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Static and Kinetic Friction Forces (Finding Net Force)

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data

    The coefficient of static friction between a block and a horizontal floor is 0.414, while the coefficient of kinetic friction is 0.138. The mass of the block is 5.24 kg. A horizontal force is applied to the block and slowly increased. What is the value of the applied horizontal force at the instant that the block starts to slide?

    b) What is the net force on the block after it starts to slide?

    2. Relevant equations


    3. The attempt at a solution

    a)Okay, so the answer to the first part is 21.3 N.

    F = uN where us is .414 and N is mg.

    b) I don't get how to find the net force. So the mg and Normal force cancel out, so I calculated the horizontal forces using F=uN. I subtracted the static force from the kinetic force and the answer was wrong. What am I doing wrong?
  2. jcsd
  3. Jan 25, 2009 #2


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    No information is given about the acceleration, which affects the answer. Some assumption must be made. You could assume the object is moving in uniform motion with no acceleration, in which case the net force is zero from F = ma. However, I suspect you are supposed to assume that the force you found in part (a) continues to be applied. Since the friction force is now reduced, the object will accelerate horizontally so there is a nonzero net force acting on it. Just add up the horizontal forces (not including F = ma which tells what acceleration the net force causes). The vertical forces add to zero as you said.
  4. Jan 25, 2009 #3
    Yea what I'm not sure about is what forces they are.

    Is it the static Friction force + the Kinetic Friction force? I really don't know what the horizontal forces are
  5. Jan 25, 2009 #4
    bump sry but its due tomorrow
  6. Jan 26, 2009 #5
    doesn't anyone know this?
  7. Jan 27, 2009 #6
    bump bump bump
  8. Jan 27, 2009 #7
    Please someone help me.
  9. Jan 27, 2009 #8


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    FAJISTE: For part (b), change the plus sign in post 3 to a minus sign.
  10. Jan 28, 2009 #9
    Yeah i tried that too and the answer was wrong :(
  11. Jan 28, 2009 #10


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    FAJISTE: You said you tried that, but what number did you try? Show your work so we can make sure we are talking about the same equation and numbers. How many tries are you allowed? I agree with the post by Delphi51. In other words, the only other logical tries for part (b) I can think of (so far), if the above fails (and show your work for that), are zero, and "unknown." Do they subtract points from you for multiple tries? Also, are you sure you did not omit any words from the given problem statement in post 1?
  12. Jan 28, 2009 #11
    n/m got it lol, i was putting 14.9 into the computer instead of 14.19

    Thanks though!
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