Question

Crew members attempt to escape from a damaged submarine 100 mbelow the surface. What force must be applied to a pop-out hatch, which is 1.2 mby, 0.60 mto push it out at that depth? Assume that the density of the ocean water is 1024 kg /m3and the internal air pressure is at.

Short answer

The force which must be applied to a pop-out hatch to push it out at that depth is$7.2\times {10}^{5N}$

Step-by-step solution

The given data

1. Depth of the surface, d = 100m
2. Density of the ocean water, $\mathrm{p}=1024kg/m^3$
3. Dimension of the pop-out hatch are$1.2m\times 0.60m$
4. Internal pressure,$P_i=1.00atm$

Understanding the concept of pressure

Using the formula of pressure, we can find the pressure at the given depth. Pressure is force per unit area. Using this, we can find the force required that must be applied to open the hatch.

Formulae:

Pressure applied on a fluid surface,$P_{gauge}=\mathrm{p}gd$ (i)

Pressure applied by force F on area A,

$p=\frac{F}{A}$ (ii)

Calculation of the force to be applied

Comparing equations (i) & (ii), we get

$$\begin{gathered} P_{gauge}=\frac{F}{A} \\ F=pgd\times A \\ =\left(1024\frac{kg}{m^3}\right)\left(9.8\frac{m}{s^3}\right)\left(100m\right)(1.2\times 0.60m)(\because p=1024\frac{kg}{m^3}\&A=1.2m\times 0.60m) \\ =7.2\times {10}^{5}N \end{gathered}$$

Therefore, the force which must be applied to a pop-out hatch, which is by to push it out at that depth is$7.2\times {10}^{5}N$