# 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.