System of particles and Rotational motion NCERT Solutions Class 11 Physics - Solved Exercise Question 7.9

Question 7.9:

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Solution:
Mass of the car, m = 1800 kg

Distance between the front and back axles, d = 1.8 m
Distance between the C.G. (centre of gravity) and the back axle = 1.05 m
The various forces acting on the car are shown in the following figure.

R_f and R_bare the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium:
R_f + R_b = mg
= 1800 X 9.8
= 17640 N   ....(i)
For rotational equilibrium, on taking the torque about the C.G., we have:
R_f(1.05) = R_b(1.8 - 1.05)
R_b / R_f  =  7 / 5
R_b = 1.4 R_f    ......(ii)
Solving equations (i) and (ii), we get:
1.4R_f + R_f = 17640
R_f = 7350 N
∴ R_b = 17640 – 7350 = 10290 N
Therefore, the force exerted on each front wheel = 7350 / 2  =  3675 N, and
The force exerted on each back wheel = 10290 / 2  =  5145 N