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Dynamics coefficient of friction

  • Thread starter polly06
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3
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My question is:
If a car is moving at 33m/s^-1 down a 20 degree slope, the friction between car tires and the road surface is 0.58, what is the shortest distance and time in which it can stop?

I assume that Fmax= u R is correct formula?
(where Fmax=force max, u=coeff of friction, and R= normal reaction)

I also assume that R = mgsin20
(where m=mass, g=9.8, sin 20 for angle of slope)

Any help would be much appreciated!
thanks
-Polly
 
Last edited by a moderator:

Doc Al

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polly06 said:
I assume that Fmax= u R is correct formula?
(where Fmax=force max, u=coeff of friction, and R= normal reaction)
The maximum static friction will be [itex]\mu R[/itex], but friction is not the only force acting on the car.

I also assume that R = mgsin20
(where m=mass, g=9.8, sin 20 for angle of slope)
Careful. The angle is usually measured from the horizontal (an 0 degree slope is perfectly horizontal) so check your trig.

To find the maximum acceleration of the car, find the net force acting parallel to the slope.
 
3
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...ok, thank you for your prompt reply Doc Al.
Where I am confused is:
Max static friction = uR
........so how can R be calculated when the mass is unknown?

........so I deduct that it should have been R= mgsin70

........so I now need to determine the net force acting parallel to the slope by deducting the friction forces from the the force produced by the car's mass and acceleration?

I have refered back to my study notes, and to online tutorials, but seem to be missing something here!
Part time studies by correspondance was never meant to be easy!
thanks
-Polly
 

Doc Al

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1,078
polly06 said:
Where I am confused is:
Max static friction = uR
........so how can R be calculated when the mass is unknown?
Just express it symbolically as you've been doing. You'll find that you don't need the mass to find the car's acceleration. (The mass will cancel in your final formula.)
........so I deduct that it should have been R= mgsin70
Right. The normal force must balance the component of the car's weight perpendicular to the slope.

........so I now need to determine the net force acting parallel to the slope by deducting the friction forces from the the force produced by the car's mass and acceleration?
There are two forces acting on the car parallel to the slope: The friction, as we've discussed, and a component of the car's weight. The sum of these two (one is negative, of course) gives the net force. Then use Newton's 2nd law to find the acceleration.
 

HallsofIvy

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polly06 said:
........so how can R be calculated when the mass is unknown?
It can't- but you don't need that, you only need the net acceleration. If you carry the mass as "m" in the formula you should see that it cancels out at the end.

........so I deduct that it should have been R= mgsin70
Yes, or mg cos(20).

........so I now need to determine the net force acting parallel to the slope by deducting the friction forces from the the force produced by the car's mass and acceleration?
Well, you don't want to determine the force "produced by the car's mass and acceleration", you want to determine the acceleration produced by the car's mass and force (force causes acceleration, acceleration does not cause force!).

Actually you are correct that the gravitational force is F= mg where g is the acceleration due to gravitational force. I think that's what you meant but I still wouldn't say that the acceleration "produced" the force. Notice that you will also want to find the component of that down the hill- a little more trig.
 
3
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..thanks for the replies.
I have booked in to see a phyics tutor this week.
I shall reply with my findings
thanks again
-polly
 

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