A liquid of specific gravity 0.9 is filled in a container, shown in Figure, upto a depth of 2.4 m. Determine magnitude and direction of hydrostatic pressure force per unit length of container exerted on its vertical face MN and curved corner NQ.

SOLUTION:

Here,

For vertical face MN,

Pressure force, `F=rho*g*A*bar (h )`

`=900**9.81**(1.2**1)**1.2/2`

`=6356.88 N`

And, the point of application of this force is,

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

`=(1**1.2^3)/(12**(1.2**1)**0.6)+0.6`

`=0.8 m`

Similarly, for curved surface NQ,

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

`=900**9.81**(1.2**1)**(1.2+1.2/2)`

`=19070.64 N`

And,

`F_v=` weight of fluid in curve PNQ + PNMV

`=rho*g*Wi d th** Area\....Show More

SOLUTION:

Here,

For vertical face MN,

Pressure force, `F=rho*g*A*bar (h )`

`=900**9.81**(1.2**1)**1.2/2`

`=6356.88 N`

And, the point of application of this force is,

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

`=(1**1.2^3)/(12**(1.2**1)**0.6)+0.6`

`=0.8 m`

Similarly, for curved surface NQ,

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

`=900**9.81**(1.2**1)**(1.2+1.2/2)`

`=19070.64 N`

And,

`F_v=` weight of fluid in curve PNQ + PNMV

`=rho*g*Wi d th** Area\ of [PNMV + PNQ]`

`=900**9.81**(1.2**1.2+90/360**pi**1.2^2)`

`=22699.124 N`

Now,

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

`F=29646.9 N`

And,

`theta =tan^-1 (F_y/F_x)=49.97^0`