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Mechanics - Pulley system finding coefficient of friction

  1. Jun 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Mechanics Q36a.png


    2. Relevant equations

    I believe:
    F = m . a
    Ffriction = μs . N
    Ffriction = μs . mg . cosθ

    3. The attempt at a solution
    I don't quite get how the equations above would give me an answer to the minimum friction required so that the farmer doesn't slip. I know Fl = (70x9.81)cos(40) = 526N But there is now a variable Ffriction that I don't know how to find...

    I'm terrible at physics. If someone could explain or give me a head start, it would be appreciated.
     
  2. jcsd
  3. Jun 10, 2012 #2
    Hi Roaku!! Welcome to PF :smile:

    Start by making the free body diagram of the problem. You will have the forces balanced as, for the vertical direction, of the farmer...

    [tex]Tsin\theta - mg = N[/tex]

    where N is the normal reaction, and T is tension in the rope.

    Similarly, you will have an equation for the horizontal direction for the farmer, involving frictional force. Find the normal reaction from the above equation and use it in the one you made to get the minimum static friction.
     
  4. Jun 10, 2012 #3
    Use only general equation.
    Then find value of N.
    You also need to know not only the magnitude but also the direction of all the forces involve.
     
    Last edited: Jun 10, 2012
  5. Jun 10, 2012 #4

    ehild

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    Gold Member

    The free-body diagram refers to the man. N is the normal force exerted on his feet by the ground. It points upward. The ground can not exert downward force. The vertical component of the tension also points upward. The weight of the man points downward. So the equation for the vertical force components is Tsin(theta)+N-mg=0, that is, N= mg-Tsin(theta)

    ehild
     
  6. Jun 10, 2012 #5
    Whoops, I mis-posted the equation.

    [tex]Tsin\theta - mg + N = 0[/tex]

    Thanks for observing that, ehild :smile:
     
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