A 6 m ** 2 m rectangular gate is hinged at base and is inclined at an angle of 60° with horizontal. The upper end of gate is kept in position by a weight of 60 kN acting at angle of 90° as shown in figure. Neglecting weight of gate, find level of water when gate begins to fall.

SOLUTION:

We know, the pressure force on the gate is,

`F=rho*g*A*bar (h)`... (i)

Now,

Area of gate,`=6**2=12m^2`

Now,

Length of gate submerged in water = AD

From figure,

`AC=AD*sin60`

or,`AD=(AC)/(sin60)`

`AD=2/(root ()(3))*h`

Now, the area of the gate immersed in water is,

`A=AD**WIDTH`

`A=2/(root ()(3))*h*2`

`A=4/(root ()(3))*h`

Also,depth of C.G. of the i....Show More

SOLUTION:

We know, the pressure force on the gate is,

`F=rho*g*A*bar (h)`... (i)

Now,

Area of gate,`=6**2=12m^2`

Now,

Length of gate submerged in water = AD

From figure,

`AC=AD*sin60`

or,`AD=(AC)/(sin60)`

`AD=2/(root ()(3))*h`

Now, the area of the gate immersed in water is,

`A=AD**WIDTH`

`A=2/(root ()(3))*h*2`

`A=4/(root ()(3))*h`

Also,depth of C.G. of the immersed area is,

`bar (h)=h/2`

Thus,the  pressure force on the gate is,

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

`=9.81**1000**4/(root ()(3))*h*h/2`

`F=19620/(root ()(3))*h^2`

Also,

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

Where,`I_G=`moment of inertia

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

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

`I_G=(4h^3)/(9root ()(3))`

Thus,

`h^**=((4h^3)/(9root ()(3))*(root ()(3)/2)^2)/((4h)/(root ()(3)*h/2)+h/2`

or,`h^**=(2h)/3`

Also, the distance of centre of pressure from the hinge along the length of the gate` = AE`

We know,

`h-h^**=AE*sin60`

or,`AE=(h-h^**)/(sin60)`

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

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

Now taking the momentum about the hinge,

`60000**6-F**AE=0`

or,`60000**6=19620/(root ()(3))*h^2*(2h)/(3root ()(3))`

or,`h=4.36 m`

Note: This type of problem can be solved easily by plotting pressure diagram as explained in next problem.