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Pulley accelerations problem

  1. Jun 13, 2005 #1
    Hi. I have a homework problem I've been trying to solve for three days I think.

    There is a block that lies on a horizontal table with mass = m1 it is connected to a cord which goes over a pulley on the edge of the table. The cord goes down and around another pulley (lets call it p2) the cord goes up again and is connected to a ceiling a little higher than the table. To the p2 pulley there is connected another block with mass= m2. The m2 block has an downward acceleration a2, and the m1 block has an acceleration a1,to the right . The problem is to find these accelerations a1 and a2 in terms of m1, m2 and g.
    I hope you will be able to picture this.

    So far my work is this:

    The string connected to block m1 has tension: T=m1a1
    For block m2 : m2*g - 2*T = m2*a2

    The right answers are :

    a1 = 2*m2*g/(4*m1 + m2)

    a2 = m2*g/(4*m1 + m2)

    I have no clue how to reach to this conclusion, could someone please give me a hint.
  2. jcsd
  3. Jun 13, 2005 #2


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    Your work so far is good. What relationship exists between a1 and a2?
  4. Jun 13, 2005 #3
    I see that the acceleration a1 is twice the acceleration a2.
    I that the key. I wouldn't have seen that if I didn't have the answers.

    Still in question. :confused:
  5. Jun 13, 2005 #4
    ***Find enclosed Attachment ***

    In this kind of Question Remember that

    no of equations == no of unknowns

    basically u might miss some equations which can be obtained from constraint relation.
    For eg here we know the length of string is constant.

    here how we use it

    here x1 + (x3 - x2) + x3 = l
    now x1+2x3 -x2 =l
    now diff twice w.r.t time we have

    now a2=0 as pulley is fixed
    and you get constarined relation
    a1 = -2a3 sign signifies direction

    Attached Files:

  6. Jun 13, 2005 #5
    I'm sorry but I don't get what you are trying to tell me. Couldn't open the attachment either.
  7. Jun 13, 2005 #6


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    There is an acceleration constriant on the system. Meaning that the acceleration of the block in the x direction is equal to the negative acceleration of the block in the y direction.

    thus: knowing that both blocks have the same acceleration denote the acceleration term in the question by a general acceleration and solve from there using newtons laws.

    Ax= -Ay which evidentenially represents a negative ma
  8. Jun 13, 2005 #7
    That's not what's he's saying, and that's not correct Ax = -Ay.

    Swatch, himanshu is trying to explain to you how the magnitude of the acceleration in the x direction is equal to twice the magnitude of the acceleration of the second object in the y direction.
  9. Jun 13, 2005 #8


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    The diagram is visible now. It might be easier to follow the logic using a different set of lables. Let x1 be the length of string from the block on the table to the first pulley, and x3 be the distance from the ceiling to the movable pulley as shown in the diagram. Then let x2 be the length of string between the two pulleys. The length of the string is constant, so x1 + x2 + x3 = constant, or x1 = constant - x2 - x3. As the lower block falls, x2 and x3 both increase by the distance the lower block moves, so x1 must decrease by the sum of the increases of x2 and x3, which is twice as much as either of them. This 2:1 ratio of distances means that there will also be a 2:1 ratio of velocities and accelerations.
  10. Jun 14, 2005 #9
    Thank you very much, himanshu 121 and OlderDan. Now I understand it. Didn't think of this at all.

    Thanks. :biggrin:
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