Question

Question 4

The sea water (RD=1.03) level on one side of a sea lock is 3.5m high.

1. Determine the force on the 5m wide sea lock and also the location of this force from the bottom.
2. If a 300mm diameter hole is cut out and a cover plate bolted in place, determine the force on the cover plate and also the location to the center of the hole. The tops of the hole is 2.35m below the level of the sea water.

Answer 1

a) The pressure at a depth d in a liquid is given by

Consider a small region oh height dh and width 5 m at a depth of h from the top of the water.

The pressure due to this region is

Since pressure = Force/Area

So, the total force on the sea lock is given by

Since relative density is given to be

RD = 1.03,

= 1.03*1000 = 1030 kg/m³

g = 9.8 m/s²

So,

b) The diameter of the hole is 300 mm = 0.3 m

The top is at 2.35 m

So, the bottom will be at 2.65 m

The area element in this case is given by

The length l for any height x (considering the center of the circle to be zero) is given by

The center of the circle is at h = 2.35+0.15 = 2.5 m

So,

the distance from top to x is

h(x) = 2.5+ x

So, x = (h-2.5)

So,

Thus,

the integral becomes

The integral can be evaluated to be

0.129126.

So,

F = 20188*0.129126 = 2606.796 N