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Inclined force problem

  • Thread starter ace123
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  • #1
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Hi it's been a while since I posted here so I will avoid the latex on this question and hope you understand.

The coefficient of kinetic friction for a 22kg bobsled on a track is .10. What force is required to push it down a 6.0 degree incilne and achieve a speed of 60km/h at the end of 75m?

So this is what I did:

I found the acceleration of the sled to be 1.85 m/s^2
Then I found the normal force= mg times cos 6= 214N
I then found the force of friction to be 21N
I then set Net Force equal to ma= 22 times 1.85 which equaled 40.7
Then I set 40.7 equal to forcex minus force of friction
and solved for forcex

Can anyone point out what I did wrong?
Thanks
 

Answers and Replies

  • #2
Doc Al
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What forces acting on the sled parallel to the track?
 
  • #3
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You mean like force of friction and Sin (6) times mg which is the horizontal component of Force of gravity
 
Last edited:
  • #4
Doc Al
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Sin (6) times mg
That's the one you missed. (Three forces act, including the applied force which you are trying to find.)

So set up an equation for the net force.

You mean like force of friction and Sin (6) times mg which is the horizontal component of Force of gravity
Yes.
 
  • #5
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O I was confused by that. So wouldn't then net force just equal the applied force?
 
  • #6
Doc Al
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So wouldn't then net force just equal the applied force?
Why would you say that? The net force is the sum of all the forces acting on the sled--the applied force is just one of those forces. (Pay attention to direction--sign--when you add the force components.)
 
  • #7
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Well I figured the ohter forces cancel out since the object was at rest in the beginning and needed an applied force to move.
 
  • #8
Doc Al
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No reason to assume so. Especially when you can calculate the forces and know for sure.
 
  • #9
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Yea I forgot it could be moving but just not accelerating. Thanks for your help.
 
  • #10
Doc Al
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We must assume (lacking information to the contrary) that it starts from rest. But that doesn't mean it was just sitting there waiting to be pushed.
 
  • #11
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I'm solving the same problem. I don't understand how to find the friction force. Can someone explain in this particular problem?
 
  • #12
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There is a formula for finding the force of friction. It's the coefficient of friction multiplied by the normal force. If you can find the normal force then I don't see your problem.
 

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