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Torques, Moments, Levers

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  1. Dec 27, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-12-27_19-50-1.png

    2. Relevant equations
    Torque=one of the forces x the perpendicular distance
    moment=force x perpendicular distance
    3. The attempt at a solution
    moment of pulling the weight up around the disc:
    1.2/2 = 0.6 m
    0.6-0.2 = 0.4 m
    moment= 0.4 x 900 = 360 Nm
    To lift it, the torque of the rod has to be 360 Nm too
    So,
    F x 1.2 = 360
    F = 300 N

    But the answer is 150N
     
  2. jcsd
  3. Dec 27, 2015 #2
    Hi Priyadarshini:

    Your description of your solution omits some details, so I am not sure where you went astray. I suggest thinking about the problem in terms of mechanical advantage.
    What is the mechanical advantage of the 2F force (1F on each end of the lever) applied at its radius compared with the 900N force applies at the radius if the disc?

    Hope this helps.

    Regards,
    Buzz
     
  4. Dec 27, 2015 #3
    Hi,
    What exactly do you mean by mechanical advantage?
     
  5. Dec 27, 2015 #4

    CWatters

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    Try calculating the torque on the vertical shaft due to the 900N force. The two forces F must create a torque at least as great.
     
  6. Dec 27, 2015 #5
    Hi, finding the minimum force is the same as saying that the torque of the disk has the same magnitude of the torque of the lever

    T (lever) = F*d where d= 1.20 m
    T (disk)= W*R where R is the radius of the disk and W is the weight

    T (lever)= T (disk)
     
    Last edited: Dec 27, 2015
  7. Dec 27, 2015 #6

    haruspex

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    This does not calculate anything useful. What physical law are you planning to use this in?
    It's the ratio of the two distances that's interesting, not the difference.
     
  8. Dec 27, 2015 #7
  9. Dec 28, 2015 #8
    Isn't that 0.4*900?
     
  10. Dec 28, 2015 #9
    why do we take the radius of the disc instead of half the rod minus the radius of the disc? Isn't the turning point on the circumference of the disc where the string starts to rotate around the disc?
     
  11. Dec 28, 2015 #10
    I thought that 0.4 was the perpendicular distance from the pivot for calculating the torque of the disc, which is why I subtracted.
     
  12. Dec 28, 2015 #11

    haruspex

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    No, both rod and disc rotate about an axis through their centres.
    I suppose you could consider that the not-yet-wound string instantaneously rotates about the point of contact with the disc, but since it exerts no torque about that point it is quite irrelevant.
     
  13. Dec 28, 2015 #12
  14. Dec 28, 2015 #13
    How do you know that the string at the contact point of the disc doesn't exert any torque?
     
  15. Dec 28, 2015 #14

    haruspex

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    I meant its rotation about the contact point exerts no torque. It is a flexible string, not a rigid lever. Only the tension in the string exerts a torque, but that has no moment about the contact point: torque = force x perpendicular distance, and the distance from the string to the contact point is zero.
    The torque of interest exerted by the tension is at the disc's axis.
     
  16. Dec 28, 2015 #15
    The string exerts a torque at the contact point of the disk (F*R), the torque is because of the 900 N force, to balance this torque you need the torque of the lever that is needed to be minimum equal to the torque at the disk
     
  17. Dec 28, 2015 #16
    Oh, I see! Thank you. It works now.
     
  18. Dec 28, 2015 #17

    CWatters

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    0.2 * 900
     
  19. Dec 28, 2015 #18
    Thanks.
     
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