A 3.6 m square gate provided in an oil tank is hinged at its top edge. This tank contains gasoline (s=0.7) upto a height of 1.8 m above top edge of plate. The space above oil is subjected to a negative pressure of 8250 N/`m^2`. Determine necessary vertical pull to be applied at lower edge to open gate. [GATE]

SOLUTION:

Here, head of oil equivalent to negative pressure of 8250 N/`m^2` is,

`h=p/(rho*g)`

`=8250/(0.7**9810)`

`=1.2 m`

This negative pressure will reduce the oil head above the top edge of the gate from 1.8 m to 1.2 m i.e by 0.6 m.

This free surface is known as imaginary free surface. 

Now,....Show More

SOLUTION:

Here, head of oil equivalent to negative pressure of 8250 N/`m^2` is,

`h=p/(rho*g)`

`=8250/(0.7**9810)`

`=1.2 m`

This negative pressure will reduce the oil head above the top edge of the gate from 1.8 m to 1.2 m i.e by 0.6 m.

This free surface is known as imaginary free surface. 

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

`bar (h)=0.6+(3.6/2*sin45)`

`bar (h)=1.873 m`

And,

Area,`A=3.6**3.6=12.96 m^2`.

The pressure force is,

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

`=700**12.96**9.81**1.873`

`=166690N`

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

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

`I_G=`moment of inertia

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

Thus,

`h^**=((3.6**3.6^3)/12*sin^2(45))/(12.96**1.873)+1.873`

`h^**=2.16 m`

Now, taking moments about hinge A,we get,

`F**3.6sin45=P**(2.16-0.6)/(sin45)`

or, `F=144465 N`

This is the vertical force to be applied at the lower end of the gate to open the gate.