Answer
The crew must exert a downward force of $2.27\times 10^5~N$.
Work Step by Step
The external-internal pressure difference is equal to the gauge pressure at a depth of 30 meters. We can find the gauge pressure at this depth.
$P_G = \rho~g~h$
$P_G = (1030~kg/m^3)(9.80~m/s^2)(30~m)$
$P_G = 3.03\times 10^5~N/m^2$
We can find the total force required to open the door.
$F = P_G~A$
$F = (3.03\times 10^5~N/m^2)(0.75~m^2)$
$F = 2.27\times 10^5~N$
The total force on the door is the weight of the door plus the applied force. We can find the applied force $F_a$.
$F_a+mg = F$
$F_a = F - mg$
$F_a = (2.27\times 10^5~N)-(300~N)$
$F_a = 2.27\times 10^5~N$
The crew must exert a downward force of $2.27\times 10^5~N$.
