Jet Transport Weight Distribution and Normal Force Calculation

[PianoMan]

A jet transport has a weight of 1.00x10^6 N and is at rest on the runway. The two rear wheels are 15m behind the front wheel, and the plane's center of gravity is 12.6m behind the front wheel. Determine the normal force exerted by the ground on the front whell and each of the two rear wheels.



Ok, so I drew a sketch. I've got front wheel, rear wheels, in between the two i have the center of gravity, pushing down with 1x10^6 N. Behind are the rear wheels pushing up with an FN and in front of it the front wheel is pushing up with an FN. There's 15 m of distance between them. The distance from the center of gravity to the front wheel is 12.6 m, to the back wheels it is 2.4 m. But now, I don't know what to do. I can't find Torque (r * F) because even though I have distances from the center of gravity (r) I don't know an F. That's what I'm trying to find...which I can't find without Torque. lol. Any help?

 

[jamesrc]

First, let's call the normal force on the front wheel F_f and the normal force on one of the rear wheels F_r (assume each rear wheel has the same normal force exerted on it, but not equal to the normal force of the front wheel). All you have to do is sum the moments about the front wheel to find the force on each of the rear wheels.

0 = 12.6*W - 2*F_r*15

(where W is the weight of the transport)

If you need to find F_f, you could check the sum of the forces or do a moment balance about the rear wheel, e.g. 0 = F_f + 2*F_r - W

 

[M98Ranger]

First, we need to calculate the total weight distribution of the jet transport. We know that the weight of the jet is 1.00x10^6 N, so we can divide this by the number of wheels to find the weight distribution per wheel. Since there are three wheels, each wheel will have a weight distribution of 1.00x10^6 N / 3 = 3.33x10^5 N.

Next, we can use the concept of moments to calculate the normal force exerted by the ground on each wheel. The moment is the product of the force and the distance from the point of rotation. In this case, the point of rotation is the front wheel, so we can calculate the moment for each wheel using the following formula:

Moment = Distance from front wheel * Weight distribution per wheel

For the front wheel, the moment is:

12.6m * 3.33x10^5 N = 4.19x10^6 Nm

For the rear wheels, the moment is:

2.4m * 3.33x10^5 N = 7.99x10^5 Nm

Now, we can use the concept of equilibrium to find the normal force for each wheel. Since the jet transport is at rest, the sum of all the forces acting on it must be equal to zero. This means that the normal forces exerted by the ground on each wheel must balance out the weight of the jet.

For the front wheel, the equation is:

FN - 3.33x10^5 N = 0

Solving for FN, we get:

FN = 3.33x10^5 N

For the rear wheels, the equation is:

2FN - 7.99x10^5 N = 0

Solving for FN, we get:

FN = 3.99x10^5 N

Therefore, the normal force exerted by the ground on the front wheel is 3.33x10^5 N, and the normal force exerted by the ground on each of the rear wheels is 3.99x10^5 N. This ensures that the jet transport remains at rest on the runway and the weight is evenly distributed among all three wheels.