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Masses hanging on massless meter stick problem

  1. Apr 21, 2006 #1
    Two masses (mA = 3 kg, mB = 6 kg) are attached to a (massless) meter stick, at the 0 and 75 cm marks, respectively.

    a. Now, if mass B was removed, how much force would need to be exerted at the 100 cm mark in order to keep the meter stick level?

    b. Now, if mass B was removed, and no additional force was supplied, calculate the size of the angular acceleration of the meter stick at that instant.

    For a I was clueless on how to approach the problem, but for b, I used angular acceleration = torque/inertia, and got 0.8889, but got it wrong. I tried it again, plugging in different values for r and F, and got 13.066, and still got the problem wrong. I thnk my problem is that I am not using the correct values to solve this problem. Anybody know exactly which values I should use? Thanks!
     
  2. jcsd
  3. Apr 21, 2006 #2

    nrqed

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    as it stands, the question does not make any sense. there is information missing. The ruler is resting on a pivot somehwere? Is it supported at the center??
     
  4. Apr 21, 2006 #3
    Sorry...I took out the parts I solved already...maybe this will help...

    The system is then hung from a string, so that it stays horizontal. The string is placed 50 cm, from mass A.
     
  5. Apr 21, 2006 #4

    nrqed

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    Ok. Without that information it was impossible to answer.

    You must impose that the net torque is zero. What is the torque exerted by mass A? The other torque must be minus that. Knowing the force, it is easy to find how far it must be from where the string is attached
     
  6. Apr 21, 2006 #5
    Wait, you confused me...which part are you on, a or b?
     
  7. Apr 21, 2006 #6

    nrqed

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    well, I was doing part a first :approve:
     
  8. Apr 21, 2006 #7
    Okay, so part a haha. Torque exerted by mass A is 22.05, right?
     
  9. Apr 21, 2006 #8

    nrqed

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    ?? How did you get this? torque = force * distance from the fulcrum and force = mg
     
  10. Apr 21, 2006 #9
    distance = 0.75
    mass = 3
    gravity = 9.8

    I probably have the distance wrong, should it be 0.5?
     
  11. Apr 21, 2006 #10

    nrqed

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    Yes. torque = F d (in magnitude..the sign tells you the direction) The distance there is always the distance between where the force si applied and the axis of rotation (if the force is perpendicular to the line connecting the fulcrum and the point where the force is applied)
     
  12. Apr 21, 2006 #11
    Okay, so 14.7 N for torque.
     
  13. Apr 21, 2006 #12

    nrqed

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    ok. The sign depends on whether th emass is to the left or right of the suspension point. Let's say it is to the left and creates a counterclockwise torque, then it is positive.

    The other force will create a torque of - F * 0.50 m (since it is also 50 cm from the axis of rotation) Imposing that the sum of the two gives a net torque euqal to zero gives you the value of F (in that case, it is trivial because the two forces act at the same distance from the axis of rotation! so of course you will find a value equal to mg !)
     
  14. Apr 21, 2006 #13
    So I'm correct?
     
  15. Apr 21, 2006 #14

    nrqed

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    yes, your torque is correct
     
  16. Apr 21, 2006 #15
    Thank you! So what do I do from there lol?
     
    Last edited: Apr 21, 2006
  17. Apr 21, 2006 #16

    nrqed

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    My post #12 gives the way to answer part a.

    I have to log off soon. Good luck
     
  18. Apr 21, 2006 #17
    Oh okay, I got part a, lemme know when you can work on part b!
     
  19. Apr 21, 2006 #18
    Hmm...okay, for this one I tried...

    I = mr^2 = (3)(0.75)^2 = 1.6875
    angular acceleration = torque/I = 1.47/1.6875 = 8.711 rad/s^2.

    Is this correct?
     
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