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?

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-05-10T05:38:30+02:00

Answer:

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

Explanation:

From the question we are told that

The center of gravity of the plane to its nose is [tex]z = 2.58 m[/tex]

The distance of the front wheel of the plane to its nose is [tex]l = 0.800\ m[/tex]

The distance of the main wheel of the plane to its nose is [tex]e = 3.02 \ m[/tex]

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

[tex]mg (z- l) - G_B *(e - l) = 0[/tex]

Where m is the mass of the airplane

[tex]G_B[/tex] is the weight of the airplane supported by the main wheel

So

[tex]G_B =\frac{mg (z-l)}{(e - l)}[/tex]

Substituting values

[tex]G_B =\frac{mg (2.58 -0.8 )}{(3.02 - 0.80)}[/tex]

[tex]G_B = 0.8018 mg[/tex]

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

[tex]G_F = mg - G_B[/tex]

Substituting values

[tex]G_F = mg - 0.8018mg[/tex]

[tex]G_F = (1 - 0.8018) mg[/tex]

[tex]G_F = 0.1982 mg[/tex]

Now the weight of the airplane is = mg

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