Deriving equation for force from pressure that varies with water level

AI Thread Summary
To determine the resultant force exerted by water on a dam, the pressure at a depth y is expressed as P = pg(H-y), where p is the water density. The differential force dF is calculated as dF = PdA = pg(H-y)wdy, leading to the total force F being the integral of dF from 0 to H. The integration simplifies to F = (1/2)pgwH^2, which is derived by taking the antiderivative of (H-y)dy. Clarification on integration steps was provided, emphasizing that constants can be factored out, making the integration process straightforward. Understanding these steps is crucial for solving similar problems in fluid mechanics.
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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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