Question

A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wheel (nose wheel) is 0.800 m behind the nose, and the main wheels are 3.02 m behind the nose. What percentage of the airplane's weight is supported by the nose wheel?

Answer 1

The percentage of the weight supported by the front wheel is A= 19.82 %

Step-by-step explanation:

From the question we are told that

The center of gravity of the plane to its nose is
`z = 2.58 m`

The distance of the front wheel of the plane to its nose is
`l = 0.800\ m`

The distance of the main wheel of the plane to its nose is
`e = 3.02 \ m`

At equilibrium the Torque about the nose of the airplane is mathematically represented as

`mg (z- l) - G_B *(e - l) = 0`

Where m is the mass of the airplane

`G_B` is the weight of the airplane supported by the main wheel

So

`G_B =(mg (z-l))/((e - l))`

Substituting values

`G_B =(mg (2.58 -0.8 ))/((3.02 - 0.80))`

`G_B = 0.8018 mg`

Now the weight supported at the frontal wheel is mathematically evaluated as

`G_F = mg - G_B`

Substituting values

`G_F = mg - 0.8018mg`

`G_F = (1 - 0.8018) mg`

`G_F = 0.1982 mg`

Now the weight of the airplane is = mg

Thus percentage of this weight supported by the front wheel is
`A = 0. 1982 *100 =` 19.82 %