# 10. Force Due to Liquid Pressure by Integration

by M. Bourne

The force *F* on an area *A* at a depth *y* in a liquid of density *w* is given by

`F = wyA`

The force will increase if the density increases, or if the depth increases or if the area increases.

So if we have an unevenly shaped plate submerged vertically in a liquid, the force on it will increase with depth. Also, if the shape of the plate changes as we go deeper, we have to allow for this.

So we have:

Now, the **total force on the plate** is given by

`F=wint_a^bxy\ dy`

where

*x* is the length (in m) of the element of area (expressed in terms of *y*)

*y* is the depth (in m) of the element of area

*w* is the density of the liquid (in N m^{-3})

(for water, this is

w= 9800 N m^{-3})

*a* is the depth at the top of the area in question (in m)

*b* is the depth at the bottom of the area in question (in m)

### Example 1

Find the force on one side of a cubical container `6.0` cm on an edge if the container is filled with mercury. The weight density of mercury is `133` kN/m^{3}.

Answer

This is the same as having a square plate of sides 6.0 cm submerged in mercury.

This is a very basic example where the width of the plate does not change as we move down the plate.

It is always `x = 6`.

Also, the depth of the top of the plate is 0, so `a = 0`.

To apply the formula, we have:

`x = 0.06\ "m"``y` is the variable of integration

`w = 133\ "kN m"^-3= 133,000\ "N m"^-3`

`a = 0``b = 0.06`

So we have:

`"Force"=w int_a^b xy\ dy`

`=133000 int_0^0.06 0.06y\ dy`

`=7980 int_0^0.06\ y\ dy`

`=7980[(y^2)/(2)]_0^0.06`

`=14.4\ text[N]`

### Example 2

A right triangular plate of base `2.0` m and height `1.0` m is submerged vertically in water, with the top vertex `3.0` m below the surface.

Find the force on one side of the plate.

Answer

Before we can proceed, we need to find *x* in terms of *y*.

Now when `x = 0`, `y = 3` and when `x = 2,` `y = 4`.

So we have:

`y=mx+c`

`=1/2x+3`

This gives us `x = 2y − 6`.

To apply the formula, we have:

`x = 2y − 6 = 2(y − 3) ``y` is the variable of integration

`w = 9800\ "N m"^-3``a = 3`

`b = 4`

So we have:

`"Force"=wint_a^bxy\ dy`

`=9800 int_3^4 2y(y-3)\ dy`

`=19600 int_3^4 (y^2-3y) text[d]y`

`=19600 [(y^3)/(3)-(3)/(2)y^2]_3^4`

`=19600[((4^3)/(3)-3/2(4)^2)` `{:-((3^3)/(3)-3/2(3)^2)]`

`=19600[-2.667 - (-4.5)]`

`=35900\ text[N]`