# Hydrostatic Pressure and Force

Suppose that a thin horizontal plate with area `A` square meters is submerged in a fluid of density `rho` kilograms per cubic meter at a depth `d` meters below the surface of the fluid.

The the volume of fluid above plate is `V=Ad`, so its mass is `m=rho V=rho Ad`.

The force exerted by the fluid on the plate is `F=mg=rho g Ad` where `g=9.8 m/s^2`.

The **pressure** on the plate is defined to be the force per unit area: `P=F/A=rho g d`.

The SI unit for measuring pressure is newtons per square meter, which is called a pascal (`1 N/m^2=1Pa`). Since this is a small unit, the kilopascal (kPa) is often used.

For example, since density of water is `rho=1000 (kg)/m^3`, the pressure at the bottom of a swimming pool 3 meters deep is `P=rho g d=1000 (kg)/m^3xx9.8 m/s^2xx 3 m=29400Pa=29.4\ kPa`.

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 `x_i` and we choose `x_i^(**)in[x_(i-1),x_i]`. The i-th horizontal strip of the dam is approximated by a rectangle with height `Delta x` and width `w_i`,

To find `w_i` we need to find `a` first.

From similar triangle we have that `a/(10-x_i^(**))=5/15` or `a=10/3-1/3 x_i^(**)`.

Therefore, `w_i=2(10+a)=2(10+10/3-1/3x_i^(**))=80/3-2/3 x_i^(**)`.

If `A_i` is the area of the i-th strip then `A_i~~w_iDelta x=(80/3-2/3x_i^(**))Delta x`.

If `Delta x` is small then pressure `P_i` is almost constant, so `P_i~~rho gx_i^(**)=1000*9.8x_i^(**)=9800x_i^(**)`.

The hydrostatic force acting on the i-th strip is the product of the pressure and the area: `F_i~~P_iA_i~~9800x_i^(**)(80/3-2/3x_i^(**))Delta x`.

Adding these forces and taking the limit as `n->oo`, we obtain the total hydrostatic force on the dam: `F=lim_(n->oo)sum_(i=1)^n9800x_i^(**)(80/3-2/3x_i^(**))Delta x=9800int_0^(10)x(80/3-2/3x)dx=9800int_0^(10)(80/3x-2/3x^2)dx=`

`=9800(40/3x^2-2/9 x^3)|_0^(10)~~1.09xx10^7\ N`.

**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 `Delta y` and choose sample point `y_i^**` from i-th strip.

We can approximate area of i-th strip by rectangle with height `Delta y` and width `2 sqrt(16-(y_i^**)^2)`: `A_i~~2sqrt(16-(y_i^**)^2)dy`.

i-strip is `7-y_i^**` meters under water the pressure on each strip is `P_i~~rho*g*(7-y_i^**)=9800(7-y_i^**)`.

Now force that acts on each strip is `F_i~~P_iA_i=19600(7-y_i^**)sqrt(16-(y_i^**)^2)dy`.

If we now add force that acts on each strip and take limit `n->oo` then we will obtain that `F=19600int_(-4)^4(7-y)sqrt(16-y^2)dy=19600(int_(-4)^4 7 sqrt(16-y^2)dy-int_(-4)^4 ysqrt(16-y^2))dy`.

First interval requires trigonometric substitution `y=4sin(u)`, second requires substitution `v=16-y^2`.

So, `F=1097600pi`.