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Friction problem of a sled

  • Thread starter sonic91
  • Start date
  • #1
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Homework Statement


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.



Homework Equations


Fnet=ma


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??
 

Answers and Replies

  • #2
21
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i got the same as you.
 
  • #3
9
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did i use the correct method though?

thanks for the replying btw.
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
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491
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).
 
  • #5
21
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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?
 
  • #6
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,146
491
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.
yes
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?
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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