Find resultant force on surface ABCDEF of 10 m width of surface shown below.

SOLUTION:

Given,

Width,`=2m`

Gauge Pressure,`P=-117.72 KPa`

Thus pressure head,`h=P/(rho*g)=-117.72/(1.2**9.81)`

`h=-10 m`

Thus the free surface of the tank will be at a height of 10m further below the top of the given oil surface.

This free surface is known as imaginary free surface. 

Figure below shows the equivalent free surface of water.

Now, 

The total horizontal component of force acting on ....Show More

SOLUTION:

Given,

Width,`=2m`

Gauge Pressure,`P=-117.72 KPa`

Thus pressure head,`h=P/(rho*g)=-117.72/(1.2**9.81)`

`h=-10 m`

Thus the free surface of the tank will be at a height of 10m further below the top of the given oil surface.

This free surface is known as imaginary free surface. 

Figure below shows the equivalent free surface of water.

Now, 

The total horizontal component of force acting on the curved surface ABCDEF   is equal to the force on the projected area of the curved surface on the vertical plane

I.e,

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

Where,

`bar (h)=`distance of the c.g of the area from free water surface

`bar (h)=(3+2+2**1.6)/2=4.10 m`

and projected area,`A=(3+2+2**1.6)**10=82 m^2`

Thus,

`F_x=1.2**9.81**82**4.10`

or,`F_x=3957.7 KN->`

and this acts at a distance of,

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

`I_G=`moment of inertia

`I_G=(10**8.2^3)/12=459.47 m^4`

Thus`I_G=459.47/(82**4.10)+4.10`

`=5.47 m` from the imaginary surface.

Vertical force

 In portion C'D = 0 KN

In portion DD',

`F_(y1)=rho*g*10*Area`

`=9.81**1.2**3**2**10`

`=706.32 KN`(downward)

In portion D'E =0 KN

In portion EF,

`F_(y2)=rho*g*h*Area`

`=9.81**1.2**10**pi**1.6^2/2`

`=473.38 KN`(upward)

Thus, the net vertical force is,

`F_y=473.38-706.32+0`

`=232.94 KN`(Downward)