A car weighs 1800 , kg. The distance between | vedclass

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(A) Mass of the car,$m = 1800 \, kg$.Distance between the front and back axles,$d = 1.8 \, m$.Distance of the centre of gravity $(C.G.)$ from the fron

Vedclass · Accepted Answer

$5145$

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(A) Mass of the car,$m = 1800 \, kg$.Distance between the front and back axles,$d = 1.8 \, m$.Distance of the centre of gravity $(C.G.)$ from the front axle $= 1.05 \, m$.Distance of the $C.G.$ from the back axle $= 1.8 - 1.05 = 0.75 \, m$.Let $R_f$ be the total force exerted by the ground on the two front wheels and $R_b$ be the total force exerted by the ground on the two back wheels.For translational equilibrium:$R_f + R_b = mg = 1800 	imes 9.8 = 17640 \, N \dots (i)$For rotational equilibrium,taking the torque about the $C.G.$:$R_f 	imes 1.05 = R_b 	imes 0.75$$R_f = R_b 	imes \frac{0.75}{1.05} = R_b 	imes \frac{5}{7} \dots (ii)$Substituting $(ii)$ into $(i)$:$R_b 	imes \frac{5}{7} + R_b = 17640$$R_b 	imes \frac{12}{7} = 17640$$R_b = \frac{17640 	imes 7}{12} = 10290 \, N$.This is the total force on the two back wheels.Therefore,the force exerted on each back wheel $= \frac{10290}{2} = 5145 \, N$.

$A$ 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. The force exerted by the level ground on each back wheel is ....... $N$.

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