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Physics Ramp Problem

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data
    A student pulls on a box up an incline of 10 degrees. The student pulls with a force of 180 N directed at 30 degrees above the line of the incline. The box has a mass of 28 kg, and the coefficient of friction between the box and floor is 0.299. The acceleration of gravity is 9.8 m/s^2.

    What is the acceleration of the box?


    2. Relevant equations
    Fnet = ma


    3. The attempt at a solution
    From Fnet = ma I get:
    Fp - Ffr = ma, where Fp is the force of the person pulling and Ffr is the force of friction resisting.

    Ffr = MkFn
    Ffr = (.299)(Fn)

    I then solved for Fn:
    Fn = mg/cos10
    Fn = (28)(9.8) / cos10
    Fn = 278.63

    So Ffr = (.299)(278.63)
    Ffr = 83.31 N

    So we have Fp - Ffr = ma,
    so Fp - 83.31 = (28)a

    We need Fp.
    Fp = cos30 * 180
    Fp = 155.88

    So Fp - 83.31 = (28)a,
    so 155.88 - 83.31 = (28)a

    Solving for a i get a = 2.59 m/s^2

    I submitted this to my online homework system and it said the answer was wrong. Where did I go wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Oct 7, 2007
  2. jcsd
  3. Jul 2, 2009 #2
    ...you forgot mgsin(theta) in your original "Fnet" equation. Include mgsin(theta) and you should get the acceleration. --------> ma = T - F - mgsin(theta)
     
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