An opening in a dam is covered by the use of a vertical sluice gate. The opening is 2 m wide and 1.2 m high. On the upstream of the gate the liquid of specific gravity 1.45 lies upto a height of 1.5 m above the top of the gate, whereas on the downstream side the water is available upto a height touching the top of the gate. Find:

• The resultant force acting on the gate and position of centre of pressure
• The force acting horizontally at the top of the gate which is capable of opening it.

Assume that the gate is hinged at the bottom. [Rajasthan University][A.M.I.E,May 1975]

SOLUTION:

Given,

Width of the gate, `b = 2 m `

 Depth of the gate, `d = 1.2 m`

 Area, `A = b ** d = 2 ** 1.2 = 2.4 m^2`

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SOLUTION:

Given,

Width of the gate, `b = 2 m `

 Depth of the gate, `d = 1.2 m`

 Area, `A = b ** d = 2 ** 1.2 = 2.4 m^2`

 Specific gravity of liquid`S = 1.45`

Let,

`F_1` = Force exerted by the liquid of sp. gravity 1.45 on the gate, and

` F_2` = Force exerted by water on the gate.

`bar (h_1)` = the depth of C.G of the gate from free surface of liquid.i.e,

`bar (h_1)=1.5+1.2/2=2.1 m`

`bar (h_2)` = the depth of C.G of the gate from free surface of water.i.e,

`bar ( h_2)=1.2/2=0.6 m`

Now,the resultant force on the gate is

`F=F_1-F_2`

`=rho*g*A*bar (h_1)-rho*g*A*bar (h_2)`

`=1.45**1000**9.81**2.1-1000**9.81**0.6`

`F=57565 N`

We get,

`F_1=71691 N` and `F_2=14126 N`

Let the force `F_1` and `F_2` acts at a distance of `(h_1)^**` from the liquid surface and at a distance of `(h_2)^**` from the water surface respectively.

Now,

`(h_1)^** =I_G/(A*bar (h_1))+bar (h_1)`

Where,

`I_G=`moment of inertia

`I_G=(b*d^3)/12=(2**1.2^3)/12=0.288 m^4`

Thus,

`(h_1)^**=0.288/(2.4**2.1)+2.1=2.1571 m`

Now, the distance of point of application of `F_1` from the hinge is,

`=1.5+1.2-(h_1)^**`

`=1.5+1.2-2.1571`

`=0.5429 m`

Similarly,

`(h_2)^** =I_G/(A*bar (h_2))+bar (h_2)`

`(h_2)^**=0.288/(2.4**0.6)+0.6=0.8 m`

Now, the distance of point of application of `F_2` from the hinge is,

`=1.2-0.8`

`=0.4 m`

Thus, the resulting force `F` will be acting at a distance  of, 

`=(71691**0.5429-14126**0.4)/57565`

`=0.578 m` above the  hinge.

Let `F ` is the force required on the top of the gate to open it. Now, taking moment about `F,F_1,F_2`, we get,

`F**1.2+F_2**0.4=F_1**0.5429`

Thus,`F=27725.5 N`