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How can I find the critical mu of friction?

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A lady pushes a box with mass (m) with force (F) at a horizontal angle of θ. There is a static friction (μ)

    2. Relevant equations

    ƩFy = Fn - m*g - F*sinθ = 0

    ƩFx = F*cosθ - μFn = 0, where μFn is the force of friction.

    3. The attempt at a solution

    F*cosθ - μ*Fn = 0

    μ*Fn = F*cosθ; and Fn = m*g + F*sinθ

    So, μ(m*g + F*sinθ) = F*cosθ

    μ = (F*cosθ)/(m*g + F*sinθ)

    But the textbook's answer is μ = 1/tanθ. Is it wrong? I almost get their answer except for the m*g. And when I use dummy values I don't get the same answer as the book, so I am pretty sure that my problem is from not simplifying the equation enough.
     
  2. jcsd
  3. Apr 8, 2013 #2

    rude man

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    The answer must have F in it somewhere! So yeah, the book answer is wrong. I believe.
    My answer has tan(theta) in it, but it also has mg and F.
    (Ooh, can't wait to be corrected on this one by some smart ME! :smile: )
     
  4. Apr 8, 2013 #3
    Is your answer equivalent to mine?
     
  5. Apr 8, 2013 #4

    rude man

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    'Fraid not.

    Don't even bother to write separate equations in x and y.

    The lady has to overcome two forces as she's pushing the mass uphill. What are those two forces (along the ramp)?
     
  6. Apr 8, 2013 #5

    haruspex

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    Is she pushing down at angle theta or up at that angle? I'll guess it's up.
    You haven't said what the question asks. I'll guess again: it asks for the angle which minimises F.
     
  7. Apr 9, 2013 #6
    She pushes down at an angle θ below the horizontal.
     
  8. Apr 9, 2013 #7

    haruspex

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    You're still not saying what the book actually asks. Pls post the whole question verbatim.
     
  9. Apr 9, 2013 #8
    I am not sure if I have to do this, but I am going to reference the question just in case I am not committing some kind of plagiarism. So just ignore the reference.

    "A large crate with mass m rests on a horizontal floor. The coefficients of friction between the crate and the floor are μ(static) and μ(kinetic). A woman pushes downward at an angle θ below the horizontal on the crate with force F. If μ(static) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of μ(static)" (P167).

    Referenced from Sears and Zemansky's University Physics (Young and Freedman 13th Edition)
     
  10. Apr 9, 2013 #9

    SammyS

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    Showing the complete problem does help very much.

    Start by drawing a free body diagram for the crate.
     
  11. Apr 9, 2013 #10

    haruspex

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    That's the piece of information that was missing. You obtained this equation correctly (rude man seems to have misread the set-up as being a ramp):
    μ = (F*cosθ)/(m*g + F*sinθ)​
    With the understanding that F is not a specific value and is unlimited in magnitude, can you now derive the desired result from your equation?
     
  12. Apr 9, 2013 #11
    The only way that I can get rid of F is by using ƩFy = Fn - m*g - F*sinθ, where Fn = (F*cosθ)/μ. Then I isolate F which equals (μ*m*g)/(cosθ - μ*sinθ), and substitute this equation in for F from our answer μ = (F*cosθ)/(m*g + F*sinθ). But it turns into a massive mess that I can't simplify to 1/tanθ, and these questions never expect us to do that kind of math. So I know that I must be missing a way that is much more simple.
     
  13. Apr 9, 2013 #12

    haruspex

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    You're not grasping the implication of "no matter how hard she pushes".
    Consider your equation μ = (F*cosθ)/(m*g + F*sinθ) as defining μ as a function of F. What happens to μ as F increases?
     
  14. Apr 9, 2013 #13

    SammyS

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    A free body diagram should help you get an expression for the normal force -- which does depend upon F .
     
  15. Apr 9, 2013 #14
    After trying some values of F, I noticed that μ increases less and less as F increases. So, now I do understand how F increases so does μ therefore so does friction pushing against her push. But I can't make the connection as to how they get μ = 1/tanθ.

    It all makes good sense except for how they get μ = 1/tanθ.
     
  16. Apr 9, 2013 #15
    I have made a FBD. In it Fn = m*g + F*cosθ, and Fn = (F*cosθ)/μ. I can't find anymore useful relationships to Fn.
     
  17. Apr 9, 2013 #16

    mukundpa

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    for maximum force F applied toslide depends on angle θ. for critical value of coefficient of friction dF/dθ must be equal to zero for some angle θ.
     
    Last edited: Apr 9, 2013
  18. Apr 9, 2013 #17

    SammyS

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    What equation do you get from the horizontal component of force?
     
  19. Apr 9, 2013 #18
    ƩFx = F*cosθ - μFn = 0, where μFn is the force of friction, and Fn = m*g + F*cosθ, and Fn = (F*cosθ)/μ
     
  20. Apr 9, 2013 #19

    SammyS

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    That should be Fn = mg + F sin(θ) .


    Put those together.
     
  21. Apr 10, 2013 #20

    haruspex

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    What you might also notice is that μ does not go on increasing without limit. What value does it tend to as F tends to infinity?
     
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