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

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  $z = 2.58 m$

    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  

             $G_B =\frac{mg (z-l)}{(e - l)}$

Substituting values

            $G_B =\frac{mg (2.58 -0.8 )}{(3.02 - 0.80)}$

           $G_B = 0.8018 mg$

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 %