A 1050-kg car rounds a curve of radius 72 m banked at an angle of 14°. If the car is traveling at 85 km/h, will a friction force be required? If so, how much and in what direction.

The friction force is always required to continue the motion of the body. The friction force acts opposite to the direction of motion of the body.

The mass of car is \(m = 1050\;{\rm{kg}}\).

The radius of curve is \(r = 72\;{\rm{m}}\).

The angle of bank is \(\theta = 14^\circ \).

The speed of car is \(v = 85\;{\rm{km/h}}\).

The free-body diagram of car is shown below,

The friction force is required to continue the motion of the car, and the direction of friction is toward the center of rotation.

The net force on the car be written as,

\(\begin{aligned}mg\sin \theta + f &= {F_C}\\mg\sin \theta + f &= \frac{{m{v^2}}}{r}\\f &= \frac{{m{v^2}}}{r} - mg\sin \theta \end{aligned}\)

Here, f is the friction force.

Substitute the given values in above equation,

\(\begin{aligned}f = \frac{{\left( {1050\;{\rm{kg}}} \right){{\left( {85\;{\rm{km/h}} \times \frac{{\frac{5}{{18}}\;{\rm{m/s}}}}{{1\;{\rm{km/h}}}}} \right)}^2}}}{{72\;{\rm{m}}}} - \left( {1050\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\sin 14^\circ \\f = 5638\;{\rm{N}}\end{aligned}\)

Thus, the magnitude of friction force is 5638 N.