6.System of Particles and Rotational Motion
hard

A car wetghs $1800\; kg$. The distance between its front and back axles is $1.8\; m$. Its centre of gravity is $1.05\; m$ behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Option A
Option B
Option C
Option D

Solution

Mass of the car, $m=1800 kg$

Distance between the front and back axles, $d=1.8 m$

Distance between the $C.G.$ (centre of gravity) and the back axle $=1.05 m$

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

$R_{ f }(1.05)=R_{ b }(1.8-1.05)$

$R_{ f } \times 1.05=R_{ b } \times 0.75$

$\frac{R_{ f }}{R_{ b }}=\frac{0.75}{1.05}=\frac{5}{7}$

$\frac{R_{ b }}{R_{ f }}=\frac{7}{5}$

$R_{ b }=1.4 R_{ f } \ldots(i i)$

Solving equations ($i$) and ($i i$), we get:

$1.4 R_{t}+R_{ f }=17640$

$R_{ f }=\frac{17640}{2.4}=7350 N$

$\therefore R_{ b }=17640-7350=10290 N$

Therefore, the force exerted on each front wheel $=\frac{7350}{2}=3675 N$

The force exerted on each back wheel $=\frac{10290}{2}=5145 N$

Standard 11
Physics

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