# Amount of force on a trapezoidal surface need help following solution

1. Mar 17, 2012

### IntegrateMe

A trash container is located outside a building. It starts to rain, causing water to enter the vessel. The trapezoidal side of the vessel is cracked and can only support 6000 newtons of force; eventually the water pressure causes the wall to break.

Container

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

The solution says:

F = δ*g*depth*A
depth = (H-h)
A = (1+h)ΔH

And so F = ∫1000(9.8)(H-h)(1+h)dh ; from 0 to H

I don't understand how they got (H-h) to be the depth nor how the area = (1+h)ΔH. Anyone care to explain?

Thank you!

2. Mar 17, 2012

### IntegrateMe

Anyone?

3. Mar 18, 2012

### LCKurtz

H-h is not the depth of the water. It is the distance below the surface that the horizontal dh element where you are calculating the force is. And I also think that (1+h) is not the correct length of that horizontal dh element of area and dH should be dh in his expression for A.