91Ó°ÊÓ

Frictional force acts when there is relative motion between two bodies or the possibility of relative motion arises. Its direction is such that it tries to stop that relative motion.

The acceleration of the aircraft,$a=3.8\mathrm{m}/{\mathrm{s}}^2$

The angle above the horizontal at which the aircraft climbs,$\mathrm{θ}=18°$

the mass of the pilot,$m=75\mathrm{kg}$

Let the frictional force exerted by the seat on the pilot be f, the normal reaction force acting on the pilot be N, and the net force acting on the pilot be represented by F.

The FBD of the pilot on his seat is given below.

Here, mg is the gravitational pull of the earth on the pilot.

Applying Newton’s second law in the direction perpendicular to the incline seat, you get:

$N=mg\cos 18°…\left(i\right)$

(As there is no component of acceleration in this direction)

Applying Newton’s second law in the direction along the incline, you get:

$f-mg\sin 18°=ma$

From the above equation, you can write:

role="math" localid="1645684430653" $f=ma+mg\sin 18°…\left(ii\right)$

The magnitude of the net force applied by the seat will be:

$F=\sqrt{N^2+f^2}$

Substituting the values from (i) and (ii), you get:

$F=\sqrt{{\left(mg\cos 18°\right)}^2+{\left(ma+mg\sin 18°\right)}^2}$

Substituting the values in the above equation, you get:

$\begin{array}{c}F=\sqrt{{\left(75×9.8×\cos 18°\right)}^2+{\left(75×3.8+75×9.8×\sin 18°\right)}^2}\\ =859.26\mathrm{N}\end{array}$

Thus, the net force exerted by the seat on the pilot is 859.26 N.