91Ó°ÊÓ

$$\begin{gathered} \mathrm{The}\mathrm{distance}\mathrm{below}\mathrm{the}\mathrm{surface}\mathrm{of}\mathrm{the}\mathrm{ocean}\mathrm{is}h=30\mathrm{m}. \\ \mathrm{The}\mathrm{area}\mathrm{is}A=0.75{\mathrm{m}}^2 \\ \mathrm{The}\mathrm{weight}\mathrm{is}W=300\mathrm{N}. \\ \mathrm{The}\mathrm{pressure}\mathrm{inside}\mathrm{is}P=1\mathrm{atm}. \end{gathered}$$

In this problem, the total force on the door hatch can be calculated by subtracting the net force required to open the hatch with weight of the hatch.

The relation of gauge pressure can be written as:

${\mathrm{P}}_{\mathrm{g}}=\mathrm{pgh}$

Here, $p$is the density of ocean water and g is the gravitational acceleration.

localid="1657176177192" $$\begin{gathered} \mathrm{Substitute}30\mathrm{m}\mathrm{for}h,1030\mathrm{kg}/{\mathrm{m}}^3\mathrm{for}\mathrm{ÒÏ},\mathrm{and}9.80\mathrm{m}/{\mathrm{s}}^2\mathrm{for}g\mathrm{in}\mathrm{the}\mathrm{above} \\ \mathrm{relation}. \\ {P}_g=\left(1030\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(30\mathrm{m}\right) \\ {P}_g=3.02×{10}^5\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

The relation of net force on the hatch can be written as:

localid="1657176194143" $F=P_{g}A$

localid="1657176241037" $$\begin{gathered} \mathrm{Substitute}3.02×{10}^5\mathrm{N}/{\mathrm{m}}^2\mathrm{for}{p}_g,\mathrm{and}0.75{\mathrm{m}}^2\mathrm{for}A\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{relation}. \\ F=\left(3.02×{10}^5\mathrm{N}/{\mathrm{m}}^2\right)\left(0.75{\mathrm{m}}^2\right) \\ F=2.26×{10}^5\mathrm{N} \end{gathered}$$

The relation to calculate the downward force exerted by the crew can be written as:

$F_d=F-W$

Substitute $2.26×{10}^5\mathrm{N}\mathrm{for}F,\mathrm{and}300\mathrm{N}\mathrm{for}W$in the above relation.

$$\begin{gathered} F_d=\left(2.26×{10}^5\mathrm{N}\right)-\left(300\mathrm{N}\right) \\ {F}_d=2.25×{10}^5\mathrm{N} \end{gathered}$$

Thus, the required downward force is$2.25×{10}^5\mathrm{N}.$