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Push force up an incline

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A 5.00-kg box sits at rest at the bottom of a ramp that is 8.00m long and that is inclined at 30.0° above the horizontal. The coefficient of kinetic friction is μk = 0.40, and the coefficient of static friction is μs = 0.50. What constant force F, applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of 4.00s?


    g = 9.8 in my class


    2. Relevant equations


    Fg = m(g)
    FN = m(g)(cos(θ))
    Ffk = μk(FN)
    FNET = m(a)



    3. The attempt at a solution

    Fg = 5(9.8) = 49N

    FN = 5(9.8)(cos(30)) = 42.44N

    Ffk = .40(42.44) = 16.98N

    This is where I get stuck. I am able to find the minimum amount required to get it to start and keep moving but not what the question is asking. My textbook does not explain this topic very well so I was getting frustrated.
     
  2. jcsd
  3. Jun 26, 2012 #2
    Hi Zarrey!

    You know the forces that act on the block along(parallel) the incline. These are constant, and so is your force F. Now the sum of these forces need to produce an acceleration a, such that the distance traveled in t seconds is the length L of the incline. As an equation, you can write that as

    [tex]L = \frac{1}{2}at^2[/tex]
     
  4. Jun 26, 2012 #3
    so then

    L = .5a(t2)
    8 = .5a(42)
    8 = .5a(16)
    8 = 8a
    1m/s = a

    Fll = 49(sin(30))
    Fll = 24.5N

    FNET = Fp - Fll - Ffk

    And then newtons law:

    Fp - Fll - Ffk = m(a)
    Fp - 24.5 - 21.22 = 5(1)
    Fp = 50.72N

    is this correct? (i substituted Fp as the push force since that what we use in class)
     
  5. Jun 27, 2012 #4
    Bump.

    Sorry for this bump but this question is kind of important as it is on a take home test. I'm kind of freaking out because I am already turning it in late. A validation response would be much appreciated. And thanks for the help in advance.
     
  6. Jun 27, 2012 #5

    Doc Al

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    Staff: Mentor

    Looks good except that you used static friction instead of kinetic friction.
     
  7. Jun 27, 2012 #6
    Hi, sorry for the late reply.

    The bolded term seems wrong. You need to use the kinetic friction coefficient.
     
  8. Jul 3, 2012 #7
    thanks a lot, i appreciate everyone's help in getting me to understand this frustrating topic :).
     
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