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Calculating Kinetic Friction

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    An 85-N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward.


    2. Relevant equations
    Calculate the coefficient of kinetic friction between the box and the floor.


    3. The attempt at a solution
    I have no idea how to solve this problem, here's what i wrote down on the test and only missed 7 points on a 17 point problem:

    A = -0.90m/s^2

    Fx = µxN
    Fx = -0.90 * 20
    Fx = -18 N
     
  2. jcsd
  3. Dec 8, 2008 #2
    F=ma is your best friend :)
     
  4. Dec 8, 2008 #3
    so using F=ma i would get....

    F = 85N * -0.90 m/s
    F = -76.5N

    Would that be a correct solution?
     
  5. Dec 8, 2008 #4
    on which axis is the box accelerating, (or deccelerating)?
    does the box have any y-direction movement?

    think about those, and remember
    [tex]\sum[/tex]F=ma, this means all the forces on the same axis are equal to the mass of the body times its acceleration..

    breaking F=ma into components we get...
    [tex]\sum[/tex]Fx=max, and
    [tex]\sum[/tex]Fy=may
    on which axis is the box accelerating? is there any acceleration in the y-direction? if you read the question it says the box is being pushed across the floor horizontally which means no y-acceleration, do you see that? so now we have,
    [tex]\sum[/tex]Fx=max, and
    [tex]\sum[/tex]Fy=0


    here is your part now draw out your FBD, your second best friend, and ALL the forces acting on it.... then decide which force goes in which equation....
     
  6. Dec 8, 2008 #5
    so we can just ignore the vertical component of the force?

    and get:

    20 N - Friction Force = 85N * A(x)

    and would Ax be -0.90 m/s?
     
  7. Dec 8, 2008 #6
    yes, you got it now!
     
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