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turtle777rescuer Let's try it..
In a car at a steady speed going in a circle what's the friction coefficient.
First the distance traveled don't matter .
What forces do we have going here..
We have gravity pulling the car down. Then we have the force of the road pushing up on the car..
That's what's called Normal Force or N.
And since the car is balanced up and down..that means that N and gravity cancel out. Which means
N=mg, with g being gravity acceleration...9.8 m/ssquared
N and gravity cancel out because the car is not going up or down.
Then there's the horizontal force that pulls the car to the center..
That force comes from the dude turning the steering wheel, which turns the wheel, which makes the tires push against the asphalt which in turns pushes back and brings the car inward.
That means that the only unbalanced force is the friction force pushing that car inwards . Yes??
I know that by some kind of law I can't remember...Ff = μN (this the equation for maximum friction. Before it starts skidding)
That's what we are looking for ...that little μ
Which is the Friction coefficient.
Where am I going to get that from ??
I have to find another equation that I can equal to that so I can get that N to the other side and keep μ.
I also know that at any point that Ff is also
Ff = MV2/R. ( The 2 means square)
Thats the friction force at any time..but wait!??
I know what that force is right at the skidding point because I have the equation for maximum friction force which is. Ff = μN
Therefore I can say that I want the coefficient just before it starts skidding..
So I make the two equations equal.
Friction force at the moment just before it skids=
μN = Mv2/R
Solving for μ = MV2/RN
We know that N. Normal force is mg from earlier
So my equation becomes..
μ = mv2/Rmg....crossing out m, that leaves...
μ = v 2 (square)/Rg
Where g is the acceleration of gravity 9.8 m/s(square)
I think