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प्रश्न
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.
उत्तर १
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.
Rf and Rb are the forces exerted by the level ground on the front and back wheels respectively.
At translational equilibrium:
`R_f + R_b = mg`
= 1800 × 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_f xx 1.05 = R_b xx 0.75`
`R_f/R_b = 0.75/1.05 = 5/7`
`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 = 17640/2.4 = 7350 N`
∴Rb = 17640 – 7350 = 10290 N
Therefore, the force exerted on each front wheel = `7350/2 = 3675 N` and
The force excerted on each back wheel = `10290/2 = 5145 N`
उत्तर २
Let F1 and F2 be the forces exerted by the level ground on front wheels and back wheels respectively.
Considering rotational equilibrium about the front wheels,F2 x 1.8 = mg x 1.05 or F2 = 1.05/1.8 x 1800 x 9.8 N =10290 N
Force on each back wheel is =10290/2 N or 5145 N.
Considering rotational equilibrium about the back wheels.
F1 x 1.8 = mg (1.8 – 1.05) = 0.75 x 1800 x 9.8 or F1=0.75 x 1800 x 9.8/1.8 = 7350 N
Force on each front wheel is 7350/2 N or 3675 N.
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