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.

Asked by Pragya Singh | 1 year ago |  123

##### Solution :-

Given,

Mass of the car, m= 1800 kg

Distance between the two axles, d = 1.8 m

Distance of the C.G. (centre of gravity) from the front axle = 1.05 m

The free body diagram of the car can be drawn as :

Let Rf and Rbe are the forces exerted by the level ground on the front and back wheels, respectively.

At translational equilibrium:

Rf + Rb = mg

= 1800 × 9.8

= 14700 N . . . . . . . . . . . ( 1 )

For rotational equilibrium  about the C.G., we have:

R(1.05) = Rb  (1.8 – 1.05)

$$\dfrac{R_f}{R_b}=\dfrac{0.75}{1.05}=\dfrac{5}{7}$$

Rf = ($$\dfrac{5}{7}$$) Rb  . . . . . . . . . . . . ( 2 )

Using value of equation ( 2 ) in equation ( 1 ), we get:

($$\dfrac{5}{7}$$) R+  Rb= 14700

Rb = 10290 N

Rf=($$\dfrac{5}{7}$$) Rb  = ($$\dfrac{5}{7}$$)10290 = 7350 N

Answered by Abhisek | 1 year ago

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