What Is the Hydrostatic Force on a Vertical Gate in an Irrigation Canal?

[SUchica10]

A gate in an irrigation canal is in the form of a trapezoid 3 feet wide at the bottom, 5 feet wide at the top, with the height equal to 2 feet. It is placed vertically in the canal with the water extending to its top. For simplicity, take the density of water to be 60 lb/ft cubed. Find the hydrostatic force in pounds on the gate.

I am having problems setting this problem up. It looks like its really easy but I am just not sure how to start it.

I know F = density x gravity x area x depth

 

[chanvincent]

Because the force acts on the gate is not constant, i.e. force at the bottom is larger than that at the top , we have
dF = density x gravity x depth x d(area)
Do this integration over the trapezoid will yield the correct answer.

 

[HallsofIvy]

Imagine the gate being divided into many narrow horizontal bands of width "$\Delta y$". If y is the depth of a band, and $\Delta y$ is small enough that we can think of every point in the band as at depth y, then the force along that band is the pressure, 60(y) [NOT "times gravity"! The density of the water is weight density, not mass density!], times the area: the length of the band times $\Delta y$. Of course, the length of the band depends on y: it is a linear function of y since the sides are straight lines, length(2)= 3 and length(0)= 5 so length(y)= -y+ 5. The force on that narrow band is 60y(5-y)$\Delta y$. The total force on the gate is the sum of those,$\Sum 60y(5-1y)\Delta y$, as y goes from 0 to 2. In the limit, that becomes the integral
$$60\int_0^5 y(5-y)dy$$