**Question: **

Finding the fluid force on the vertical side of the rectangular tank where the dimensions are 5 feet by 6 feet ( length is 5 feet and width is 6 feet ). Assume that tank is full in water ( weight-density of water is 62.4 lb/cubic foot).

I have got a few answers but still very lost, a step by step of this problem would be very helpful thanks

**Answer: **

If the tank is rectangular in horizontal section, with perpendicular sides, the tank can be divided up into horizontal segments the full length (horizontal measure) of a side of the tank and width, height (vertical measure) of each segment. The force on each segment is the pressure at the depth of the segment, multiplied by the area of the segment. The sum of the forces of all segments is equal to the total force on the side of the tank. Pressure on each segment,

p = rho x g x d. where rho is the density of water, g is accel. of gravity and d is the depth of the segment from surface of the water. The force on the segment is the pressure times length (L) times height (h) of the segment:

The total pressure of the side = L x rho x g x integral[h (dh)] with limits between top and bottom of the tank. The solution is:

F = L x rho x g x h^2 / 2, calculated from 0 to depth of tank.