Deriving equation for force from pressure that varies with water level

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SUMMARY

The discussion focuses on deriving the equation for the resultant force exerted by water on a dam, specifically when water is filled to a height H. The key equation utilized is F = 1/2pgwH^2, where p represents the density of water, g is the acceleration due to gravity, and w is the width of the dam. The integration process involves calculating the antiderivative of the pressure function P = pg(H-y) over the height of the water, leading to the final force equation. Participants clarified the integration steps necessary to arrive at the solution.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly hydrostatic pressure.
  • Familiarity with calculus, specifically integration techniques.
  • Knowledge of the variables involved: density (p), gravitational acceleration (g), and dam width (w).
  • Basic understanding of the concept of resultant force in physics.
NEXT STEPS
  • Study the principles of hydrostatic pressure in fluids.
  • Learn integration techniques, focusing on antiderivatives in calculus.
  • Explore applications of the resultant force in engineering contexts, particularly in dam design.
  • Review examples of similar fluid mechanics problems to reinforce understanding.
USEFUL FOR

Students in engineering or physics courses, particularly those studying fluid mechanics, as well as professionals involved in civil engineering and dam design.

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Homework Statement



Water is filled to a height H behind a dam of width w. Determine the resultant force exerted by the water on the dam.

Homework Equations



P=pgh = pg(H-y) (where p is density, greek rho)

dF = PdA = pg(H-y)wdy (where dA is a narrow horizontal strip of the dam with area wdy and dy is delta y, the change in height)

F = antiderivative (PdA) which is [pg(H-y)wdy] bounded by H and 0.

The Attempt at a Solution



The solution is 1/2pgwH^2, but I don't understand how to do the integration. In fact, it's not even a problem, it's an example (and I'm on break between semesters) but the fact that I can't remember how to take the integral is bothering me. If anyone can make sense of my ascii attempt to copy the steps, and then explain it, I'll really appreciate it.

Thanks.
 
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p,g,w are constants, you can take them outside the integral. Then you are left with (H-y)dy. That's pretty straightforward, just take the antiderivative of each term, with respect to y. Does that help?
 
Yes, thanks hage567.
 

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