Question

A car is having weight W is moving in the direction as shown in the figure. The centre of gravity (CG) of the car is located at height h from the ground, midway between the front and rear wheels. The distance between the front and rear wheels, is l. The acceleration of the car is a, and acceleration due to gravity is g. The reactions on the front wheels $\left(R_f\right)$ and rear wheels $\left(R_r\right)$ are given by

• A. $R_f=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a;R_r=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a$
• B. $R_f=R_r=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a$
• C. $R_f=R_r=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a$
• D. $R_f=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a;R_r=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a$

Answer 1

The correct option is D

$R_f=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a;R_r=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a$

Pseudo force, $F_{\text{pseudo}}=$ma in opposite direction of motion of car.

$\sum F_v=0$

$R_r+R_f=W....\left(1\right)$

Taking moment about A

$\left(F_{\text{pseudo}}\right)h+\left(R_f\right)l=\frac{Wl}{2}$

$m_{a}h+R_{f}l=\frac{Wl}{2}$

$R_f=\frac{W}{2}-\frac{W}{g}\left(\frac{h}{l}\right)a......\left(2\right)$

from (1) and (2)

$R_f=\frac{W}{2}+\frac{W}{g}\left(\frac{h}{l}\right)a$