"Calculate Resultant Force on Vertical Dam Wall

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Homework Help Overview

The problem involves calculating the resultant force on a vertical dam wall due to water pressure, with specific reference to the height of water behind the dam and the width of the wall.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply pressure equations to find the resultant force, while some participants question the assumption of uniform pressure and suggest the need for integration to account for varying pressure with depth.

Discussion Status

Participants are exploring different approaches to the problem, with some guidance offered regarding the necessity of integration to accurately determine the net force. There is no explicit consensus on the correctness of the original poster's method.

Contextual Notes

The discussion includes considerations of fluid mechanics principles and the implications of varying pressure in the context of the problem. The original poster notes this is their first fluid question involving integration.

etothey
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Homework Statement


Pressure_2.jpg: http://www.imageupload.org/thumb/thumb_8609.jpg
A vertical dam wall has a width w. Water is filled to a height H behind the dam. Calculate the resultant force on the dam wall.



Homework Equations


P=F/A
P= Pa + p*g*h





The Attempt at a Solution


Using picture to solve the question
P=F/A
A=H*W
Thus P=Pa g*h*p where p is density and P is pressure.
Thus F=P*A=H*W*(Pa + gHp) where g is gravity.

Is this correct?
 
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Your solution, as I've understood it, would work if the pressure were uniform, that is, constant. However, it changes as you go down, so you would need to perform an integration to determine the net force.
 
Metaleer said:
Your solution, as I've understood it, would work if the pressure were uniform, that is, constant. However, it changes as you go down, so you would need to perform an integration to determine the net force.

Aha, I see. This is first fluid question I had involving integration.
Thus, I integrated \intPo + pgh dh from 0 to H.
Getting P=PaH + pgH^2/2.
Thus
PA = HW(PaH + pgH^2/2).
Better now?
 
anyone?
 

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