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Minimum Force

  1. Aug 6, 2006 #1
    This one's killing me:

    A box with a mass of 22 kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficients of friction between the box and the ramp are u(s) = 0.78 and u(k) = 0.65.

    Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.

    Here's what I'm coming up with:

    F(s) = mgsin theta
    = (22 kg)(9.8 m/s^2)sin 45 degrees
    = 152 N

    F(n) = F(s)*u(s)
    = (152 N)(0.78)
    = 119 N

    F(a) = F(n) - Fgcos theta
    = 119 N - 152 N
    = -33 N

    Does this look correct?
     
  2. jcsd
  3. Aug 6, 2006 #2
    Actually, after looking this over once more, I'm thinking:

    F(n) = F(g) cos theta + F(a)

    Since, F(s) = u(s)F(n), and I've determined F(s) to be 152 N, then

    152 N = u(s) (F(g) cos theta + F(a))
    = 0.78(152 N + F(a))
    = 119 N + 0.78(F(a))
    152 N - 119 N = 0.78(F(a))
    33 N = 0.78(F(a))

    Therefore F(a) = 33 N/0.78
    = 42 N

    Can anyone confirm this?
     
    Last edited: Aug 6, 2006
  4. Aug 7, 2006 #3
    Seems correct to me. I got the same answer.
     
  5. Aug 7, 2006 #4

    Gokul43201

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    Gold Member

    Haven't looked through all the steps in your working, but I have 43N as the force required.
     
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