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 18.4 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 the front wheel.
Determine the normal force exerted by the ground on each of the two rear wheels.
To solve this problem, we need to calculate the normal forces exerted by the ground on the front wheel and each of the two rear wheels of the cargo plane.
First, let's find the total torque provided by the rear landing gears. The torque is calculated by multiplying the weight by the distance from the center of gravity to the rear landing gears. In this case, the weight of the cargo plane is 1.00 * 10^6 N, and the distance from the center of gravity to the rear landing gears is 12.0 m.
Torque provided by rear landing gears = weight * distance
Torque = 1.00 * 10^6 N * 12.0 m = 1.20 * 10^7 N·m
Now, let's determine the torque that needs to be countered by the front landing gear to maintain equilibrium. The total torque provided by the rear landing gears is counteracted by the torque due to the weight of the cargo plane and the torque due to the normal force acting on the front wheel.
Torque due to weight = weight * distance from center of gravity to front wheel
Torque due to normal force on the front wheel = normal force on front wheel * distance from center of gravity to front wheel
Since the plane is at rest on the runway, the total torque must be zero. Therefore, we can set up an equation:
Torque due to weight + Torque due to normal force on front wheel = Torque provided by the rear landing gears
(weight * distance from center of gravity to front wheel) + (normal force on front wheel * distance from center of gravity to front wheel) = 1.20 * 10^7 N·m
Substituting the given values:
(1.00 * 10^6 N * 12.0 m) + (normal force on front wheel * 12.0 m) = 1.20 * 10^7 N·m
Now, we can solve for the normal force on the front wheel.
normal force on front wheel = (1.20 * 10^7 N·m - (1.00 * 10^6 N * 12.0 m)) / 12.0 m
normal force on front wheel = (1.20 * 10^7 N·m - 1.20 * 10^7 N·m) / 12.0 m
normal force on front wheel = 0 N
The normal force exerted by the ground on the front wheel is 0 N.
For the two rear wheels, since the torque provided by the rear landing gears must be countered by the torque due to the weight of the cargo plane and the torque due to the normal forces on the rear wheels, we can set up the equation:
Torque due to weight + Torque due to normal forces on rear wheels = Torque provided by the rear landing gears
(weight * distance from center of gravity to rear wheels) + (normal force on each rear wheel * distance from center of gravity to rear wheels) = 1.20 * 10^7 N·m
Substituting the given values:
(1.00 * 10^6 N * 18.4 m) + (normal force on each rear wheel * 18.4 m) = 1.20 * 10^7 N·m
Now, we can solve for the normal force on each of the two rear wheels.
normal force on each rear wheel = (1.20 * 10^7 N·m - (1.00 * 10^6 N * 18.4 m)) / 18.4 m
normal force on each rear wheel = (1.20 * 10^7 N·m - 1.84 * 10^7 N·m) / 18.4 m
normal force on each rear wheel = -0.565 * 10^6 N
The normal force exerted by the ground on each of the two rear wheels is approximately -0.565 * 10^6 N.
Note: It's important to interpret the negative sign in front of the normal force on the rear wheels. It indicates that the direction of the force is opposite to the assumed direction. In this case, it means that the ground is exerting an upward force on the rear wheels instead of a downward force.