1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A little Force help
  1. Little help (Replies: 2)

  2. Atom of argon charge (Replies: 5)

  3. A little help (Replies: 2)

Loading...