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Forces/Friction: Penguin Sled

  1. Sep 29, 2007 #1
    A sled weighing 54.0 N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is 0.100. A penguin weighing 71.0 N rides on the sled, as in Fig. P4.76. If the coefficient of static friction between penguin and sled is 0.700, find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.
    [​IMG]

    I figured the friction on the sled with Fk = .1(71 N + 54 N) = 12.5 N
    Static friction on the penguin is Fs <= .7(71 N) <= 49.7

    So I figured that that the point where the penguin is just about to slip off is where the pulling force minus the kinetic friction equals the static friction of the penguin. So I did

    F - fk = fs
    F = fs + fk = 62.2

    This isn't right however... Any help would be appreciated..
     
    Last edited: Sep 29, 2007
  2. jcsd
  3. Sep 29, 2007 #2

    learningphysics

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    What is the acceleration of the system in terms of F?

    Examining the freebody diagram of the penguin... what is the force that keeps the penguin moving forward? Solve for this force in terms of F...

    slipping occurs when the friction on the penguin reaches 49.7N.
     
  4. Sep 30, 2007 #3
    The force moving the penguin forward is static friction...

    Ok here's what I tried.
    F - fk = ma
    F = ma + fk where m = 125/9.8 = 12.8 and fk = 12.5
    so aSystem = (F - 12.5)/12.8
    and then aPenguin = (-F + 49.7)/7.24

    I don't really know what to do with these because the a's aren't the same... Are they equal in magnitude but opposite in direction?
    If you take them as the same and solve and then plug them back to find F you get the wrong answer.
    I'm still stuck.
     
  5. Sep 30, 2007 #4
    Bump... Need this solved tonight.... :-(
     
  6. Sep 30, 2007 #5

    learningphysics

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    You got the acceleration of the system is:

    (F - 12.5)/12.8

    This is also the acceleration of the penguin.

    so using the freebody diagram of the penguin.

    friction = ma

    friction = (71/9.8)(F-12.5)/12.8

    we know that friction < = 0.700(71) (this is just uk*mg)

    hence (71/9.8)(F-12.5)/12.8 <= 0.700(71)

    So what is the maximum value for F?
     
  7. Sep 30, 2007 #6

    learningphysics

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    this is where you're making the mistake. F is not acting on the penguin.
     
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