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Sliding egg I think the books wrong, could someone please verify?

  1. Jun 17, 2010 #1
    ***EDIT***Sorry for a second there, I tried to convince myself that FN was not equal to Fg, but it must.

    Given the coefficient of static friction (.4) between an egg and a pan, what is the smallest angle from the horizontal taht will cause the egg to slide?

    Here's my drawing:

    Here's my work:
    x, implying that it's along the x-axis
    Fs is static friction force
    Fg is the force of gravity
    m = mass
    a = acceleration
    (mue) = .4, the coefficient of friction

    Fs(max) = (mue)FN Equation 1
    FN = mg
    Substitute to get:
    Fs(max) = (mue)(mg) Equation 2

    F(net)x = max
    Fs - Fgx = max

    Since it is stationary, a = 0

    Fs - Fgx = m(0)
    Fs = Fgx Equation 3

    Now substitute (mue)(mg) for Fs,
    (mue)(mg) = Fgx, and substitute the Fgx from the diagram,
    (mue)(mg) = Fgsin(theta), and substitute mg for Fg
    (mue)(mg) = (mg)sin(theta), mg cancels out
    (mue) = sin(theta), and substitute the known .4 for (mue),
    .4 = sin(theta)
    theta = sin-1(.4)
    theta = about 23.6 degrees

    The book says 2... any ideas on where I may have gone wrong?
    Last edited: Jun 17, 2010
  2. jcsd
  3. Jun 17, 2010 #2

    D H

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    The magnitude of the normal force is not equal to mg.
  4. Jun 17, 2010 #3


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    To elaborate on DH's remark. If the egg is just about to slide, it is still at rest and the sum of all the forces on it must be zero. Look at your drawing. For the sum of the three vectors to be zero, the magnitudes of the vectors must form a closed triangle. Fg is the hypotenuse and Fn one of the right sides. Can they be equal? What is it that must be equal instead?
  5. Jun 17, 2010 #4
    FN = Fgcos(theta)... that seems to make more sense.
  6. Jun 17, 2010 #5


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    Can you finish the problem now?
  7. Jun 19, 2010 #6
    Ya, I figured it out... finally thanks again.
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