Figure below shows cross section of a tank full of water under pressure. The length of tank is 2m. An empty cylinder lies along length of tank on one of its corner as shown in figure. Find vertical and horizontal components of force acting on curved surface ABC of cylinder.

SOLUTION:

Given,

Radius of the cylinder,`R=1 m`

Length,`l=2m`

Gauge Pressure,`P=0.2 (kg_f)/(cm)^2`

`=0.2**9.81**10^4`

`P=1.962**10^4`N/`m^2`

And pressure head,`h=P/(rho*g)=(1.962**10^4)/(9810)`

`h=2 m`

Thus the free surface of the tank will be at a height of 2m from the top of the tank.

This free surface is known as imaginary free surface. 

Figure below shows the equivalent free surface of water.

SOLUTION:

Given,

Radius of the cylinder,`R=1 m`

Length,`l=2m`

Gauge Pressure,`P=0.2 (kg_f)/(cm)^2`

`=0.2**9.81**10^4`

`P=1.962**10^4`N/`m^2`

And pressure head,`h=P/(rho*g)=(1.962**10^4)/(9810)`

`h=2 m`

Thus the free surface of the tank will be at a height of 2m from the top of the tank.

This free surface is known as imaginary free surface. 

Figure below shows the equivalent free surface of water.

Now, 

The total horizontal component of force acting on the curved surface ABC of the cylinder is equal to the force on the projected area of the curved surface on the vertical plane

I.e,

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

Where,

`bar (h)=`distance of the c.g of the area from free water surface

`bar (h)=(2+1.5/2)`

Thus,

`F_x=9810**(1.5**2)**(2+1.5/2)`

`F_x=80932.5N`

Also, the vertical component of the force is,

`F_y=`weight of water enclosed by the curved surface upto the free surface  (real or imaginary )

`F_y=` weight of water in portion `CODEABC`

`=`Weight of water in CODFBC - Weight of water in AEFB

But,

Weight of water in CODEFBC = Weight of water on [COB + BFDOB]

`=rho *g ((pir^2)/2+BO**OD)*Wi dth `

`=9810(pi/4*1^2+1**2.5)**2`

`=64458.5 N`

Again, the weight of water in AEFB = Weight of water in  [AEFG + AGBH - AHB] 

`=rho g*2(2BH+0.5BH-AHB)`

From figure,

`sintheta=(AH)/(AO)=0.5/1=0.5`

And, `BH=BO-HO=1-AO*cos30`

`BH=1-1cos30=0.134`

And the area of `triangle ABH`=Area of [ABO - AHO]

`=pir^2*30/360-(AH**HO)/2`

`=(pir^2)/12-(0.5**0.866)/2`

`=0.0453`

Thus using all these values,we get,

Weight of water in AEFB `=9810**2 (2**0.134+0.134**0.5-0.0453`

`=5684 N`

Hence,

`F_y=64458.5-5684=58774.5 N`