A gate supporting water is shown in figure. Find height h of water so that gate tips about hinge. Take width of gate as unity.

SOLUTION:

The gate would tip about the hinge point B when the line of action of the resultant pressure force lies from B to C anywhere on the gate; the limiting condition being the situation when the resultant force passes through the hinged point B. Hence for the given condition point B becomes the centre of pressure.

Thus,from figure, the depth of centre of pressure is,

`h^**=h-3`..... (i)

But, we know that, `h^**` is also given by, 

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

Where, `I_G=`moment of inertia

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

`=b*(AC)^3/12`

SOLUTION:

The gate would tip about the hinge point B when the line of action of the resultant pressure force lies from B to C anywhere on the gate; the limiting condition being the situation when the resultant force passes through the hinged point B. Hence for the given condition point B becomes the centre of pressure.

Thus,from figure, the depth of centre of pressure is,

`h^**=h-3`..... (i)

But, we know that, `h^**` is also given by, 

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

Where, `I_G=`moment of inertia

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

`=b*(AC)^3/12`

`=b/12*((2h)/(root ()(3)))^3`

`=(2h^3)/(9root ()(3))`

Taking the width of gate as unity,we get,

Area,`A= AC**1=(2h)/(root ()(3)`

And `bar (h)=h/2`

Thus,

`h^**=((2h^3)/(9root ()(3))*sin^2 (60))/((2h)/root ()(3)*h/2)+h/2`

`h^**=(2h)/3`... (ii)

Now, equating (i) and (ii), we get,

`(h-3)=(2h)/3`

or,`h=9m`

Hence,the height of water for tripping the gate is 9 m.