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Help with net force!

  1. Dec 25, 2006 #1
    1. The problem statement, all variables and given/known data
    A decorative paperweight of mass 0.50 kg. sits on a desktop. The coefficient of friction is 0.60.

    a) How much force in the horizontal direction is needed to make the paperweight just move?

    b) What is the net force on the paperweight at the instant it begins to move (with negligible acceleration)?

    2. Relevant equations

    coefficient of friction = Force to push paperweight / Weight of paperweight
    weight = (mass)(gravity)

    3. The attempt at a solution
    a) w = mg = .50 kg. (9.81 m/s^2) = 4.905 N

    coefficient of friction = Force to push paperweight / Weight of paperweight
    Derivation: Weight of paperweight x coefficient of friction = Force to push paperweight
    4.905 N x 0.60 = 2.943 N

    b) I'm not really sure with this, but I think the net force is still 2.943 N because I read somewhere that "If there is just one force on an object, then that force is the net force."

    Can someone help me with this? And please check if my answers are correct?
     
  2. jcsd
  3. Dec 25, 2006 #2

    radou

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    Homework Helper

    Drawing a free body diagram should answer your question immediately. :wink:
     
  4. Dec 26, 2006 #3
    That's another problem, I dunno how to make the free body diagram... Can anybody help me? And see if my answers are right?
     
  5. Dec 26, 2006 #4
    When the paperweight is in equilibrum on the table, there is no net force on it. When you exert force on it (i.e below static frictional force), an equal and opposite force will be produced balancing each other. When u gradually increase the force the opposite force will also increase but till a certain limit. For a certain value of exerted force the frictional force will be the maximum, when yr exerted force is equal to this maximum value the paperweight will still remain unmoved , but any force greater than this will accelerate the paperweight.
    In your problem the maximum value of the mentioned force is 2.943 N.
    Anything above 2.943 N will make the paperwight move.
     
  6. Dec 26, 2006 #5
  7. Dec 26, 2006 #6
    So does that mean the force of 2.943 N in the horizontal direction will make the paperweight move? And if it does, then there's just one force acting on the paperweight, making it the net force? The net force is the sum of all forces acting on the object, so that means 2.943 N is the net force? Can anyone please clarify...
     
  8. Dec 26, 2006 #7
    a) The force of friction is F_s = uN, where u = coefficient of friction and N = normal force.

    You figured out the normal force when computing the weight mg. So now you have F_s. F_s actually doesn't show up until we start pushing the paperweight, so let's pretend we're pushing the paperweight horizontally to the right.

    So now we have our applied force, F_ext, and the frictional force F_s. When F_ext exceeds F_s, then we'll have a non-zero F_net = F_ext - F_s and the object will accelerate in the direction we're pushing the paperweight. So your answer is right, as the point where F_ext = F_s is the point just when the weight will move.

    b) Looking at the justification for part a), this should be simple. We've calculated where F_ext = F_s and we know that F_net = F_ext - F_s, so what must F_net be at that point?
     
  9. Dec 28, 2006 #8
    ^ Oh ok, so that means the net force is zero? Is that right? Please correct me... And F_ext and F_s have the same value, 2.943 N?
     
  10. Dec 28, 2006 #9

    Hootenanny

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    Gold Member

    Correct:smile:
     
  11. Dec 29, 2006 #10
    Yay! Thanx a lot guys! ^^
     
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