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Pulley problem

  1. Jun 24, 2010 #1
    pulley problem :D

    1. The problem statement, all variables and given/known data
    Find the acceleration and tension of the system.
    there were no values given so i assumed that our teacher wants the answer in terms of m, M, g, a, T, F.
    http://img202.imageshack.us/img202/8615/picture1gz.th.jpg [Broken]

    Uploaded with ImageShack.us

    2. Relevant equations
    F=ma


    3. The attempt at a solution
    F=ma
    T = F ----> where the F = (mass of hanging block)g
    i still dunno about the acceleration though
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 24, 2010 #2
    Re: pulley problem :D

    No, T is not equal to mg.

    The mass 'm' has force 'mg' acting downwards, T acting upwards, and it accelerates downwards.

    So,
    mg - T = ma (net F = ma)

    Mass M has just one force, T acting on it.

    So,
    T = Ma.

    You can solve these two equations for T and a in terms of m, M and g.
     
  4. Jun 24, 2010 #3
    Re: pulley problem :D

    I recently had a question simillar to this on an exam and got it wrong for some reason. I figure the the EQ of motion for the hanging block would be T-mg=-ma, the answer is obviously the one which graphene gave but I don't understand why. The way I visualized this problem was there is tension acting opposite to gravity but my system is accelerating downward so I should, or so I assumed, have a negative acceleration component. Why is this not the case? I see that if I divide my answer through by a negative I get the correct answer, but this is merely an algebraic trick, it doesn't appear to change anything. Thanks in advance.

    Joe
     
  5. Jun 24, 2010 #4
    Re: pulley problem :D

    mg-T=ma

    this is because the blocks are accelerating and not decelerating.
    If you understand this then you can see that acceleration of the block is positive because if is negative. Also mg > T because it is moving downwards.. With knowns your equation would be positive, and just isn't simplified.
     
  6. Jun 25, 2010 #5
    Re: pulley problem :D

    Do not put any sign on 'a'. Let it just be 'a'. Assume a direction for 'a' and write down Newton's 2nd law. (Net force in the direction of 'a' = ma).
    In case you assumed the correct direction for 'a', you'll get a positive answer, else you'll get a negative answer, implying that 'a' points the other way round.
     
  7. Jun 25, 2010 #6
    Re: pulley problem :D

    Thanks for the explanation. I will remember to leave ma alone until the problem is fully simplified as much as possible. Also as Joshmdmd said since gravity and the direction of motion are in the same direction, it makes sense that it would be mg-T=ma. Take care.

    Joe
     
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