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A book is pushed against a vertical wall at an angle

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Albert is pushing a 1.8 kg book against a vertical wall at an angle of 30.0 degrees. If the coefficient of static friction is 0.47 and kinetic friction is 0.35, how hard does Albert need to push the book to allow it to slide down the wall at a constant speed?

    2. Relevant equations

    Fnet = ma
    uK = Fk/Fn (coefficient of kinetic friction)
    uK = Fs/Fn (coefficient of static friction)

    3. The attempt at a solution

    If there was no angle involved, I know that I'd first calculate Fk (force of kinetic friction) as it is equal to Fg (force of gravity). Then using the formula for the coefficient of kinetic friction, I would rearrange to solve for Fn (normal force) since Fn = Fa (force applied) which would get me the answer.

    However, I have no idea how to factor in the angle given in this question to solve the problem since I have not done any forces questions involving angles.

    Thank you in advance!
     
  2. jcsd
  3. Apr 12, 2013 #2

    haruspex

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    Just resolve Albert's push into two forces, a horizontal and a vertical. Do you know how to do that? (Is he pushing down at 30 degrees or up at 30 degrees? Maybe it'll become clear as you solve the problem.)
     
  4. Apr 12, 2013 #3
    Thank you for your response haruspex.

    The question doesn't state if he's pushing up or down at 30 degrees, so I will assume he is pushing up. By a horizontal and vertical forces, do you mean that the force applied (Fa) and the normal force (Fn) are the horizontal forces and the force of gravity (Fg) and Fk (kinetic friction) are the vertical forces? Because that's how I drew my free body diagram and I have the horizontal forces equalling each other and the vertical forces equalling each other.
     
  5. Apr 12, 2013 #4

    haruspex

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    No, I meant that if Albert is pushing with force Fa at an angle θ above horizontal then you can resolve that into a horizontal force and a vertical force. I.e. it is equivalent to the sum of a horizontal and a vertical force. Do you not know how to write those in terms of Fa and θ?
     
  6. Apr 12, 2013 #5
    Are you saying to break up Fa into x and y components so that Fax = (Fa)(sin 30°) and Fay = (Fa)(cos 30°)?

    If not, then I'm afraid I don't know how to write it in terms of Fa and θ.
     
  7. Apr 12, 2013 #6

    haruspex

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    Yes. Having done that, what are your two equations relating all the forces?
     
  8. Apr 13, 2013 #7
    Vertical: Fk + Fay = Fg

    Horizontal: Fn = Fax

    Are these the correct equations?
     
  9. Apr 14, 2013 #8

    haruspex

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    Yes. You know what Fax and Fay are in terms of Fa, and you know the relationship between Fk and Fn. Put all that together.
    Note that if you assume Albert is pushing down at 30 degrees you get an invalid solution.
     
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