Question

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BRAINLYYY Intell's

$\sf\huge{\underline{\mathfrak{Question!!}}}$

A car weighs 1800kg. The distance between its front and back axle is 1.8m . It's centre of gravity is 1.05m behind the front axle . Determine the force exerted by the level ground on each front wheel and each back wheel!!..

##Solve it
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Answer 1

Here is your answer mate...✍✍

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

$\huge\blue{Answer}$

Mass of car m=1800kg

Distance between front and back axles d=1.8m 

Distance between center of gravity and front axle=1.05m

Let Rb and Rf be the forces exerted by the level ground on the back and front wheels respectively.

At translational equilibrium:

Rf+Rb=mg=17640 N.....(i)

For rotational equilibrium, on taking the torque about the C.G., we have:

Rf(1.05)=Rb(1.8−1.05)

⟹Rb=1.4Rf.......(ii)

Solving (i) and (ii) gives

Rf=7350 N

Rb=10290 N

The force exerted on each front wheel =7350/2=3675 N

The force exerted on each back wheel =10290/2=5145N