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Pulley attached to two objects with a rope

  • Thread starter zm500
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  • #1
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Homework Statement



In the system shown in the figure , block A has mass = 2.00 , block B has mass = 0.360 , and the rope connecting them has a nonzero mass 0.240 . The rope has a total length 1.10 and the pulley has a very small radius. Let be the length of rope that hangs vertically between the pulley and block B. If there is friction between block A and the table top, with mu_Kinetic = 0.203 and mu_Static= 0.250, find the minimum value of the distance such that the blocks will start to move if they are initially at rest. Ignore any sag in the horizontal part of the rope.

http://img651.imageshack.us/img651/5571/capturejup.jpg [Broken]

Homework Equations



F= ma
Torque = F*R
Work = F * ds
dx = Vot + .5at^2
Vf^2 = Vo^2 + 2ax
f = mu * N

The Attempt at a Solution



The attempt at a solution
I know how to draw Free body diagram, but i got confused when the problem mentioned the mass of the pulley. What's the significance of it? Do I apply work in this? I literally no idea how to start this problem! >.<
 
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Answers and Replies

  • #2
gneill
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I know how to draw Free body diagram, but i got confused when the problem mentioned the mass of the pulley. What's the significance of it? Do I apply work in this? I literally no idea how to start this problem! >.<
I don't see the mass of the pulley being mentioned in the problem.
 
  • #3
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I don't see the mass of the pulley being mentioned in the problem.
Well, my professor told me to always draw the Free Body Diagram, but i can't seem to find the relationship [tex]\Sigma[/tex]F = ma with finding the distance.
 
  • #4
collinsmark
Homework Helper
Gold Member
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Well, my professor told me to always draw the Free Body Diagram,
That's very good advice. :approve: Do that with this problem too! :smile:
but i can't seem to find the relationship [tex]\Sigma[/tex]F = ma with finding the distance.
As gneill points out, no information was given about the pulley except that it has a negligible radius. You should assume that the pulley is mass-less and frictionless.

But the rope has mass!

The force due to gravity (the pulling force) is not only a function of mass B's weight, But also the mass of the vertical portion of the rope that happens to be hanging off the pulley.

And it actually gets a little more complicated than that too. Block A is also supporting a portion of the rope's weight (the section of rope that is horizontal). Half of the horizontal portion of the rope is supported by the pulley, and the remaining half of the horizontal part is supported by block A. So the normal force is a function of the weight of the horizontal portion of the rope (and that is in addition to the weight of block A).
 

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