There is an opening in a container shown in figure below. Find force F and reaction at hinge.

SOLUTION:

Given,

Gauge pressure = 23.5 KN/`m^2`

`=23.5/(9.81**0.8)`

`=3 m` of oil.

Hence the free liquid surface may be considered to be 3 m further above the hinge as shown in figure. This free surface is known as imaginary free surface of oil.

Now, the distance of centroid of plate from the free oil surface is,

`bar (h)=3+(0.6 sin30)`

`=3.3....Show More

SOLUTION:

Given,

Gauge pressure = 23.5 KN/`m^2`

`=23.5/(9.81**0.8)`

`=3 m` of oil.

Hence the free liquid surface may be considered to be 3 m further above the hinge as shown in figure. This free surface is known as imaginary free surface of oil.

Now, the distance of centroid of plate from the free oil surface is,

`bar (h)=3+(0.6 sin30)`

`=3.3 m`

Now, total pressure force on the plate AB is,

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

`=0.8**1.2**1.2**9.81**3.3`

`=37.29 KN`

Let, this force acts at a depth  of `h^**` from the imaginary oil surface.

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

`I_G=`moment of inertia

`I_G=(b*d^3)/12`

Thus,

`h^**=((1.2**1.2^3)/12*sin^2(30))/(1.2**1.2**3.3)+3.3`

`=3.309m`

Now, taking moments about hinge A,we get,

`F**1.2-P*(AM)/(sin30)=0`

or,`F**1.2-37.29**(h^** -3)/(½)=0`

or,`F**1.2-37.29**2 (3.309-3)=0`

or,`F=19.2 KN`

Let, `R_A` be the reaction at the hinge.

Thus,

`R_A=P-F`

`=37.29-19.2`

`=18.09 KN`