Find horizontal and vertical component of force exerted by water on horizontal cylinder of 10m length (perpendicular to paper ) as shown in figure below. The water to left of cylinder is in a closed tank at a pressure of 35 KN/`m^2`.

SOLUTION:

Here, 

Pressure head corresponding to 35 KN/`m^2` is,

`h=P/(rho g)`

`=35/9.81=3.57 m`

Hence the imaginary free liquid surface is 3.57 m above the point A.

Now, the arrangement can be redrawn as,

SOLUTION:

Here, 

Pressure head corresponding to 35 KN/`m^2` is,

`h=P/(rho g)`

`=35/9.81=3.57 m`

Hence the imaginary free liquid surface is 3.57 m above the point A.

Now, the arrangement can be redrawn as,

Here, the net projection on the vertical plane of the cylindrical plane of the cylindrical surface under consideration is,

`=2+2cos30`

`=3.73 m`

Thus the projected area is,

`=3.73**10`

`=37.3 m^2`

Now, the horizontal pressure force is,

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

`=9.81**37.3 (3.73/2+3.56)`

`=1987.9 KN ->`

Similarly,Vertical force is,

`F_y=`weight of water enclosed by the abcdef

`F_y=rhog**10**Area [abcdo + triangle odd'+od'ef]`

`=9.81**10(pi*2^2(180+30)/360+1/2**2cos60**1+1**5.57)`

`=1349.9 KN`