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Springs UGH

  1. Dec 28, 2004 #1
    An apparatus to determine coefficients of friction is shown above. The box is slowly rotated counterclockwise. When the box makes an angle (Theta) with the horizontal, the block of mass m just starts to slide, and at this instant the box is stopped from rotating. Thus at angle(Theta), the block slides a distance d, hits the spring force constant k, and compresses the spring a distance x before coming to rest. In terms of the given quantities, derive and expression for each of the following.

    (a) u(s). The coefficient of static friction
    (b) (change in)E, the loss in total mechanical energy of the block-spring system from the start of the block down the incline to the moment at which it comes to rest on the compressed spring.
    (c) u(k). The coefficient of kinetic friction
  2. jcsd
  3. Dec 28, 2004 #2


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    Can you show some of your work? Where are you getting stuck?

    For part a, remember that when the block starts to move, the maximum static frictional force is just barely overcome... so acceleration can be taken to be zero.
  4. Dec 28, 2004 #3
    all i know is that F=uN
  5. Dec 28, 2004 #4


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    Yes, but just remember
    [tex]f_{static max}=u_{static}N[/tex]

    Reason I've put in the static max subscript is because, when you try to move the block, the frictional force will increase to prevent the block from moving, until a max value when it starts to move. It's this max value that's used for the equation above.

    If you can find f and N when it just starts moving then you can find [tex]u_{static}[/tex]. Draw a free body diagram of the mass m. Draw all the forces. Let's take x to be parallel to the plane, and y to be perpendicular to the plane. Write out these equations:

    [tex]\Sigma F_{x}=ma_{x}[/tex]
    [tex]\Sigma F_{y}=ma_{y}[/tex]

    Just as the block starts moving, the acceleration along the plane is 0. And since the block isn't moving perpendicular to the plane, the acceleration perpendicular to the plane is 0.

    So for part a the equations are:
    [tex]\Sigma F_{x}=0[/tex]
    [tex]\Sigma F_{y}=0[/tex]

    So substitute in all the forces, get and then get f and N. Then you have [tex]u_{static}=f/N[/tex]
  6. Dec 28, 2004 #5
    um.... what?
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