In a cargo plane, the center of gravity must be between the front and rear landing gears in order to maintain a torque equilibrium. The front landing gears help counteract the torque provided by the rear landing gears in this case. A jet transport has a weight of 1.00 ✕ 106 N and is at rest on the runway. The two rear wheels are 16.8 m behind the front wheel, and the plane's center of gravity is 10.8 m behind the front wheel.
(a) Determine the normal force exerted by the ground on the front wheel.
(b) Determine the normal force exerted by the ground on each of the two rear wheels.
To solve this problem, we need to use the concept of torques and equilibrium. The torque produced by a force about a point is the product of the force and the perpendicular distance between the point and the line of action of the force.
(a) To determine the normal force exerted by the ground on the front wheel, we need to consider the torque equilibrium. Since the aircraft is at rest, the sum of the torques about any point must be zero.
Let's choose the point of reference to be the rear landing gears. The torque produced by the weight of the aircraft about this point is given by:
Torque due to weight = Weight * Distance
= 1.00 * 10^6 N * (16.8 + 10.8) m
= 1.00 * 10^6 N * 27.6 m
In order to maintain torque equilibrium, the torque produced by the normal force on the front wheel must be equal in magnitude but opposite in direction to the torque produced by the weight. Since the front wheel is located at the pivot point, the lever arm for the normal force is zero. Therefore, the torque due to the normal force is zero.
Torque due to normal force on the front wheel = Normal force * Distance
= Normal force * 0 m
= 0
Setting the torques equal to zero, we can write the equation:
Torque due to weight = Torque due to normal force on the front wheel
1.00 * 10^6 N * 27.6 m = 0
Simplifying the equation, we find that the normal force on the front wheel is zero.
Hence, the normal force exerted by the ground on the front wheel is zero.
(b) To determine the normal force exerted by the ground on each of the two rear wheels, we again consider torque equilibrium.
Torque due to weight = Torque due to normal force on each rear wheel
Since the two rear wheels are identical and the weight of the aircraft acts at its center of gravity, the torque due to weight will be evenly divided between the two rear wheels.
Torque due to weight = 1/2 * Weight * Distance
= 1/2 * 1.00 * 10^6 N * 27.6 m
Setting the torques equal to each other, we get:
1/2 * 1.00 * 10^6 N * 27.6 m = Normal force * 16.8 m
Simplifying the equation, we find that the normal force exerted by the ground on each of the two rear wheels is:
Normal force = (1/2 * 1.00 * 10^6 N * 27.6 m) / 16.8 m
You can now calculate the numerical values to get the final answers for each part.
Please note that this explanation assumes an idealized scenario and does not take into account additional factors like friction, tire characteristics, or other real-world considerations that may affect the actual forces exerted on the wheels.