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Work and energy

  1. Mar 23, 2006 #1
    Hello. I need some help with the following problem.

    There is a cart on an incline and a pulley with a suspended block.

    Assuming that (the force of friction)= (mu)(N) symbolically show that the coefficient of rolling friction for the car moving down the incline plane with a constant speed is given by mu = tan(theta) - m/[(M*cos(theta))]

    M is the mass of the car and
    m is the mass of the suspended weight over the incline

    Any help would be appreciated

    Thanks
     
  2. jcsd
  3. Mar 23, 2006 #2

    berkeman

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    I don't get the part with the pulley and the block. Is there a rope from the pulley to the cart or something? The main part of the question just sounds like the rolling friction is in balance with the acceleration of gravity....
     
  4. Mar 23, 2006 #3
  5. Mar 23, 2006 #4

    Hootenanny

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  6. Mar 23, 2006 #5
  7. Mar 23, 2006 #6

    Hootenanny

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    The formula you placed at the bottom of the diagram is incorrect. You need to resolve all forces so that they are parallel and perpendicular to the inclined plane (with the exception of the weight of the block)
     
  8. Mar 23, 2006 #7
  9. Mar 23, 2006 #8

    Doc Al

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    There's nothing wrong with that equation if you understand what it means. It's an expression of the equilibrium condition. But that's just the starting point. You need to figure out (in terms of known quantities):
    What's F(parallel)?
    What's f?​

    If you understand what the equation is saying, these should be easy questions to answer.
     
  10. Mar 23, 2006 #9
    F(parallel) = mgsin(theta)
    f = mgsin(theta) - mg

    right?
     
  11. Mar 23, 2006 #10
    What exactly does the F stand for? I know the f is the force of friction.

    Thanks
     
    Last edited: Mar 23, 2006
  12. Mar 23, 2006 #11

    Doc Al

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    Right.
    Yes. But f is the force of friction (rolling friction, in this case); express it in terms of the normal force.

    The way I read the diagram and the equation is that:
    f is the friction force
    F is the tension in the string pulling the cart
    F(parallel) is the component of the cart's weight parallel to the incline​
     
  13. Mar 23, 2006 #12
    ok thanks:smile: i figured it out now

    Thanks
     
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