Horizontal resistance to sliding
$\Sigma F_H = 0$
$R_x + F \cos 30^\circ = 1000$
$R_x + 600 \cos 30^\circ = 1000$
$R_x = 480.38 \, \text{ kN}$ answer
$\Sigma F_V = 0$
$R_y = W + F \sin 30^\circ$
$R_y = 2400 + 600 \sin 30^\circ$
$R_y = 2700 \, \text{ kN}$
Righting moment
$M_R = 11(2400) + 4(600)$
$M_R = 28\,800 \, \text{ kN}\cdot\text{m}$
Overturning moment
$M_O = 6(1000)$
$M_O = 6000 \, \text{ kN}\cdot\text{m}$
$\Sigma M_B = 0$
$\bar{x} \, R_y = M_R - M_O$
$\bar{x}(2700) = 28\,800 - 6000$
$\bar{x} = 8.44 \, \text{ m to the left of B}$
Eccentricity
$e = \frac{1}{2}B - x = \frac{1}{2}(18) - 8.44$
$e = 0.56 \, \text{ m}$
Foundation pressure (See Analysis of Gravity Dam for more information)
$p = \dfrac{R_y}{B} \left( 1 \pm \dfrac{6e}{B} \right)$
$p_1 = \dfrac{2700}{18} \left[ 1 - \dfrac{6(0.59)}{18} \right] = 122 \, \text{ kN/m}$ answer
$p_2 = \dfrac{2700}{18} \left[ 1 + \dfrac{6(0.59)}{18} \right] = 178 \, \text{ kN/m}$ answer
