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Static equilibrium and torque

  1. Jul 17, 2006 #1
    There's a question that I'm having trouble with that I do not know exactly how to set up.

    Here it is.....

    A 320g mass and a 400g mass are attached to the two ends of a string that goes over a pulley with a radius of 8.70 cm. Because of friciton, the pulley does not begin to rotate. What is the magnitude of the frictional torque on the bearing of the pulley if the system is in static equilibrium.

    Can someone help, I know since it's in static equlibrium that the sum of Fy=0 and sum of torque = 0.
     
  2. jcsd
  3. Jul 17, 2006 #2

    Office_Shredder

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    Each mass is being pulled down by the force of gravity. From there, calculate the torque on the pulley from each mass (knowing that torque = F*R), and you should be able to find how much friction is necessary
     
  4. Jul 17, 2006 #3

    Doc Al

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    What forces do the strings exert on the pulley and what is the resultant torque? Hint: Since the system is in equilibrium, you should be able to find the tension in the string on each side of the pulley.
     
  5. Jul 17, 2006 #4

    Andrew Mason

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    Do a free body diagram of each mass and determine the tension in each end of the string. What is the torque on the pully in terms of the two tensions?

    AM
     
  6. Jul 17, 2006 #5
    Alright, I calculate each tension in both of the strings and calculated 3.14 and then 3.924. Can I add those two tensions together and use the equation torque = F*R? (7.1 * .087 = .61)?
     
  7. Jul 17, 2006 #6
    i got the answer as 0.068208 n.m.
    is it correct....
    the sum is very easy i suppose....
    first find the gratational force possesed by each of the weight....you ll find that one of the force is greater tha the other. now minus the greater force by the smalller force...you get a value...multiply this value with the radius of the circle.....voila...A+!
     
  8. Jul 17, 2006 #7

    Doc Al

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    No, you can't just add the tensions since they exert torques in opposite directions. Each tension exerts its own torque: one clockwise, the other counterclockwise. (Thus one torque is positive, the other negative.)
     
  9. Jul 17, 2006 #8

    Office_Shredder

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    Why are you adding the two forces together? If you hang one mass off each side of a pulley, do they oppose each other, or do they work together to make the pulley spin?
     
  10. Jul 17, 2006 #9
    Thanks Everyone!!! :)
     
  11. Jul 17, 2006 #10
    I have another question,

    A 15.0 kg child is sitting on a plyground teeer-totter, 1.50m from the pivot. What force, applied 0.300m on the other side of the pivot, is need to make the cild lift off the ground?

    I got the Force to be 735.8N, is this correct?
     
  12. Jul 17, 2006 #11

    Andrew Mason

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    Right, although the answer should be rounded to 736 N.

    AM
     
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