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Normal force, friction, Newton's law

  1. Sep 29, 2011 #1
    While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction between the box and the floor is 0.41. The pushing force is directed downward at an angle θ below the horizontal. When θ is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find that value of θ.

    2. Relevant equations
    FN = mg + FsinƟ

    Ffriction = 0.41(mg + FsinƟ)

    Ffriction = FcosƟ in our problem.
    0.41(mg+FsinƟ) = FcosƟ
    0.41mg + 0.41FsinƟ = FcosƟ

    I think I am correct and that my logic is sound so far but I just do not see how I can find Ɵ? Hints please?

    3. The attempt at a solution
  2. jcsd
  3. Sep 29, 2011 #2


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    Homework Helper

    The second equation gives the horizontal component of the pushing force, say, Fx, which has to be equal to the force of friction.
    Fx = FcosƟ, Fx=Ffriction. But the logic is correct otherwise.

    You need to apply some trigonometry to find the angle. Express sin and cosine with the same function: cos with sine, or both cosine and sine with tangent...

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