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Homework Help: A little Force help

  1. Oct 3, 2004 #1
    A force of 21.2 N is required to start a 6.65 kg box moving across a horizontal concrete floor. What is the coefficient of static friction between the box and the floor?

    I've drawn my free body diagram and found that the y motion is N^F,B-W^E,B=0 I also know that the Force is larger than the friction
    21.2 N - friction = 6.65a acceleration is not zero in this case right cause if it were zero than the friction would be 21.2
  2. jcsd
  3. Oct 3, 2004 #2
    Speaking in components parallel to the floor, u*Fn + Fapplied = 0 for a box to start moving at a constant (but slow) velocity. Or..

    u * m * g <= -21.2N in this case because the force applied needs to be a teeny bit larger than static friction where u is the coefficient of static/kinetic friction.
  4. Oct 3, 2004 #3
    really don't understand ur problem or possibly the notation u have used
    N^F,B-W^E,B=0 ??

    Anyways ...
    F <= mu_s*N
    and this resists ur applied force
    consider the equilibrium and finish ur work

    -- AI
  5. Oct 3, 2004 #4
    Excellent, thanks for the help. What I was missing was the propper equation for the problem. This is the other half of the problem and I know it uses a slightly different equation but I don't know what that is.

    If the 21.2 N force continues, the box accelerates at 0.400 m/s2. What is the coefficient of kinetic friction?

    (sorry about the notation thing I should've used lower case letters instead of coefficents.)
  6. Oct 3, 2004 #5
    same equation except sum of the forces isn't 0 anymore:

    Fapplied + Ffriction = Fresultant

    21.2N + u*Fnormal = m*0.4

    careful with the sign on Fnormal.
  7. Oct 3, 2004 #6
    Isn't Fnormal just 21.2N*-9.8?
  8. Oct 3, 2004 #7
    R U still around? =(
  9. Oct 3, 2004 #8
    If the box isn't moving horizontally, then the normal force N is equal to the weight of the box mg.
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