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Coefficient of Kinetic Friction on Inclined Plane? [FORCES]

  1. Oct 26, 2013 #1
    1. The problem statement, all variables and given/known data

    There's no specific problem statement but I have to explain how I would find out the coefficient of kinetic friction given the angle (θ) of the inclined plane and mass of the object sliding down, nothing else.

    2. Relevant equations

    ƩF = ma
    μk = Fk (kinetic friction) / Fn (normal force)

    3. The attempt at a solution

    I split the Fg (Fg = mg; weight) vector into x- and y- components so I could determine Fn (normal force).
    As a result, Fn = mgcosθ.
    And the force that moves the object downhill (I don't know what to call this) = mgsinθ.

    Now, to determine μk I know I must find Fk (kinetic friction) first. Using μk = Fk/Fn, Fk = μk*Fn which = μkmgcosθ.

    So, ƩF = (force that moves object downhill) - Fk, ƩF = mgsinθ - μkmgcosθ ; mg's cancel each other.
    Therefore ƩF = sinθ - μkcosθ, and ƩF = ma, so... ma = sinθ - μkcosθ.
    I then isolated μk (*I don't know if I did this correctly)
    * μk = ma - sinθ / -cosθ ?

    After doing all of this, I found out that I have all the variables except for ACCELERATION...
    and a = ƩF / m which I can not solve for!
    Can I just say that a = 0? I have no idea what to do.
     
  2. jcsd
  3. Oct 26, 2013 #2

    Doc Al

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    The mg's don't cancel. (But mass will when you apply Newton's 2nd law.)

    You need something additional, measured or given. Just knowing the mass and the angle of the board tells you very little. (I can readily swap boards so that it has a different μ, right?)

    Unless there's something special about the angle, you haven't enough information to solve for μ.

    Perhaps this is an experimental question and you can measure the acceleration?
     
  4. Oct 26, 2013 #3
    Oh, right...
    Yes, it is an experimental question but it says the only objects I have in hand is the track, object sliding down the track, and a protractor (that measures the angle).
    I'm still having trouble understanding... a = ƩF/m and ƩF = gsinθ - μkgcosθ
    Aren't I still missing two variables?
     
  5. Oct 26, 2013 #4

    haruspex

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    It's easily proved you need more information. You could put two objects the same mass but very different surfaces on the same incline. You need information about what happens when you do this. And as Doc Al says, the mass isn't useful anyway. Have you quoted the problem exactly as it was given to you?
     
  6. Oct 26, 2013 #5
    "You find a block and cart, a track, and a protractor attached to the track. You find nothing else, no stopwatch, string, spring-scale or masses for hanging. The protractor shows the inclination of the track and was used for another experiment where steepness of the slope was found. Describe how you could measure the coefficient of kinetic friction using the items you found, in detail and with a formula."

    This is the full question.
    Oh, and the block is beneath the friction-less(?) cart, which is why there is friction.
    The mass of the block+cart is 0.734 kg.
     
    Last edited: Oct 26, 2013
  7. Oct 26, 2013 #6

    haruspex

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    There is a crucial difference from what you originally posted. It says "Describe how you could measure". That is, you are allowed to conduct an experiment, but using the given equipment only.
     
  8. Oct 26, 2013 #7
    I'm still a bit confused. So could I possible say that I adjusted the track where the cart+block moved at constant velocity (∴ no acceleration)?
     
  9. Oct 26, 2013 #8

    haruspex

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    Yes indeed. You should supply a little more detail on how you would do that (think about static friction here), what you would measure, and how you would deduce the coefficient of kinetic friction.
     
  10. Oct 26, 2013 #9
    Static friction? When the amount of force surpasses the amount of static friction, that's when it starts moving right? But where and why would I need to include that?

    If I just say that I would adjust the track, and then repeat the stuff above, is that applicable?
     
  11. Oct 26, 2013 #10

    haruspex

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    You want kinetic friction, which is generally somewhat less than static friction. If you adjust the angle until the block moves, what will happen subsequently?
     
  12. Oct 26, 2013 #11
    Um, the frictional resistance decreases? and force is applied to move it downhill?
     
  13. Oct 26, 2013 #12

    haruspex

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    Yes, the resistance decreases, but you have raised the angle far enough to overcome the static friction. So what does the FBD tell you about how it will move once it has started?
     
  14. Oct 26, 2013 #13
    It will move with constant velocity and frictional force will be less (shorter arrow) than the downhill(?) force? I don't know if I'm fully understanding the question, sorry.
     
  15. Oct 27, 2013 #14

    haruspex

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    The second part of that is correct. You adjusted the angle high enough to overcome static friction, but immediately the frictional force will reduce to be only kinetic friction. Thus the downhill force will now exceed the frictional force by some margin. So now think again about the first half of your answer.
     
  16. Oct 27, 2013 #15
    Oh, its initial velocity will be 0 and then it will move at a constant velocity?
    Because it wouldn't be accelerating and it wouldn't have been moving from its initial position, that's the whole idea, right?
     
    Last edited: Oct 27, 2013
  17. Oct 27, 2013 #16

    haruspex

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    Why not? You set the angle so that the downslope component of gravity matched/overcame static friction: ## mg \sin(\theta_s) = \mu_s mg\cos(\theta_s) ##. Once it starts moving we have friction ## \mu_k mg\cos(\theta_s) < \mu_s mg\cos(\theta_s) ##, but the downslope force hasn't changed. So why will it not accelerate?
     
  18. Oct 27, 2013 #17
    I'm sorry but I'm really confused now. When I asked before if I could say that I adjusted the track so that the object moved at a constant acceleration (so no acceleration), didn't you say yes? But then why would there have to be acceleration.

    In other words, can't I say I set the track to a minimum angle so that it went downhill at a constant speed?
     
  19. Oct 27, 2013 #18
    Here, this is what I said (readjusted):

    I would first adjust the track to an inclination in which the block and cart moves at a constant speed (so, at its minimum with no acceleration) then measure angle θ. After, I would decompose vector Fg into its x- and y- components to calculate normal force which = mgcosθ. I will also determine downhill force by mgsinθ. Then, I will calculate kinetic friction by Fk = μkFn; = μkmgcosθ.

    So, total net force would be ma = mgsinθ - μkmgcosθ; divide m on both sides
    a = gsinθ - μkgcosθ; a = 0
    gsinθ = μkgcosθ; divide both sides by g
    sinθ = μkcosθ
    μk = sinθ/cosθ
    μk = tanθ
    ---
    Is there something wrong with this method?
     
  20. Oct 27, 2013 #19

    haruspex

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    But there's a practical difficulty in doing that. You need to spell out the details of exactly how you would achieve this. If you simply raise the angle until the block moves, it will accelerate down, so that's too high. It would not be practical to try to readjust it as this is happening.
     
  21. Oct 27, 2013 #20
    Oh, so it simply can't be eyeballed?
    This is the only way I have learned to measure the coefficient of kinetic friction in the lab; I know of no other way.
     
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