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

IntegrateMe
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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?
 
\int_0^H (1000)(9.8)(1-h)(A)Δx

I don't know how to do the area, though...at least the part without Δx.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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