The amount of force on one trapezoidal side of a water vessel

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SUMMARY

The discussion focuses on calculating the force exerted on one trapezoidal side of a water vessel, given the height of the water (H meters), water density (1000 kg/m³), and gravitational force (9.8 N/kg). The relevant equations include F = P * A and P = ρ_water * g * depth, where the pressure P varies with depth. The user attempts to express the force as an integral, specifically \(\int_0^H (1000)(9.8)(1-h)(A)Δx\), but seeks clarification on determining the area A as the water depth changes.

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I need help understanding certain aspects of this problem...


Problem statement
Write an integral expressing the amoutn of force on one trapezoidal side of this water vessel when the height of the water is H meters. The density of water is 1000 kg/m3, and the force of gravity is 9.8 N/kg.

Relevant Equations

F = P * A
P = ∂water*g*depth

Attempt at solution

From what I understand, the depth of the water is a height (1-h) from the top of the trapezoidal side. Thus, P = (1000)(9.8)(1-h).

Now, I'm trying to solve for the area, and I know h is changing as the depth increases, but I'm unsure on how I can find a precise value. Any help on this one?
 
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You can probably break that thing up into 3 integrals, two of which should be identical right?...

Can you try writing the integral?
 
[tex]\int_0^H (1000)(9.8)(1-h)(A)Δx[/tex]

I don't know how to do the area, though...at least the part without Δx.
 

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