Chapter 4, Problem 79GP

To determine

The total force exerted by the cockpit seat on the pilot.

Answer to Problem 79GP

The total force exerted by the cockpit seat on the pilot is $939.14\text{N}$ .

Explanation of Solution

Given:

Acceleration of the jet aircraft, $a=3.5\text{m}/{\text{s}}^2$

Angle of the jet aircraft with horizontal, $\theta =45°$

Mass of the pilot, $m=75\text{kg}$

Formula Used:

Draw the free body diagram of the system.

Here, $F_x$ is the horizontal force component and $F_y$ is the vertical force component. The weight of the pilot is $mg$ .

The expression to calculate the horizontal component of the force is,

  $F_x=ma\cos \theta$

The expression to calculate the vertical component of the force is,

  $\begin{array}{l}{F}_y=mg+ma\sin \theta \end{array}$

  $\begin{array}{c}{F}_y=m\left(g+a\sin \theta \right)\end{array}$

The expression to calculate the total force exerted by the cockpit seat on the pilot is,

  $\begin{array}{c}F=\sqrt{F_x^2+F_y^2}\end{array}$

  $\begin{array}{c}F=\sqrt{{\left(ma\cos \theta \right)}^2+{\left(m\left(g+a\sin \theta \right)\right)}^2}\end{array}$

  $\begin{array}{c}F=m\sqrt{{\left(a\cos \theta \right)}^2+{\left(g+a\sin \theta \right)}^2}\end{array}$

Where,

  $F$ is the total force.

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}F=\left(75\text{kg}\right)\sqrt{{\left(\left(3.5\text{m}/{\text{s}}^2\right)\left(\cos 45°\right)\right)}^2+{\left(9.8\text{m}/{\text{s}}^2+\left(3.5\text{m}/{\text{s}}^2\right)\left(\sin 45°\right)\right)}^2}\end{array}$

  $\begin{array}{c}=939.14\text{N}\end{array}$

Conclusion:

Thus, the total force exerted by the cockpit seat on the pilot is $939.14\text{N}$ .