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Coefficient of Friction of a piano

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Two men loading a 900 kg piano onto a truck lose control and the piano slips, rolling down the loading ramp. The ramp is a 12 degree incline that is 3.2m long. If the piano accelerates down the ramp at 0.4m/s^2, find the coefficient of rolling friction.

    2. Relevant equations
    F=ma


    3. The attempt at a solution
    a=g(cos)(theta)-(mu)g(sin)(theta
    mu=g(cos)(theta)=9.8(cos12)+0.4
    mu=9.9

    This looks wrong. Where did I mess up?

    Thanks!
     
  2. jcsd
  3. Feb 6, 2014 #2
    Looks like your calculations for weight force and normal force are off, on account of switching the sin and cos. Try drawing it out to see which trig function to use -- weight force acceleration being parallel to the ramp, normal force being normal to it.
     
  4. Feb 7, 2014 #3

    haruspex

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    Not only is the first equation wrong, as jackarms points out, the second equation does not follow from the first. What happened to the sin, and the sign?
     
  5. Feb 7, 2014 #4
    Edit: Too much information

    Sketch out the scenario. Think about how you would find the acceleration of the piano if you had its resultant force and the co-efficient of friction given. Then flip the script and find the co-efficient instead. What do you do?

    The resultant force Fr is the sum of friction and the slope-directional pull of gravity - whatchamacallit.
     
    Last edited: Feb 7, 2014
  6. Feb 7, 2014 #5
    Not only is the first equation wrong, as jackarms points out, the second equation does not follow from the first. What happened to the sin, and the sign?

    (I think I wrote out a lot more, but didn't want to be annoying, so I didn't include all the steps, just the vital ones. I'll include all of them this time. )

    Edit: Too much information.(Ah, I thought that may be the case. My teacher is starting to throw in extra information to trip me up.)

    Sketch out the scenario. Think about how you would find the acceleration of the piano if you had its resultant force and the co-efficient of friction given. Then flip the script and find the co-efficient instead. What do you do?
    (Okay, I was going off a diagram in my lesson for a nearly identical situation, but it may help to draw a new one)

    The resultant force Fr is the sum of friction and the slope-directional pull of gravity - whatchamacallit.

    I'll work on this when I have a moment and hopefully post my new answer soon.
     
  7. Feb 7, 2014 #6
    a=g(sin)(theta)-(mu)g(cos)(theta)
    a+(mu)g(cos)(theta)=g(sin)(theta)
    mu=g(sin)(theta)-a+g(cos)(theta)
    mu=(9.8sin12)-0.4+(9.8cos2)
    mu=2.04-9.99
    mu=7.95

    Looks closer, but still wrong! I've done problems like these before, I'm not sure what's messing me up.
     
  8. Feb 7, 2014 #7

    haruspex

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    Check that last step.
     
  9. Feb 7, 2014 #8
    What is the force of friction? Ff = μmgcosΘ
    What is the slope directional component of the pull of gravity? Fs = mgsinΘ
    The resulting force Fr = Fs + Ff , where Ff is directed the other way hence its value is negative.
    If you write it out with the given data then:
    ma = mgsinΘ - μmgcosΘ -> ma - mgsinΘ = -μmgcosΘ
    gsinΘ - a = μgcosΘ -> μ = (gsinΘ - a)/gcosΘ

    I find it to be roughly around 0.17 , quite realistic since the piano has wheels.
     
  10. Feb 7, 2014 #9
    Do the g's cancel out? If so, I think I could do this:

    a+m(cos)(theta)=sin(theta)
    m=sin(theta)-a+cos(theta)

    But plugging it in and solving yields 0.2-1.38, which is better, but still too large!

    Edit: What if I divide?
    m=sin(theta)/a+cos(theta)
    m=.208/1.28
    m=.16 !

    That looks right!


    Could someone refresh my memory on how I know whether to subtract or divide?
    Thanks for all your help!
     
    Last edited: Feb 7, 2014
  11. Feb 7, 2014 #10

    haruspex

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    The rule is that you must do the same to both sides of the equation.
    From
    a+(mu)g(cos)(theta)=g(sin)(theta)
    you subtracted a from each side to give:
    (mu)g(cos)(theta)=g(sin)(theta)-a
    But then to arrive at
    mu=g(sin)(theta)-a+g(cos)(theta)
    you divided the left by g(cos)(theta) but added it to the right.
     
  12. Feb 7, 2014 #11
    Ah, okay. I remembered the rule, but for some reason I didn't see that I was going against it. Is my answer correct? Thanks so much for your help!
     
  13. Feb 7, 2014 #12

    haruspex

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    No, this is still wrong:
    Take (mu)g(cos)(theta)=g(sin)(theta)-a and divide both sides by the same thing to leave just mu on the left.
     
  14. Feb 7, 2014 #13
    (mu)g(cos)(theta)=g(sin)(theta)-a
    (mu)g(cos)(theta)/g(cos)(theta)=g(sin)(theta)-a/g(cos)(theta)

    Thanks! Like above?
     
  15. Feb 7, 2014 #14

    haruspex

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    Yes, if you put the parentheses in correctly. What you have written is g sin(θ)-(a/g)cos(θ), which is wrong. (And why do you put sin and cos in parentheses?)
     
  16. Feb 10, 2014 #15
    Oops, forgot to reply! I put sin and cos in parentheses because I hoped it would make it easier to read. How's this?

    (mugcostheta)/(gcostheta)=(gsintheta-a)/(gcostheta)
     
  17. Feb 10, 2014 #16
    You don't have to use theta for the angle if it is uncomfortable for you to type in the smybol for it. You could simply say "theta" = B, for example, and write it as
    mu gcos(B) / gcos(B) = (g sin(B) - a) / g cos(B)

    The value of a function is taken from the product, you can add parenthesis to make it clear. There is no need to add parenthesis around the name of a function.

    Mostly you add parenthesis based on the hierarchy of operations, but that's for another thread.
     
  18. Feb 10, 2014 #17

    haruspex

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    No, you should put the arguments in parentheses: cos(theta), etc. For the rest, you could use spaces: mu g cos(theta), or dots or asterisks: mu.g.cos(theta), mu*g*cos(theta). Better still, select "Go Advanced" and pick out the Greek letters from the Quick Symbols panel: μg cos(θ). (That takes a little getting used to because the symbol appears in the typing area as selected text, so if you now continue typing it will replace your symbol. You just need to click to the right of the symbol to deselect it before continuing.)
     
  19. Feb 24, 2014 #18
    Sorry about the later reply! As I explained in my other thread I've been quite busy. Thanks for the formatting advice!

    mu*g cos(theta)/g*cos(theta)=g*sin(theta-a)/g*cos(theta)

    Is that better?

    Thanks!
     
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