Question

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Consider the Boeing 787 "Dreamliner", with a mass of 177000 kg. In this particular model, the distance from the front wheels to the rear set of wheels is 21.7 m.
(a) If the center of mass of the airplane is along a line through the center and 3.00 m in front of the rear wheels, how much force, in meganewtons, does the ground exert on each set of rear wheels when the plane is at rest on the runway?
(b) How much force, in meganewtons, does the ground exert on the front set of wheels?

Answer 1

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

Center mass of the airplane

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

Some assumptions

• The wheels under the wind do not pass through the center line.
• The position of the front wheel is constant and it is zero mark (origin).
• The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

Mass of the plane at the position of the rear wheels

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

Force exerted by the ground on each rear wheel

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

Mass of the plane at the position of the front wheel

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

Force exerted by the ground on the front wheel

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822