Static equilibrium and torque of string and pulley

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  • #1
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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.
 

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  • #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
 
  • #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.
 
  • #4
Andrew Mason
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jhrnndz1 said:
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.
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
 
  • #5
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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)?
 
  • #6
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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+!
 
  • #7
Doc Al
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jhrnndz1 said:
Can I add those two tensions together and use the equation torque = F*R?
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.)
 
  • #8
Office_Shredder
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jhrnndz1 said:
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)?

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?
 
  • #9
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Thanks Everyone!!! :)
 
  • #10
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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?
 
  • #11
Andrew Mason
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jhrnndz1 said:
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?
Right, although the answer should be rounded to 736 N.

AM
 

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