Question

A car weighs 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.

Answer 1

Weight of car = 1800 Kg
Distance of COG from front axle= 1.05 m
Distance of COG from back axle = 1.8 - 1.05 = 0.75 m
Vertical forces are balanced ,
So, R1 + R2 = mg { see figure }
R1 + R2 = 1800× 9.8 = 17640


We know ,
Angular momentum about centre of gravity is zero.
So,
R1 × 1.05 = R2 × 0.75
R1 = (5/7)R2
Use this in above equation, we get ,
R2 = 10290 N and R1 = 7350 N
Hence, force on each back wheel = R2/2 = 10290/2 = 5145 N

Force on each front wheel = 7350/2 = 3675 N

{ note :- there are two wheels in back and front axles . so, normal reaction on each wheel of back or front axle will be half of getting normal reactions }

Answer 2

Weight of car = 1800 Kg

Distance of COG from front axle= 1.05 m

Distance of COG from back axle = 1.8 - 1.05 = 0.75 m

Vertical forces are balanced ,

So, R1 + R2 = mg { see figure }

R1 + R2 = 1800× 9.8 = 17640

We know ,

Angular momentum about centre of gravity is zero.

So,

R1 × 1.05 = R2 × 0.75

R1 = (5/7)R2

Use this in above equation, we get ,

R2 = 10290 N and R1 = 7350 N

Hence, force on each back wheel = R2/2 = 10290/2 = 5145 N

Force on each front wheel = 7350/2 = 3675 N