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Sliding box question- finding applied force with friction

  1. Jan 25, 2014 #1
    1. The problem statement, all variables and given/known data

    A 90 kg box is pushed by a horizontal force F at constant speed up a ramp inclined at 28°, as shown. Determine the magnitude of the applied force.
    when the ramp is frictionless.
    when the coefficient of kinetic friction is 0.18.

    2. Relevant equations

    Ff= uk(Fn)

    3. The attempt at a solution

    I think I have the answer to part 1 complete and correct: I tilted the axis and ended up multiplying sin28degrees(90kg)(9.8m/s^2) to give me 414N for the applied force.
    Im not sure about the second part though, where I have friction involved. I tried finding my force of friction --> (0.16)(882N) = 141 N, then subtracting it from my initial applied force I found - so 414N - 141 N gives me 273 N. Not sure if this is correct though.. Help please!
     
  2. jcsd
  3. Jan 25, 2014 #2

    gneill

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    Staff: Mentor

    Keep in mind that the applied force is horizontal. Only a component of that force will be directed along the slope. Draw a free body diagram and see if you can't work out how to decompose the applied force into components along the slope and normal to the slope.

    Again, since the applied force is not parallel to the slope, but is horizontal, it's component normal to the slope will contribute to the net normal force that results in friction...
     
  4. Jan 25, 2014 #3

    PhanthomJay

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    Gold Member

    No, you are assuming that the applied force is parallel to the incline, when it is given that the applied force is horizontal.
    Since the applied force is horizontal, it has a component perpendicular to the incline, which affects the value of the normal force.
    But why in any case would the applied force be less than the no friction case? It would be harder to push , wouldn't it?
     
  5. Jan 26, 2014 #4
    Okay I see what you guys mean about the force being parallel versus horizontal . So instead, I break the Fapplied into x and y components. Since my Fnet of x is 0 d/t no acceleration, my Fgx should be equal and opposite to Fax.. So 414N I think. Then I used cos28Fax to find the applied force, so 370N?
     
  6. Jan 26, 2014 #5

    gneill

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    Staff: Mentor

    You can't get a component of a force that's larger than the force itself (414 is larger than 370) So revisit your calculation of the applied force. If F is the applied horizontal force, what expression yields Fax (its component parallel to the slope)?
     
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