A car of weight W = 10.0 kN makes a turn on a track that is banked at an angle of  θ = 20....

SKETCH:

The sketch for the given problem is shown in the below figure

RESEARCH:

The tensional force on the string can be resolved into two components, and then the horizontal component of the tension provides the centripetal force on the ornament. The centripetal force depends on the mass of the ornament () and the square of the linear speed () and the radius of the circular path of the ornament (). Hence, the horizontal component of the tension is

…… (1)

Now, the vertical component of the tensional force is balanced by the weight () of the ornament.

…… (2)

The static frictional force acting between the car and the road is the product of the coefficient of the friction () and the weight of the car (). So

Here is the mass of the car and is the acceleration due to gravity.

Here the static frictional force provides the centripetal force acting on the car. Hence

…… (3)

SIMPLIFY:

Using equations (1) and (2), we get

…… (4)

From equations (3) and (4), we get the frictional force acting between the road and the car.

…… (5)

Given data:

The weight of the car is

The banking angle of the turn of the car,

The ornament angle with the normal,

CALCULATE:

Now, substituting the numerical values in the equation (3), we get the static frictional force between the car and the road.