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Homework Help: Pulleys with masses

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data

    The system is in equilibrium and the pulleys
    are frictionless and mass less.
    Find the force T. The acceleration of gravity is 9.8 m/s2 .
    Answer in units of N.
    s5fjav.jpg
    2. Relevant equations
    F=ma


    3. The attempt at a solution
    Okay so I looked at this problem and I thought that T would be mg because thats the only force acting on the 1kg mass. But this is wrong, and I don't know what else to do or where to go to try and figure this out. I know that there is no acceleration because it is in equilibrium would this then make T = 0 because theres no acceleration? I have also drawen FDBs for all the pulleys and tried figuring out tensions to see if that would explain anything to me, but it just confused me even more =( any help would be great thanks!
     
  2. jcsd
  3. Feb 17, 2010 #2

    PhanthomJay

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    No it's not, Draw a free body diagram of the 1 kg mass. There are three forces acting on it. Please identify them.
    correct
    No, it means that the net force acting on the system of masses is 0. There are more forces than just T acting on the system of masses.
    Try first drawing a FBD of the 6kg block.
     
  4. Feb 17, 2010 #3
    You say there are three forces acting on the 1 kg mass, does that mean that I have to think of the mass and pulley two as one? so there are two tensions acting on the mass upwards and then T acting downwards? And then does that mean that I need to look at all the mass in the same way? Because I think I'm getting confused by the pulleys
     
  5. Feb 17, 2010 #4

    PhanthomJay

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    Yes, they sure do get confusing. First note that when a cord wraps around a massless frictionless ideal pulley, the tension in the cord around the pulley changes direction, but the magnitude of the tension in that continuous cord is the same, on both sides of the pulley. Now regarding your question on the 1kg mass, when you draw a FBD of that mass (draw a cloud or circle around it, and wherever the circle cuts thru a known or unknown force, identify it, and always remember the non contact weight force also), and you thus identify the contact and gravity forces acting on that mass alone. There is the weight force acting down, the applied force T acting down, and the upward tension in the rope between the 1kg mass and pulley 2 acting up. Their algebraic sum is 0. You have 2 unknowns. So try a FBD of the 6Kg mass first and work from there as your start point.
     
  6. Feb 17, 2010 #5
    Okay so I think I am understanding it a little better, after working my way from the bottom 6kg, to the 6kg above that and finally over to the 1kg I got 107.8N as the F. I am not sure if I did that right though.

    For the first 6kg there were two forces on the FBD. T1 with an upward tension and mg going downward. I got T1 = mg = 58.8N

    For the second 6kg there were three forces and I got T2=T1+mg=117.6N

    Then for 1kg I got T=T2-mg=107.8N

    I have no idea if I worked that correctly though but thats the only thing I could think of doing. Am I on the right track?
     
  7. Feb 17, 2010 #6

    PhanthomJay

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    good
    no good. there are 2 (two) T1 forces acting down, so that's 117.6 n down plus the weight down, T2 = ??? Continue....
     
  8. Feb 17, 2010 #7
    So then T2=117.6+58.8=176.4? Is the tension I missed the one that is attached to the bottom? and would that be the same as the 6kg mg?

    If the 176.4 is correct for T2 and then I have T=T2-mg=166.6 do I have the right equation for the last part?
     
  9. Feb 18, 2010 #8

    PhanthomJay

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    sorry for not responding sooner, I've been straight out....

    I guess first we should define the letter designations for the unknown tension forces, or else we'll be talking jibberish, there are so many cords. T is the applied force shown. Let t_1 = the tension in the cord between the 1 kg mass and pulley 2. Let t_2 be the tension in the cord between pulley 2 and pulley 1. t_2 is also then the tension in the cord between pulley 2 and the ceiling, and it is also the tension in the cord between pulley 1 and the upper 6 kg mass. Let t_3 be the tension in the cord between the upper 6 kg mass and pulley 3. Let t_4 be the tension in the cord between pulley 3 and the lower 6 kg mass, which is also the tension in the cord between pulley 3 and the floor. Finally, the tension in the cord between pulley 1 and the ceiling, let's call it t_5. You might want to add these designations to your image.

    Now please try again, using the letter variables noted, so we're on the same page. Note you can draw a FBD of a mass alone, or a pulley alone, or some or all of them together, whatever works easiest. Just draw a closed circle around the object or objects you wish to identify forces on, and wherever that circle's circumference cuts through a cord, there is going to be a force there; tension forces always pull away from the objects on which they act; and don't forget the mass weight, if it's applicable in your FBD. Then it's just Newton 1 for this equilibrium condition.
     
  10. Feb 18, 2010 #9
    Thanks a bunch for your help I actually got the solution today so Im just going to look over it and then it should all clear up. But thank you, you have helped me and next time I come across a problem like this I will remember everything we talked about =D!
     
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