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-wheels 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-wheels 5000 N on each back wheel

Answer 2

Here, mass of the car, M = 1800 kg

Distance between front and back axles = $1.8m$

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.

$2R_{F} + 2R_{B} = Mg$

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

$\frac{4}{5}R_{B} + R_{B} = 9000or\frac{9}{5}R_{B} = 9000$

$R_{B} = \frac{9000 \times 5}{9} = 5000N\therefore R_{F} = \frac{4}{5}R_{B} = \frac{4}{5} \times 5000N = 4000N$)