Suppose that a thin horizontal plate with area A square meters is submerged in a fluid of density ρ kilograms per cubic meter at a depth d meters below the surface of the fluid.
The volume of fluid above plate is V=Ad, so its mass is m=ρV=ρAd.
The force exerted by the fluid on the plate is F=mg=ρgAd where g=9.8s2m.
The pressure on the plate is defined to be the force per unit area: P=AF=ρgd.
The SI unit for measuring pressure is newtons per square meter, which is called a pascal (1m2N=1Pa). Since this is a small unit, the kilopascal (kPa) is often used.
For example, since density of water is ρ=1000m3kg, the pressure at the bottom of a swimming pool 3 meters deep is P=ρgd=1000m3kg×9.8s2m×3m=29400Pa=29.4kPa.
So we can determine the hydrostatic force against a vertical plate or wall or dam in a fluid. We again must use integrals, because the pressure is not constant but increases as the depth increases.
Example 1. Suppose we have a trapezoidal dam whose height is 15m. Lower base is 20 m and upper base is 30 m. Find the force on the dam due to hydrostatic pressure if the water level is 5 m from the top of the dam.
We need convention for x. Let x=0 corresponds to the surface of water and x=10 be bottom of the dam. Then water is in interval [0,10].
Now divide interval [0,10] into n subintervals of equal length with endpoints xi and we choose xi⋆∈[xi−1,xi]. The i-th horizontal strip of the dam is approximated by a rectangle with height Δx and width wi,
To find wi we need to find a first.
From similar triangle we have that 10−xi⋆a=155 or a=310−31xi⋆.
If Ai is the area of the i-th strip then Ai≈wiΔx=(380−32xi⋆)Δx.
If Δx is small then pressure Pi is almost constant, so Pi≈ρgxi⋆=1000⋅9.8xi⋆=9800xi⋆.
The hydrostatic force acting on the i-th strip is the product of the pressure and the area: Fi≈PiAi≈9800xi⋆(380−32xi⋆)Δx.
Adding these forces and taking the limit as n→∞, we obtain the total hydrostatic force on the dam: F=limn→∞∑i=1n9800xi⋆(380−32xi⋆)Δx=9800∫010x(380−32x)dx=9800∫010(380x−32x2)dx=
=9800(340x2−92x3)∣010≈1.09×107N.
Example 2. Find force due to hydrostatic pressure on circular plate of radius 4 that is 3 meters under water.
Again we need convention about x. This time we will take origin to be at the center of circle and x pointing to the right (standard rectangular system).
Now we split plate by n horizontal strips of heigth Δy and choose sample point yi⋆ from i-th strip.
We can approximate area of i-th strip by rectangle with height Δy and width 216−(yi⋆)2: Ai≈216−(yi⋆)2dy.
i-strip is 7−yi⋆ meters under water the pressure on each strip is Pi≈ρ⋅g⋅(7−yi⋆)=9800(7−yi⋆).
Now force that acts on each strip is Fi≈PiAi=19600(7−yi⋆)16−(yi⋆)2dy.
If we now add force that acts on each strip and take limit n→∞ then we will obtain that F=19600∫−44(7−y)16−y2dy=19600(∫−44716−y2dy−∫−44y16−y2)dy.