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Pulley problem

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data
    An object of mass M=10kg is being pulled up by a system of pulleys.
    a)What force do you need to exert on point A for the object to move uniformly ?
    b) What force do you need to exert on point A, if the friction in each of the pullies is 0,5N and the pulleys are not massless - the mass of the 1st one is 100g, the 2nd - 200g, the 3rd - 300g, the 4th- 400g.

    http://img148.imageshack.us/img148/5526/unledbde.jpg [Broken]

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    2. Relevant equations



    3. The attempt at a solution
    I solved the A part quite easily, got a result of F=12,5N, but i'm stuck with part B - i can't figure out the directions of the friction forces .
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 15, 2011 #2

    PeterO

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    ALL friction forces oppose the motion you are trying to cause - that gives you the direction.
     
    Last edited by a moderator: May 5, 2017
  4. Aug 15, 2011 #3
    Last edited by a moderator: May 5, 2017
  5. Aug 15, 2011 #4

    PeterO

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    Looks like too many arrows to me. The friction in/at each pulley should mean that the tension on the "pulling" side of the pulley is greater than the tension on the other side.

    2:30 am here - will check back tomorrow.
     
    Last edited by a moderator: May 5, 2017
  6. Aug 15, 2011 #5
    Last edited by a moderator: May 5, 2017
  7. Aug 15, 2011 #6

    PeterO

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    Certainly you are not trying to lift #4 so it looks like it can be ignored.

    Note that pulley #4 is the only one to rotate clockwise, so friction there shoul be in the opposite direction.

    I would probably draw the friction at the midpoint of the contact zone - so horizontal in each case.
     
    Last edited by a moderator: May 5, 2017
  8. Aug 16, 2011 #7
    Okay so this is my solution
    T is the tension force , and i'll assume g=10m/s^2
    For the 1st pulley
    [tex]2T_1=Mg+m_1 g+F_f[/tex]
    so [tex]T_1=50,75N[/tex]
    2nd pulley
    [tex]2T_2=T_1+m_2 g +F_f[/tex]
    [tex]T_2=26,625N[/tex]
    3rd pulley
    [tex]2T_3=T_2+m_3 g +F_f[/tex]
    [tex]T_3=15,0625N[/tex]

    and the 4th pulley it should like something like this i guess
    [tex]F=T_3 + F_f[/tex]
    [tex]F=15,5625N[/tex]
     
  9. Aug 16, 2011 #8

    PeterO

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    The calculations look OK except that I think the tension in the RHS of each pully should be greater than T in the LHS due to friction - so rather than 50,75 for pulley 1, I think the tensions should be 50,5 and 51,0 N. And as I think about it I can't convince myself which one is the 50.5 and which is the 51.0, but I lean toward 52 on the right.

    Similar arguments all the way through.
     
  10. Aug 16, 2011 #9

    ehild

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    We can assume that the pulleys work in the regular way - so they rotate and the string does not slip on them: The linear velocity of the rim is the same as that of the string. That is ensured by static friction.
    The other kind of friction acts in the bearings, at the axes of the pulleys. This friction causes a torque against the rotation of the pulley which is equivalent to a force of 0.5 N acting on the rim.

    To get the conditions of motion with constant velocities and angular velocities, not only the forces have to be in equilibrium, but also the torques.

    Considering the first pulley, the tension of the left string is T1' and the tension of the right string is T1. The condition for zero torque means (with R the radius of the pulley)
    T1R-T1'R-FfR=0, that is, T1'=T1-0.5 N. The condition of zero net force is

    T1'+T1-m1g-Mg=0, subbing in T1'=T1-0.5,


    2T1-m1g-Mg-0.5=0.

    This is the same equation Funoras used, with T1 the tension in the right-hand string.

    In case of the last (fourth) pulley, T4=F, and the equilibrium of torques require that T4-T3-0.5=0, so F=T3+0.5.

    Good job, Funoras!

    ehild
     
    Last edited: Aug 16, 2011
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