Damn that Dam!

• Find the net force due to the water level using calculus

*Pressure due to water level at a height y = $y$ $\rho$ $g$

• If the breadth of the water level is $l$ , the force acting at a small element at a level $dy$ from the surface,

$dF$ = $y$ $\rho$ $gl$ $dy$

• So, the net force is $F$ = $\int _{ 0 }^{ H }{ y\rho gl\quad dy }$

• Hence, the net force is $\frac { 1 }{ 2 } \rho gl\quad { H }^{ 2 }$

• Now, the torque due the force about the rightmost point of the dam tends to $overturn$ the dam.

• The small torque $d\tau$ due to force at a height $dy$ is $d\tau$ = $y$ $\rho$ $gl$ $(H - y)$ $dy$

• And the net torque is $\tau$ = $\int _{ 0 }^{ H }{ y\rho gl\quad (H-y)\quad dy }$

• $\tau$ = $\frac { 1 }{ 6 } \rho gl{ H }^{ 3 }$

• If the net force acts at a height h from the bottom level, net torque is net force multiplied by $h$ . $\tau$ = $h$ $\frac { 1 }{ 2 } \rho gl\quad { H }^{ 2 }$

• Equate the above two expressions to get $h$ = $\frac { 1 }{ 3 } H$

• Thus the height $x$ from the surface is $H$ $-$ $\frac { 1 }{ 3 } H$

• And hence the height $x$ is $\frac { 2 }{ 3 } H$ = $18$ $m$

this question is related to centre of mass, since force acts on point mass and on the center of mass of dam............so taking dam as right angled triangle so co-ordinates are (0,0) (0,27) (27,0)....and center of mass coincide with centroid so COM = (27/3 , 27/3)= (9,9) .........and question asks from surface of water so answer is 27-9 =18

• the load acting is UVL starting at 0 on water surface.
• UVL can be represented as triangle
• the cg in y direction is 2/3 rd the length in y direction from 0(water surface)
• 2/3 * 27 = 18