A square aperture in vertical side of a tank has one diagonal vertical and is completely covered by plane plate hinged along one of upper sides of aperture. The diagonals of aperture are 2.4 m long and tank contains a liquid of specific gravity 1.2. The centre of aperture is 1.8 m below free surface. Calculate:

• thrust exerted on plate by liquid
• position of its centre of pressure

[ANNA UNIVERSITY ]

SOLUTION:

Given,

Diagonal of aperture,PR=QS= 2.4 m

Area of square aperture, A= area of `triangle`PQR + area of `triangle`PSR

`A=2**1/2**2.4**2.4/2`

`A=2.88 m^2`

Now,  the thrust exerted on t....Show More

SOLUTION:

Given,

Diagonal of aperture,PR=QS= 2.4 m

Area of square aperture, A= area of `triangle`PQR + area of `triangle`PSR

`A=2**1/2**2.4**2.4/2`

`A=2.88 m^2`

Now,  the thrust exerted on the plate by the liquid is,

`F=rho*g*A*bar (h)`

`=1.2**1000**9.81**2.88**1.8`

`F=61026.05 N`

Again,

And the point of application of this force is,

`h^** =I_G/(A*bar (h))+bar (h)`

Where,

`I_G=`moment of inertia of aperture PQRS about the diagonal PR

=M.I of `triangle`PQR+M.I of `triangle`PSR about PR

`=(2.4**(1.2)^3)/12+(2.4**1.2^3)/12`

`=0.6912 m^4`

Thus,

`h^**=0.6912/(2.88**1.8)+1.8`

`h^**=1.933 m`