A car weighs $1800kg$. The distance between its front and back axles is $1.8m$. Its centre of gravity is $1.05m$ behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
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Mass of the car, m = 1800kg Distance between the front and back axles, d = 1.8m Distance between the C.G. (centre of gravity) and the back axle = 1.05m 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: $R_f+R_b=m g=1800 \times 9.8=17640 N \ldots$...(i) For rotational equilibrium, on taking the torque about the C.G., We have, $R_f(1.05)=R_b(1.8-1.05) \frac{R_b}{R_f}=\frac{7}{5} R_b=1.4 R_f \ldots$...(ii) Solving equations (i) and (ii), we get $1.4 R_f+R_f=$ $17640 R_f=7350 \mathrm{~N} \therefore R_b=17640-7350=10290 \mathrm{NT}$
Therefore, the force exerted on each front wheel $=\frac{7350}{2}=3675 \mathrm{~N}$ and, The force exerted on each back wheel $=\frac{10290}{2}=5145 \mathrm{~N}$
$R_f$ and $R_b$ are the forces exerted by the level ground on the front and back wheels respectively. At translational equilibrium: $R_f+R_b=m g=1800 \times 9.8=17640 N \ldots$...(i) For rotational equilibrium, on taking the torque about the C.G., We have, $R_f(1.05)=R_b(1.8-1.05) \frac{R_b}{R_f}=\frac{7}{5} R_b=1.4 R_f \ldots$...(ii) Solving equations (i) and (ii), we get $1.4 R_f+R_f=$ $17640 R_f=7350 \mathrm{~N} \therefore R_b=17640-7350=10290 \mathrm{NT}$
Therefore, the force exerted on each front wheel $=\frac{7350}{2}=3675 \mathrm{~N}$ and, The force exerted on each back wheel $=\frac{10290}{2}=5145 \mathrm{~N}$
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