A car weighs 1800,kg. The distance between it | vedclass

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(A) Let $N_1$ be the total normal force on the two front wheels and $N_2$ be the total normal force on the two back wheels.The total weight of the car

Vedclass · Accepted Answer

$3750$

Vedclass · Answer

(A) Let $N_1$ be the total normal force on the two front wheels and $N_2$ be the total normal force on the two back wheels.The total weight of the car is $W = mg = 1800 	imes 9.8 = 17640\,N$.Taking torque about the back axle (point $B$): $N_1 	imes 1.8 = W 	imes 0.75$$N_1 	imes 1.8 = 17640 	imes 0.75$$N_1 = \frac{13230}{1.8} = 7350\,N$.The force exerted by the ground on each front wheel is $\frac{N_1}{2} = \frac{7350}{2} = 3675\,N$.*Note: If $g = 10\,m/s^2$ is used,$W = 18000\,N$,$N_1 = \frac{18000 	imes 0.75}{1.8} = 7500\,N$,and force per wheel $= 3750\,N$. Given the options,$g = 10\,m/s^2$ is intended.*

$A$ 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. Force exerted by the level ground on each front wheel is ........ $N$.

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