Question

the face of a dam is vertical to a depth of 5 m below the water surface and then slopes at 30 degrees to the vertical. If the depth of the water is 9 m,completely specify the resultant force per metre run acting on the whole face. Use equilibrium of vector to find the final answer

Answer 1

Depth$=5m$

$\theta=30^0$

Depth$=9m$

Projected area ,$A_p=h*w=9*1=9m^2$

$h=\frac{9}{2}=4.5m$

Volume above the inclined plate, $V=[Xh_2+\frac{h_1x}{2}]w$

V$=(\frac{7.5}{tan60}*5+\frac{1}{2}*7.5*\frac{7.5}{tan60})=37.9m^3$

Horizontal component of force on the dam

$F_x=r_wAP\bar{h}=9810*9*4.5=397305$

Vertical component

$F_y=r_w*v=9810*37.9=159316.2N$

Resultant, R$=\sqrt{F_x^2+F_y^2}{}=\sqrt{397305^2+159316^2}$

$=1641955.06N$

$tan\theta=\frac{F_y}{F_x}=\frac{159316.2}{397305}=0.4$

$\theta=tan^{-1}0.4=21.8$ from positive axis in clockwise direction