Question

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
The acceleration of gravity is 9.8 .
What is the required minimum coefficient of static friction between the road and the car’s tires to keep the car from skidding?

Answer 1

Answer:

$\mu =0.75$

Explanation:

Frictional Force

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$

The normal force N is equal to the weight of the car, thus

$F_r=\mu .m.g$

Equating both forces

$\displaystyle \mu .m.g=m.\frac{v^2}{r}$

Simplifying

$\displaystyle \mu =\frac{v^2}{rg}$

Substituting the values

$\displaystyle \mu =\frac{19^2}{(49)(9.8)}$

$\boxed{\mu =0.75}$

Answer 2

Minimum coefficient of static friction is 0.75

Given that;

Velocity of car = 19 m/s

Radius of curvature = 49 meter

Acceleration of gravity = 9.8 m/s²

Find:

Minimum coefficient of static friction

Computation:

Minimum coefficient of static friction = V² / rg

Minimum coefficient of static friction = 19² / [(49)(9.8)]

Minimum coefficient of static friction = 361 / [480.2]

Minimum coefficient of static friction = 0.7517

Minimum coefficient of static friction = 0.75

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