Center of pressure?

Question

An oil storage tank has one side sloping at 60 degrees to the horizontal. A circular opening of diameter 300mm is cut into the side of the tank and is covered by a plate bolted into place. The level of oil in the tank is 1.2m above the top of the opening find:
a)the force acting on the plate
b)the depth of the center if pressure for the plate below the surface of the oil

Oil has a density of 0.9
k value is D/4

I have my answer at 0.841 and 1.37 but believe this to be wrong and the correct answers are 0.83 KN and 1.33 metres.How do I do this correctly???

This isnt my homework....im just studying for an exam and using old questions with given answers so any help would be really appreciated

Answer 1

The total force on a submerged vertical or inclined plane surface is equal to the area of the surface times the depth of the centroid, (zbar).

(zbar) = 1.35 m, A = 0.0225 π m^2, ρ = 0.9 * 1000 kg/m^3

I get 0.8425 KN

I think you might just be experiencing rounding error.

ζ = depth to center of pressure

The force on the strip of area dA at depth z is, as we have seen, ρgzdA, so the first moment of that force is ρgz^2dA. The total moment is therefore ρg ∫ z^2dA, which is, by definition of radius of gyration k, (see chapter 2), ρgk^2A. The total force, as we have
seen, is ρg(zbar)A, and the total moment is to be this times ζ. Thus the depth of the centre of pressure is

ζ = k^2/0.15 with k = 0.075 m

ζ = 0.031 m, thus I get 1.381 m

I don't buy the "correct" answer since the center of pressure HAS to be below the center of the plate.

See this website.
http://orca.phys.uvic.ca/~tatum/classmechs/class16.pdf

Answer 2

Circular or otherwise the pressure on the plate will be the average of the varying height and the centre of the varying mass when rotated around its own axis.