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# 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:

Continues below

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/m3.

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.

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]