Magnitude and the location of the resultant force?

Question

A tank has a plan area of 5mx5m and a depth of 3.8m.A vertical partion divides the tank into two equal volumes,on one side of the partion the tank is filled with glycerine to a depth of 3.5m and the other side of the tank contains water to a depth of 1m. The specific density of glycerine is 1.26. Calculate the magnitude and the location of the resultant force on the vertical partion.

Answer 1

The unit pressure exerted by the glycerin at the bottom of the tank partition is:

Pg = 1.26 x 1000 x 3.5 = 4410 kg/m^2

Therefore the force exerted by the glycerin on the partition is:

Fg = Pg x 3.5 x 5/2 = 4410 x 3.5 x 5/2 = 38587.5 kg.

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

Yg = 3.5/3 meters

The unit pressure exerted by the water at the bottom of the partition is:

Pw = 1 x 1000 x 1 = 1000 kg/m^2

The force exerted by the water on the partition which is opposite that exerted by the glycerin is:

Fw = 1000 x1 x 5/2 = 2500 kg

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

Yw = 1/3 meters

The resultant force therefore is:

R = Fg - Fw = 38587.5 - 2500 = 36087.5 kg

This resultant force is acting in a direction towards the water side at a point where it creates the same amount of moment that the two individual forces makes at the bottom of the partition. Hence:

R x Yr = Fg x Yg - Fw x Yw

36087.5Yr = 38587.5 x 3.5/3 - 2500 x 1/3 = 44185.417

Yr = 1.2244 meters from the bottom