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Coefficient of static friction and kinetic friction problem!

  1. Nov 17, 2012 #1
    Hi all,

    I'm in an algebra based physics class so I apologize if the way I approach the problem is different from what you might be used to. A buddy of mine is calc based physics always gets confused about how I set the problems up. Also, this is my first post! Thanks in advance for the help.

    You are trying to slide a refrigerator down a ramp 10 degrees above the horizontal. The mass of the refrigerator is 192 kg and you need to exert a force of 500 N to make it just begin to move.

    A.) what is the coefficient of static friction between the floor at the refrigerator?

    B.) You find that after the refrigerator begins to move that you need only apply 350 N to keep it moving at a constant speed. What is the coefficient of kinetic friction?

    C.) What force would you have to apply so that the refrigerator accelerated down the ramp at 0.5 m/s^2?

    2. Relevant equations

    Coefficient of static friction (μs) = fs/n where fs is static friction force
    Normal force = y component of weight force
    force = m*a

    *for objects at rest or with no acceleration the sum of all forces must equal zero!

    3. The attempt at a solution

    A.) I drew a free body diagram and designated my axes with the x-axis parallel with the surface of the ramp and the y-axis is in the direction of the normal force and y-component of the weight force.

    weight force (w) = 192 * 9.8 = 1881.6
    y-component (wy) = cos(10)*1881.6 (which is also equal to the normal force)
    x-component (wx) = sin(10)*1881.6

    Since we have to find friction and the friction is in the x-direction we have three things to account for: wx, pushing force (fp), and fs. With that in mind I said:

    wx + fp +fs = 0

    and solved for fs. So:

    sin(10)*1881.6 + 500 + fs = 0

    This gives a friction force of about 826.7 so μs = 826.7/[cos(10)*1881.6]= 0.45. The thing is that my friend got a different answer on the test and he got it right. What am I doing wrong?

    PART B:

    I got this part right by taking the x-component of the weight force as sin(10)*1881.6 + 350 and dividing that by the normal force:

    [sin(10)*1881.6 + 350]/[cos(10)*1881.6] = μk = 0.37

    PART C:

    I got this wrong on the midterm but I think I have a new solution:

    fp + wx - fk = 192 kg * 0.5 m/s^2

    fp + [sin(10)*1881.6] - 667 = 96

    fp = 436 (Approximately!)

    How does this last part look?

    *edited to take out repetitive cut and pastes
  2. jcsd
  3. Nov 17, 2012 #2

    Simon Bridge

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    You had rounded off your values too early in the calculation.
    So your method was correct but your arithmetic was wanting.

    The way to do problems like this is to do the algebra before you substitute in the numbers ... so, keep the variable names.

    Off your free body diagram, you have and applied force ##F## pointing down the ramp, a normal force ##N## normal to the ramp, friction force ##\mu N## pointing up the ramp and a weight force ##mg## pointing "downwards". The ramp has angle ##\theta## to the normal.

    Using your axis:
    in the +x direction (pointing down the ramp)
    (1) ##F+mg\sin(\theta)-\mu_s N=0##

    in the +y direction (normal to the ramp)
    (2) ##N-mg\cos(\theta)=0##

    This tells you that:

    putting the numbers in:
    $$\frac{500+(192\cdot 9.8)\sin(10^\circ)}{(192\cdot 9.8) \cos(10^\circ)}=0.4616$$... was that what your friend got?
  4. Nov 17, 2012 #3


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    I think you typed in one '4' less. The result is 0.44616. Rounded to two significant digits, it is 0.45, the same as the OP's result.

    Flemonster: The input data are given with three significant digits, try to give the result with three decimals. Your derivation is correct, althought you have a typo in the equation in A: it should be wx + fp -fs = 0.

  5. Nov 17, 2012 #4


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    In what direction does the pushing force act? If parallel to the incline, your answer looks right.
    So i guess the pushing force is parallel to the incline, so you have both parts correct.
    Looks good!

    Note: Wheteher calc based or algebra based, the method for this problem is the same.
  6. Nov 17, 2012 #5

    Simon Bridge

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    Oh drat - so I did : that means it will probably be a difference of dp/sig.fig like you said.
    Of course, it helps to see what the "correct" answer was.

    OR the figure for the friction force should be negative.

    It is good form for the letters to stand for the magnitudes of forces though.
  7. Nov 18, 2012 #6
    So my friend got 0.46 for μs but if you all think the method I used for setting it up was correct then maybe my instructor made a mistake. She's giving us the chance to make up points on the midterm so I'll show her how I came up with the answer tomorrow and see what she says.

    And Simon, good point about setting up the problem before plugging in numbers for the variables. I sometimes start plugging in figures and get screwy calculations.

    Thanks ehild and PhantomJay for looking over the calculations too. It really helps having someone confirm that I'm on the right track.
  8. Nov 18, 2012 #7

    Simon Bridge

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    Maybe your friend and the instructor made the same mistake I did?

    In long answers, it is a very useful practice to go through the algebra like I did - that way, you demonstrate your understanding. You don't lose so many marks for an arithmetic error for instance, and you can give the examiner pause if the model answer contains a flaw.

    When I set problems like this, I hardly ever give marks for getting the "correct" number out the end.
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