Ncert Exercises & Solutions

Question 7. 9. A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its center 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.

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_b are the forces exerted by the level ground on the front and back wheels
respectively.
At translational equilibrium:

R_{f} + R_{b} = mg

= 1800 \times 9.8

= 17640 N \dots (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_{f} \times 1.05 = R_{b} \times 0.75

\frac{R_{f}}{R_{b}} = \frac{0.75}{1.05} = \frac{5}{7}

\frac{R_{b}}{R_{f}} = \frac{7}{5}

R_{b} = 1.4 R_{f} . . . . . . . . . . . (ii)

Solving equations (i) and (ii), we get:

1.4 R_{r} + R_{f} = 17640

R_{f} = \frac{17640}{2.4} = 7350 N

∴ R_{b} = 17640 - 7350 = 10290 N

Therefore, the force exerted on each front wheel

= \frac{7350}{2} = 3675 N

and
The force exerted on each back wheel

= \frac{10290}{2} = 5145 N

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