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Block resting on slop with friction - statics

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data
    a block of mass m resting on a 20° slope. The block has coefficients of friction µs = 0.80 and µk = 0.49 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass m2 = 2.0 kg.

    What is the minimum mass m that will stick and not slip?

    2. Relevant equations

    Newtons second law sum of the forces = mass * acceleration

    3. The attempt at a solution

    I used the hanging block to find the tension to be equal to 9.8*(2) = 19.6 N

    With that found I found that n = mgSin(20) and T = µs * mgSin(20)

    So 19.6 = .8*m 9.8Sin(20) and m = 7.309 kg

    The online homework says this is incorrect and I have no idea of any other way to go about this problem any help is appreciated.

    Thanks
     
  2. jcsd
  3. Oct 30, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Pullies!

    Good.

    That's an incorrect expression for the normal force (wrong component of weight). And I assume you meant friction force, not T (for tension).

    Hint: Three forces act on the block parallel to the incline.
     
  4. Oct 30, 2009 #3
    Re: Pullies!

    I don't get what you mean by wrong component of weight, when i draw my FBD i have normal force going up and mg pointing down and

    n - mg = 0 so n = mg but because its the 20 degree slope its n = mgSin20 ?
     
  5. Oct 30, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Pullies!

    While mg points down, the normal force doesn't point up--it's perpendicular to the surface, so it's at a angle.

    It's not true that n = mg. To find the normal force, consider force components perpendicular to the surface--they must add to zero. What's the component of gravity perpendicular to the surface.
     
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