Solve Fluids & Forces Problem | Physics Forum

[pianoman2700]

[SOLVED] Fluids and Forces

Hi, I'm new to this site, so I'm not sure if this is the right place to post. I found a year-old entry with the same question that I'm having trouble with, found here:

https://www.physicsforums.com/showthread.php?t=153538"

For Part a, I followed the same method that PremedBeauty followed, but can't seen to get the correct answer. Is there a force that I am missing?

For Part b, I integrated to try and find the total force, but, once again, my answer comes out incorrect. My calculations for part b are below.

F=force
(atm)= atmospheric pressure (in Pascals)
p= density of water
g=gravity
h= depth of water
dh= change in depth of water
A= area = width*h
D = depth (as defined by problem)

F=((atm) + pgh)*A
But because pressure changes with depth, we must integrate
dF = ((atm) + pgh) * (width*dh)
Integrating yields:
F= (atm)(width)(h) + (.5)(p)(g)(h^2) evaluated from 2D to 3D
Evaulating yields:
F= 3D^2((atm) + 1/5pgD) - 2D^2((atm) + pgD)

The answer that I get from this calculation does not agree with the correct answer; is there something I am missing?

 

[stewartcs]

Hi, I'm new to this site, so I'm not sure if this is the right place to post. I found a year-old entry with the same question that I'm having trouble with, found here:

https://www.physicsforums.com/showthread.php?t=153538"

For Part a, I followed the same method that PremedBeauty followed, but can't seen to get the correct answer. Is there a force that I am missing?

Not sure what you are missing since you didn't show the calculation for part a.

For part a you should have:

F_A = P_A \cdot A_A

where,

F_A is the force in Newtons
P_A is the pressure on face A in Pascals
A_A is the area of face A in square meters

If your problem is the same as the picture in the other link then you have,

P_A = P_{fluid} + P_{atm}

and

A_A = d^2

Hence the force acting on face A is,

F_A = (P_{fluid} + P_{atm}) \cdot d^2

Now with P_{fluid} = \rho gh and h = 2d you get,

P_{fluid} = \rho g \cdot 2d

Which yields,

F_A = (\rho g \cdot 2d + P_{atm}) \cdot d^2

Just plug in the value for P_{atm} in pascals, along with the other missing, but given, values and you'll have the force.


CS

 

[pianoman2700]

Not sure what you are missing since you didn't show the calculation for part a.

Just plug in the value for P_{atm} in pascals, along with the other missing, but given, values and you'll have the force.


CS

Alright, thank you. I meant to refer calculations for part 1 to those that the previous poster did (as I went through those same steps). I came up with the same answer that was displayed here, but it disagrees with the answer given. An error on the part of the book, I think...

 

[stewartcs]

Make sure your units are correct.

BTW, What did the book give for the answer?

 

[pianoman2700]

I'm not sure; it's an online homework assignment. It tells me if I'm right or wrong, but won't tell me the actual answer until I've gotten it correct.

Are you able to help on the second part? (Part b)

 

[stewartcs]

I'm not sure; it's an online homework assignment. It tells me if I'm right or wrong, but won't tell me the actual answer until I've gotten it correct.

Are you able to help on the second part? (Part b)

You'll have to integrate to find the force.

F = P_{atm} \cdot A_B + \rho g \int_{a}^{b} h \cdot dA

where,

A_B is the area of face B
h is the height
dA is the area of some small strip across the face

The limits of integration, a and b, would be the starting and ending height (depth), i.e. 2d to 3d respectively.

Hope that helps.

CS

 

[pianoman2700]

Hope that helps.

CS

Apparently it only wanted the gauge pressure, so I subtracted out the air pressure, and everything worked out great. Thanks for your help.

 

[stewartcs]

No problem. I was going to suggest trying that if you were still having problems.

CS