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.
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.
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 $=\frac{10290}{2}=5145 N$
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