Question
Question: Determine the total pressure on a circular plate of diameter 1.5 m which is placed vertically in wat...
Determine the total pressure on a circular plate of diameter 1.5 m which is placed vertically in water in such a way that the centre of plate is 2 m below the free surface of water. Find the position of centre of pressure also.
Solution
We need to find the total pressure and the position of centre of pressure of the given circular plate. To find the pressure we know the expression for pressure. By substituting the known values in the equation we get the pressure on the plate. And to find the position of centre of pressure also we know the equation. We also know the equation to find the moment of inertia about the horizontal axis through the centre of gravity. By finding this moment of inertia and substituting, we get the position of centre of pressure.
Formula used:
Pressure, F=ρghA
Position of centre of pressure, h∗=AhIG+h
Complete step by step answer:
In the question we are given a circular plate of diameter 1.5 m
It is said that this circular plate is immersed vertically in water in such a way that the centre of the plate is 2 m below the surface of water.
The given situation is depicted in the above figure.
The point ‘P’ in the figure is the centre of pressure and ‘h*’ determines the position of centre of pressure.
First let us find the total pressure experienced by the circular plate.
The total pressure is given by the equation,
F=ρghA, where ‘’F’ is the force, ‘ρ’ is density, ‘g’ is acceleration due to gravity, ‘h’ is the height and ‘A’ is area.
In the given case, we know that,
‘ρ’ is the density of water and we know that density of water is 1000kg/m3
We can take value of acceleration due to gravity as g=10m/s
‘h’ is the height from the free surface to the centre of the plate which is given as 2.0 m
Area ‘A’ is the area of the plate given as A=4πd2, were ‘d’ is the diameter of the plate which is = 1.5 m
Therefore by substituting all this known values in the equation, we get
⇒F=(1000×10×2)4π×(1.52)