Pressure and Net Force Question

[Enoch]

I'm working on my homework, and I have finished all the problems save for one. I think I am on the right track, but I'm lacking just a few bits of info. Here is the question (it has 2 parts, I have answered the first part correctly, but cannot get the second part correct).

Part 1

Engineers have developed a bathyscaph that can reach ocean depths of 7.72 mi. The acceleration of gravity is 9.8 m/s^2. If the density of seawater is about 1030 kg/m^3, what is the absolute pressure at that depth in Pa?

Answer - P = Pinitial + pgh = 1.25484 x 10 ^8

Part 2 (here is where I need help)

If the inside of the vessel is maintained at atmospheric pressure, 101300 Pa, what is the net force on a porthold of diameter 20.9 cm? Answer in units of N.

Well I know that F = PA, so I set up the equation:

PA - Pinitial(A) = density(g)(h)(A)

However, this doesn't solve for net force. I'm not sure if this is the actual equation I need, or if I can use it to find the net force. I was under the impression that PA - Pinitial(A) would equal the net force but it does not. I assume that I need the 20.9 cm in there somewhere. Any help is appreciated.

 

[anathema]

Hello there,

Thank you for reaching out for help with your homework question. I am happy to assist you in finding the solution.

First, let's review the information given in the question. We know that the bathyscaph can reach depths of 7.72 mi, the acceleration of gravity is 9.8 m/s^2, and the density of seawater is 1030 kg/m^3. We also know that the inside of the vessel is maintained at atmospheric pressure, 101300 Pa.

Now, let's focus on the net force on the porthold. The equation you have used, F = PA - Pinitial(A), is correct. However, there are a few things we need to consider in order to find the net force.

First, let's determine the area of the porthold. The diameter is given as 20.9 cm, so we can use the formula for the area of a circle, A = πr^2, where r is the radius. The radius would be half of the diameter, so r = 10.45 cm or 0.1045 m.

Next, we need to find the pressure at the porthold, which is the same as the atmospheric pressure, 101300 Pa.

Now, we can plug these values into the equation: F = PA - Pinitial(A)

F = (101300 Pa)(0.1045 m^2) - (1.25484 x 10^8 Pa)(0.1045 m^2)

F = 10568 N

Therefore, the net force on the porthold is 10568 N.

I hope this helps you to solve the second part of your homework question. Let me know if you have any further questions. Good luck with your homework!

 

[wais]

It seems like you are on the right track for part 2, but there are a few things to consider. First, make sure you convert all the units to be consistent. In part 1, you used meters for depth, so make sure to use meters for the diameter as well. Also, the pressure at the porthole will be the same as the atmospheric pressure, so you can set PA equal to 101300 Pa.

Now, for the equation, you are correct that F = PA - Pinitial(A) is the correct formula. However, you need to consider the area of the porthole as well. Since the pressure is acting over the entire area of the porthole, you need to multiply the pressure by the area. The equation would then be:

F = PA - Pinitial(A) = (101300 Pa)(0.209 m^2) - (1.25484 x 10^8 Pa)(0.209 m^2) = -1.25483 x 10^7 N

Note that the net force is negative because the pressure inside the vessel is higher than the pressure outside, creating a force pushing outwards. This is known as the "net outward force" and is a common concept in pressure calculations.

I hope this helps to clarify the equation and how to incorporate the diameter to find the net force on the porthole. Good luck with the rest of your homework!