Magnitude and position of the resultant force on the partition?

Question

A vertical partition is placed across a tank with vertical sides 1.220m apart. The tank contains water to a level of 2.130m on one side and sea water to a level of 0.915m on the

other. Find the magnitude and position of the resultant force on the partition.

Density of water (ρ) = 1000 kg/m3

Sea water – density (ρ) = 1027kg/m3

Answer 1

The unit pressure of the fresh water at the bottom of the partition is:

Pw = 1000 x 2.13 = 2130 kg/m^2

The force applied by the fresh water on the partition is:

Fw = 2130 x 2.13 x 1.22/2 = 2767.509 kg

This force is acting at 1/3 of the depth of the fresh water from the bottom which is:

Yw = 2.13/3 m

The unit pressure of the sea water at the bottom of the partition is:

Ps = 1027 x 0.915 = 939.705 kg/m^2

The force applied by the seawater on the partition is:

Fs = 939.705 x 0.915 x 1.22/2 = 524.496 kg

This is acting at 1/3 the depth of the seawater from the bottom which is:

Ys = 0.915/3

The resultant force therefore is:

R = Fw -Fs = 2767.509 - 524.496 = 2243.013 kg

This resultant is in the direction towards the seawater side.

The moment at the bottom of the partition produced by the resultant should be equal to the sum of the moments produced by the individual forces. Hence:

R x Yr = Fw x Yw - Fs x Ys

2243.013 Yr = 2767.509 x 2.13/3 - 524.496 x 0.915/3

Yr = 0.8047 m from the bottom