Find magnitude, direction and location of total pressure exerted by water on curved surface AB which is a quadrant of circular cylinder. Given that radius of surface is 2.5 m and cylinder is 3m long and water is 2 m above it.

SOLUTION:

Now,

The horizontal force is,

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

`=9.81**(3**2.5)(2+2.5/2)`

`=239.11875 KN ->`

And it acts at a distance  of,

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

`=(3**2.5^3)/(12**(3**2.5)(2+2.5/2))+(2+2.5/2)`

`=3.41m` from the top.

Similarly,the vertical force is,

`F_y=rho*g**3 ((2.5**2)+(pi/4*2.5^2))`

`=291.614 KN`(downward)

And it acts at a c.g i.e,

`=(4r)/(3pi)=1.061m` towards right from O.

SOLUTION:

Now,

The horizontal force is,

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

`=9.81**(3**2.5)(2+2.5/2)`

`=239.11875 KN ->`

And it acts at a distance  of,

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

`=(3**2.5^3)/(12**(3**2.5)(2+2.5/2))+(2+2.5/2)`

`=3.41m` from the top.

Similarly,the vertical force is,

`F_y=rho*g**3 ((2.5**2)+(pi/4*2.5^2))`

`=291.614 KN`(downward)

And it acts at a c.g i.e,

`=(4r)/(3pi)=1.061m` towards right from O.

Thus,the resulting force,

`F=root ()(F_x^2+F_y^2)`

`=377.11 KN`

AND,

`theta=tan^-1 (F_y/F_x)=50.62^0`