Question

A jet transport has a weight of 1.87 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel, and the plane's center of gravity is 12.0 m behind the front wheel. Determine the normal force exerted by the ground on (a) the front wheel and on (b) each of the two rear wheels.

Answer 1

(a). The normal force exerted by the ground on the front wheel is
`4.67*10^(5)\ N`.

(b). The normal force exerted by the ground on each of the two rear wheels is
`7.02*10^(5)\ N`

Step-by-step explanation:

Given that,

Weight of jet
`W=1.87*10^(6)\ N`

Distance = 16 m

Second distance = 12.0 m

We need to calculate the normal force exerted by the ground on the front wheel

Using formula of torque

`\sum\tau=-Wl_(W)+F_(f)l_(f)=0`

Where, W = weight of jet

`l_(f)`=lever arms for the forces
`F_(f)`

`l_(w)`=lever arms for the forces W

Put the value into the formula

`-(1.87*10^(6))*(16.0-12.0)+F_(f)*16.0=0`

`F_(f)=((1.87*10^(6))*(16.0-12.0))/(16.0)`

`F_(f)=4.67*10^(5)\ N`

(b). We need to calculate the normal force exerted by the ground on each of the two rear wheels

The sum of vertical forces equal to zero.

`\sum F_(y)=F_(f)+2F_(r)-W=0`

We using 2 for two rear wheels

`\sum F_(y)=0`

`F_(f)+2F_(r)-W=0`

`F_(r)=(F_(f)-W)/(2)`

Put the value into the formula

`F_(r)=(-4.67*10^(5)+1.87*10^(6))/(2)`

`F_(r)=7.02*10^(5)\ N`

Hence, (a). The normal force exerted by the ground on the front wheel is
`4.67*10^(5)\ N`.

(b). The normal force exerted by the ground on each of the two rear wheels is
`7.02*10^(5)\ N`