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Finding Friction

  1. Jul 11, 2015 #1
    The worker pushes downward and to the left on the box at an angle of 35°. The box has a mass of 135 kg. The box moves horizontally across the floor to the left at a constant velocity. The force applied by the worker is 2030 N∠215°. What is the value of the friction force?


    Fgravity · sin35° = Fgravityx
    Ffriction + Fgravityx = m · a
    f = μ · Fs


    sin35° =
    Fgravity · sin35° = Fgravityx
    m · g · sin35° = Fgravityx
    (135 kg)(9.81) sin35° = Fgravityx
    759.6 N = Fgravityx

    Ffriction + (759.6 N) = (135 kg)(0 m/s2)
    Ffriction - 759.6 N = 0 N
    Ffriction = 759.6 N

    I know that this isn't correct, because my study guide is multiple choice. My options are:
    a. 1164 N
    b. 1324 N
    c. 1663 N
    d. 2489 N

    I'm really not sure what I'm missing or if I am going the wrong way with finding the value of the friction force. My guess is that c. 1663 N is correct, but that is only a guess. I'd like to be able to prove which answer it is. Thank you for any help!
     
  2. jcsd
  3. Jul 11, 2015 #2

    Student100

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    Your guess is right. Your calculations are off quite a bit. What can you conclude about the friction force since the box is moving to the left at a constant velocity?
     
  4. Jul 11, 2015 #3
    Probably that it is about equal to the Horizontal force, and that the Net Force is 0
     
  5. Jul 11, 2015 #4
    Probably that it is about equal to the Horizontal force, and that the Net Force is 0
     
  6. Jul 11, 2015 #5

    Student100

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    So the forces are balanced right? So to calculate the frictional force on the horizontal plane why are you using sin and fg? What should you do instead?
     
  7. Jul 11, 2015 #6

    Student100

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    Exactly, so what is the Ff in the horizontal direction?
     
  8. Jul 11, 2015 #7
    ... about 1663 Newtons of force since it is a total of 2030N at a 215 degree angle.
     
  9. Jul 11, 2015 #8

    Student100

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    Yep. =)
     
  10. Jul 11, 2015 #9
    Awesome! Thanks a lot!
     
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