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Systems of Particles and Rotational Motion

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Question : 9 of 33
Marks: +1, -0
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:  
M = 1800 kg
Let R1R_1 and R2R_2 be the reactions on front and back wheels.
∴R1+R2=Mg=1800×9.8\therefore R_1 + R_2 = Mg = 1800 \times 9.8
Taking moment about C.G.,
R1×1.05=R2×0.75R_1 \times 1.05 = R_2 \times 0.75 or R1=57R2R_1 = \frac{5}{7} R_2
∴57R2+R2=1800×9.8\therefore \frac{5}{7} R_2 + R_2 = 1800 \times 9.8 or R2=10290 NR_2 = 10290\,\mathrm{N}
So force on each back wheel =102902=5145 N= \frac{10290}{2} = 5145\,\mathrm{N}
∴ Force on each front wheel =73502=3675 N= \frac{7350}{2} = 3675\,\mathrm{N}
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