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Coefficient of Friction, Centripetal Force problem help?
Hi folks I can't quite figure this problem out.
A car of mass 2000 kg is driving at 12 m/s. It turns the corner at an intersection. The radius of the path is 20 m. What does the coefficient of static friction have to be so that it will make the turn without sliding.
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I know that he Force centripetal is equal to mass * acceleration. Acceleration in this case would be V^2/r since it's a circle.
But I don't know how to use that to calculate the mu s.
1 Answer
- Steve4PhysicsLv 77 years agoFavorite Answer
The centripetal force = mv²/r
The weight is mg and the normal force (Fn) has the same magnitude as weight (assuming a horizontal surface).
The limiting frictional force (Fr) is Fr = μₛFn = μₛmg
The centripetal force is provided by friction. At the point of slipping the centripetal force will be equal to the limiting frictional force (maximum possible frictional force).
mv²/r = Fr
mv²/r = μₛmg
μₛ = v²/(gr)
Just stick your value into v²/(gr).
Note that m has cancelled - the mass makes no difference.
