Figure shows a cross sectional view of a masonry dam whose length perpendiclar to te diagram is `30 m`. The depth of the water behind te dam is `10 m`. the masonry of which is the dam is constructed has a density of `3000 kg//m^(3)`.
a. Find the dimensions `x` and `2x` if the weight of the dam is to be `10` times the horizontal force exerted on it by water.
b. Check the stability of the dam with respect to its overturning about the edge through point `O`.

Explanation

a. Weight of dam` =` volume `xx` density `xx g`
`=[1/2(x+2x)12xx30]3000xx10`
`=162x xx10^(5)N`
Horizontal thrust due of water (on a vertical surface)
`=` pressure at` Gxx`Area
`=5xx10^(3)xx10xx(30x10)=15xx10^(6)N`
According to the problem `162x xx10^(6)=10(15x10^(6))`
`impliesx=9.26m` and `2x=18.52m`
b. Let `G` be the centre of gravity of dam. Taking `O` as origin and the origin and the horizontal axis towards left hand side of `O` as `+x` -axis, let r be the distance of centre of gravity from `O`
`r=((12x)(3x)/2+(6x)(2x)/3)/(12x+6m)=11/9x`
Anticlockwise torque exerted by the weight of the dam is
`W(11/9x)=162xx10^(6)xx11/9x^(2)=169.78xx10^(9)Nm`
clockwise torque due to liquid thrust
`int_(0)^(10) (xrhog)(30dx)(10-x)=30rhog|5x^(2)-(x^(3))/3|_(0)^(10)`
`5000 rhog=5x10^(9)Nm`
As the torque due to the liquid thrust is much less than that due to the weight of the dam, the dam is stable.