How Much Force Must a Diving Suit Window Withstand at 150m Depth?

[jimithing]

The viewing window in a diving suit has an area of 65 cm^2. If an attempt were made to maintain the pressure on the inside of the suit of 1 atm, what force would the window have to withstand if the diver descended to a depth of 150 m. Take the specific gravity of the water to be 1.05.

So far...

$$P_{A} = P_{0} + \rho gh$$ (1)

$$P = \frac{F}{A}$$ (2)

im assuming i find the pressure in (1) and just multiply it by the area to get the F in (2). Look right?

 

[Chi Meson]

You are on the right track, just remember that the pressure inside the suit (at 1 ATM) is pushing the window back out, so the net force will be due to the difference in pressure inside and outside the suit.

Oh and remember to convert the area to square meters.

 

[M98Ranger]

Yes, that is correct. To find the pressure in equation (1), you would use the equation P = ρgh, where ρ is the density of the water (1.05 g/cm^3), g is the acceleration due to gravity (9.8 m/s^2), and h is the depth (150 m). This will give you the pressure in units of Pascals (Pa). Then, to find the force in equation (2), you would multiply the pressure by the area of the window in units of meters squared (m^2). This will give you the force in units of Newtons (N). So your final equation would be:

F = P * A = (ρgh) * A

Where ρ = 1.05 g/cm^3 = 1050 kg/m^3, g = 9.8 m/s^2, h = 150 m, and A = 0.0065 m^2 (since 1 cm^2 = 0.0001 m^2).

Plugging in these values, you would get:

F = (1050 kg/m^3 * 9.8 m/s^2 * 150 m) * 0.0065 m^2 = 1008.45 N

Therefore, the window would have to withstand a force of approximately 1008.45 N in order to maintain a pressure of 1 atm inside the diving suit at a depth of 150 m. This is a significant amount of force and highlights the importance of strong and durable materials used in diving suits to withstand the pressures of the deep sea.