Question

A car weighs $1800kg$. The distance between its front and back axles is $1.8m$. Its centre of gravity is $1m$ behind the front axle. The force exerted by the level ground on each front wheel and each back wheel is (Take $g=10m{s}^{-2})$

• A. $4000N$ on each front wheel, $5000N$ on each back wheel
• B. $5000N$ on each front wheel, $4000N$ an each back wheel
• C. $4500N$ on each front wheel, $4500N$ on each back wheel
• D. $3000N$ on each front wheel, $6000N$ on each back wheel

Answer 1

Answer: (A)

Here, mass of the car, $M=1800kg$
Distance between front and back axles $=1.8m$
Distance of gravity G behind the front axle $=1m$
Let $R_F$ and $R_B$ be the forces exerted by the level ground on each front wheel and each back wheel.

For translational equilibrium,
$2R_F+2R_B=Mg$
or $R_F+R_B=\frac{Mg}{2}=\frac{1800\times 10}{2}$
$=9000N...(i)$
(As there are two front wheels and two back wheels) For rotational equilibrium about $G$
$(2R_F)(1)=(2R_B)(0.8)$
$\frac{R_F}{R_B}=0.8$
$=\frac{8}{10}=\frac{4}{5}$
$\Rightarrow R_F=\frac{4}{5}{R}_B...(ii)$
Substituting this in Eq.$(i)$, we get
$\frac{4}{5}{R}_B=9000$ or $\frac{9}{5}{R}_B=9000$
$R_B=\frac{9000\times 5}{9}=5000N$
$\therefore R_F=\frac{4}{5}{R}_B$
$=\frac{4}{5}\times 5000N$
$=4000N$