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Fnet question.

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A 250 N force is applied at an angle of 32° above the horizontal to a 96 kg wooden box causing
    it to slide along a floor as shown.

    The coefficient of friction between the floor and the box is 0.18. What is the magnitude of the net
    force on the wooden box?

    3. The attempt at a solution

    ff=mu times N
    ff =0.18 times 9.8 times 96
    For normal force would you find the y component ie. 9.8 times 96 times sin32 or do you just do N=W?


    and if I do N as the y component "because some of the mass is being lifted" I get the wrong answer too. 122N
    it is supposed to be 67N?
  2. jcsd
  3. May 8, 2009 #2


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    To calculate the normal force, you need to consider the y component of the applied force too.
  4. May 8, 2009 #3
    I know, I just cant remember if for ff you multiply N times sin 32, I dont remember doing that. so the y component of the applied force is subtracted from the normal force?
  5. May 8, 2009 #4


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    It is subtracted from the weight. N = W - F sin(32°). Tell me if you don't see why.

    After you get the normal force, the friction force is just f = µN.
  6. May 8, 2009 #5
    Alright I got it, Its been a while and Im just doing some review. "also I was using an online calculator that was giving me some odd answers."
    Thank you alot.
  7. May 8, 2009 #6


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    No problem.
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