    If block A rises,middle rope length shortens ,this is compensated by truck moving to right and block B going down .

    What makes you deduce that block B goes up as well ?
     
  21. Aug 23, 2014 #20

    ehild

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    We got that a1=a2.

    Why?

    As it came out, a1=a2=1m/s2.


    ehild
     
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