A cylindrical gate of 4m diameter 2m long has water on its both sides as shown in figure. Determine the magnitude, location and direction of the resultant force exerted by water on the gate. Also find the least weight of the cylinder so that it may be lifted away from the floor.

Solution:

Here the force acting on the left sides of the cylinder are, 

horizontal component of force `F_(x1)`= force on the projected vertical area AOC

` =rhog*A*bar (h)`

`=9810**(4**2)**2`

`=156960 N`

and this acts at a distance of,

`(h_1)^**=(I_G sin^2theta)/(A*bar (h))`

`=(2**4^3)/(12**4**2**2)+2`

`=2.67 m` from free surface (point A)

Similarly,Vertical force is,

`F_(y1)=`weight of water enclosed by the ABCA Show More

Solution:

Here the force acting on the left sides of the cylinder are, 

horizontal component of force `F_(x1)`= force on the projected vertical area AOC

` =rhog*A*bar (h)`

`=9810**(4**2)**2`

`=156960 N`

and this acts at a distance of,

`(h_1)^**=(I_G sin^2theta)/(A*bar (h))`

`=(2**4^3)/(12**4**2**2)+2`

`=2.67 m` from free surface (point A)

Similarly,Vertical force is,

`F_(y1)=`weight of water enclosed by the ABCA

`F_(y1)=rhog*Area**L`

`=9810**2**(pi*2^2)/2`

`=123276.1 N`

Which acts at a distance,

`=(4r)/(3pi)=0.8488m`

Similarly the forces acting on the right side of the cylinder are,

horizontal component of force `F_(x2)`= force on the projected vertical area AOC

` =rhog*A*bar (h)`

`=9810**(2**2)**2/2`

`=39240 N`

and this acts at a distance of,

`(h_2)^**=(I_G sin^2theta)/(A*bar (h))`

`=(2**2^3)/(12**2**2**1)+1`

`=1.33 m` from free surface (point A)

Similarly,Vertical force is,

`F_(y2)=`weight of water enclosed by the COD

`F_(y2)=rhog*Area**L`

`=9810**2**(pi*2^2**90/360)`

`=61638.05 N`

Which acts at a distance,

`=(4r)/(3pi)=0.8488m`

Now, Resultant force in X-direction is,

`F_x=15690-39240` (force acts towards each other. So one is taken as positive and another is taken negative)

`=117720 N`

Similarly Resultant force in Y-direction is,

`F_y=123276.1+61638.05` (both are positive since acting upward )

`=184914.15 KN` (upward)

Thus, Resultant force, 

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

`=219206 N`

And, `theta=tan^-1 (F_y/F_x)=57^0` with the X-axis.

Taking moment about C,

`F_x*y=F_(x1)*(4-2.67)-F_(x2)*(2-1.33)`

or,`y=1.55 m` from bottom.

Similarly,

`F_y*x=F_(y1)**0.8848-F_(y2)**0.8848`

`x=0.282 m` from AOC.

we know, the resulting force in upward direction is,

`F_y=184914.15 N`

Thus the weight of the cylinder should not be less than the upward force `F_y`. Hence the least weight of the cylinder should be at least equal to `184914.15 N`