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Homework Help: Find acceleration of a box given the forces on it...

  1. Nov 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A(n) 895 N crate is being pushed across a level floor by a force of 397 N at an angle of 21◦ above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.24. The acceleration of gravity is 9.81 m/s.


    What is the acceleration of the box? Answer in units of m/s/s

    3. The attempt at a solution
    to find mass, Fg = mg
    895 = m9.8
    m = 91.3265

    I've tried
    ΣFy = FN - mg - FAsinθ
    ΣFy = FN - mg - FAsinθ = ma
    ΣFy = FN - mg - FAsinθ = 0
    FN = mg + 397sin21
    FN = 895 + 142.272
    FN = 1037.27

    ∑Fx = FAcosθ - Ff
    ∑Fx = 397cos21 - (0.24)(1037.27)
    397cos21 - (0.24)(1037.27) = ma
    397cos21 - (0.24)(1037.27) = (91.3265)a
    = 1.33 m/s/s but this is not correct

    can anyone tell me where i went wrong/what is the correct way? thanks
  2. jcsd
  3. Nov 16, 2016 #2

    Simon Bridge

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    Science Advisor
    Homework Helper

    Note: ##\sum\vec F = m\vec a = (mg)\vec a/g \implies \vec a/g = \frac{1}{mg}\sum\vec F## ... ie. the net force divided by the weight is the acceleration in gees.
    ie. your answer would have been a = 0.14g. Then, no need to do that 1st division risking a possible rounding error.

    I bring this up because your reasoning seems fine - so the error will be in arithmetic or rounding off or something like that... or a mistake in the problem itself.

    So redo from the start...
    W=895N, F=397N, ##\theta##=21deg, ##\mu##=0.24
    ... the task is to derive an equation for the acceleration in terms of just these 4 things.

    Off free body diagram ##\sum\vec F = m\vec a## :
    (1) ##F\cos\theta - \mu N = ma = (W/g)a## since ##W=mg##
    (2) ##N-W-F\sin\theta = 0##
    (N=normal force)

    Do all the algebra first - when you have derived the final equatio, then you can put the numerical values in.
  4. Nov 16, 2016 #3


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    Staff: Mentor

    The problem describes the applied force as having an angle of 21° above the horizontal, but the diagram seems to show it as directed below the horizontal. Could it be that the diagram is not right?
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