What is the formula for finding the force of friction on an inclined surface?

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Homework Help Overview

The discussion revolves around calculating the force required to push a bobsled down an inclined surface, specifically focusing on the forces acting on the sled, including friction and gravitational components. The subject area includes dynamics and friction on inclined planes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the forces acting on the sled, including friction and gravitational components. Questions arise regarding the net force and the role of the applied force in the context of the sled's motion.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about the forces involved and clarifying the relationships between them. Some guidance has been offered regarding the calculation of the friction force and the components of gravitational force acting on the sled.

Contextual Notes

There is a noted lack of information regarding the initial conditions of the sled, with some participants assuming it starts from rest while others suggest it may have been in motion without acceleration.

ace123
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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
 
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What forces acting on the sled parallel to the track?
 
You mean like force of friction and Sin (6) times mg which is the horizontal component of Force of gravity
 
Last edited:
ace123 said:
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.

ace123 said:
You mean like force of friction and Sin (6) times mg which is the horizontal component of Force of gravity
Yes.
 
O I was confused by that. So wouldn't then net force just equal the applied force?
 
ace123 said:
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.)
 
Well I figured the ohter forces cancel out since the object was at rest in the beginning and needed an applied force to move.
 
No reason to assume so. Especially when you can calculate the forces and know for sure.
 
Yea I forgot it could be moving but just not accelerating. Thanks for your help.
 
  • #10
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
I'm solving the same problem. I don't understand how to find the friction force. Can someone explain in this particular problem?
 
  • #12
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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