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

Answer:

(a). The normal force exerted by the ground on the front wheel is $4.67\times10^{5}\ N$.

(b). The normal force exerted by the ground on each of the two rear wheels is $7.02\times10^{5}\ N$

Explanation:

Given that,

Weight of jet $W=1.87\times10^{6}\ N$

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

$-(1.87\times10^{6})\times(16.0-12.0)+F_{f}\times16.0=0$

$F_{f}=\dfrac{(1.87\times10^{6})\times(16.0-12.0)}{16.0}$

$F_{f}=4.67\times10^{5}\ N$

(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

$\sum F_{y}=0$

$F_{f}+2F_{r}-W=0$

$F_{r}=\dfrac{F_{f}-W}{2}$

Put the value into the formula

$F_{r}=\dfrac{-4.67\times10^{5}+1.87\times10^{6}}{2}$

$F_{r}=7.02\times10^{5}\ N$

Hence, (a). The normal force exerted by the ground on the front wheel is $4.67\times10^{5}\ N$.

(b). The normal force exerted by the ground on each of the two rear wheels is $7.02\times10^{5}\ N$