Question

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

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

Answer 1

Answer: (A)

$4000 \,N$ on each front wheel, $5000 \,N$ on each back wheel

Answer 2

Here, mass of the car, $M = 1800\, kg$ Distance between front and back axles $= 1.8\, m$ Distance of gravity G behind the front axle $= 1 \,m$ 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, $2 R_F + 2 R_B = Mg$ or $R_F +R_B = \frac{Mg}{2} = \frac{1800 \times 10}{2}$ $= 9000\,N\quad ...(i)$ (As there are two front wheels and two back wheels) For rotational equilibrium about $G$ $(2R_F )(1) = (2R_B)(0.8)$ $\frac{R_F}{R_B} = 0.8$ $= \frac{8}{10} = \frac {4}{5}$ $\Rightarrow R_F = \frac{4}{5}R_B \quad ...(ii)$ Substituting this in E$(i)$, we get $\frac{4}{5} R_B = 9000$ or $\frac{9}{5}R_B = 9000$ $R_B = \frac{9000\times 5}{9} = 5000\,N$ $\therefore R_F = \frac{4}{5}R_B$ $= \frac{4}{5}\times 5000\,N$ $= 4000\,N$