91Ó°ÊÓ

Apply Newton’s second law of motion.

F_h = ma ............(1)

Here,F_his the horizontal component of force, m is the mass of the passenger, and a is the acceleration.

• Acceleration = 49.0 m/s²
• Mass of the austronaut = 75.0 kg.

(a)

Substitute 75 kg for m and 49.0 m/s² for a in the equation (i), and we get,

$\begin{array}{l}{F}_h=75\mathrm{kg}×49m/s^2\\ =3675\mathrm{kg}m/s^2\\ =3675N\end{array}$

Hence, the horizontal component of the force is 3675 N.

Calculate the weight of the passenger as:

W = mg

Here, W is the weight of the passenger, and g is the acceleration due to gravity.

Substitute 75 kg for m and 9.8 m/s² for g in the above expression, and we get,

$\begin{array}{ll}W=75\mathrm{kg}×9.8m/s^2& \\ =735kg.m/s^2& \\ =735N& \end{array}$

Determine the ratio of the horizontal component of a force and the weight of the passenger as,

$$\begin{gathered} \frac{F_h}{W}=\frac{3675N}{735N} \\ =5 \end{gathered}$$

Hence, the horizontal component of force is five times greater than the weight of the passenger.

(b)

Draw the free-body diagram of the forces acting as shown below.

Here, N is the normal reaction, $\mathrm{θ}$is the direction of the net force with the horizontal, and F_net is the net force.

The normal force is equal to the weight.

Calculate the net force as:

$F_{net}=\sqrt{\left(F_h^2+N^2\right)}$

Substitute 3675Nfor F_h and 735N for N in the above expression, and we get,

$$\begin{gathered} F_{net}=\sqrt{\left({3675}^2+{735}^2\right)N} \\ {F}_{net}=\sqrt{\left(13505625+540225\right)N} \\ {F}_{net}=\sqrt{\left(14045850\right)N} \\ {F}_{net}=3747.8N \end{gathered}$$

Hence, the magnitude of the net force exerted by the seat is 3747.8N.

Determine the direction of the net force as,

$\tan \mathrm{θ}=\frac{N}{F_h}$

Substitute 3675N for F_h and 735N for N in the above expression, and we get,

$\begin{array}{l}{\tan }^{\mathrm{θ}}=\frac{735N}{3775N}\\ \mathrm{θ}={\tan }^{-1}\left(02\right)\\ \mathrm{θ}=11.{31}^{∘}\end{array}$

Hence, the direction of the net force with the horizontal is $11.{31}^{∘}$.