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Friction problem of a sled

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A sled weighing 52 N is pulled horizontally
    across snow. A penguin weighing 67.3N rides
    on the sled,as in the figure. (sorry i can't provide the picture)

    If the coefficient of static friction between
    penguin and sled is 0.791, find the maximum
    horizontal force that can be exerted on the
    sled before the penguin begins to slide off.
    The coefficient of kinetic friction between sled
    and snow is 0.097.

    Answer in units of N.

    2. Relevant equations

    3. The attempt at a solution
    i drew a force diagram for the sled and i figured out that:

    Fnet(horizontal)= -friction of sled on penguin -friction of the snow on the sled + applied force= 0 newtons

    i solved for the friction forces:

    friction of sled on penguin= 67.3N * .791

    friction of snow on sled= (52N + 67.3N) * .097

    then i pluged those into my Fnet(horizontal) equation and solved for the applied force. i got 64.8064N as my final answer. when i entered this answer into the computer program that i use it said it was wrong. so i guess i am missing a step??
  2. jcsd
  3. Mar 25, 2010 #2
    i got the same as you.
  4. Mar 25, 2010 #3
    did i use the correct method though?

    thanks for the replying btw.
  5. Mar 25, 2010 #4


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    Gold Member

    The sled and penguin are accelerating under the applied force. You can't sum forces equal to 0. Think of it this way: if the penquin was not accelerating (with respect to the ground), there would be no friction force acting on it. You'll need two separate force diagrams (free body diagrams).
  6. Mar 25, 2010 #5
    ok, so if the sled and penguin are pulled with a force equal to the frictional force of the sled and penguin on the ice, the whole sled moves at a constant velocity, as forces are balanced, there is no net force, and so the penguin has no frictional force acting on it.

    but then if the pulling force is increased there would be a net force. and so an acceleration of the sled, as hence a frictional force on the penguin. and when this net force was equal to the frictional force of the penguin, it would move?
  7. Mar 27, 2010 #6


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    I don't think you said that quite corrrectly. When the net horizontal force on the penguin just exceeds the maximum available static friction force on the penquin, it will move (slip) relative to the sled (it is moving and accelerating with respect to the ground before that, but not moving or accelerating with respect to the sled before that). Now you must determine what the maximum horizontal applied force on the sled can be before the penguin starts to slip.
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