# Problem 266 | Resultant of Non-Concurrent Force System

**Problem 266**

Determine the resultant of the three forces acting on the dam shown in Fig. P-266 and locate its intersection with the base AB. For good design, this intersection should occur within the middle third of the base. Does it?

**Solution 266**

$R_x = 10\,000 - 6000 \sin 60^\circ$

$R_x = 4803.85 \, \text{ lb to the right}$

$R_y = \Sigma F_y$

$R_y = 24\,000 + 6000 \cos 60^\circ$

$R_y = 27\,000 \, \text{ lb downward}$

$R = \sqrt{{R_x}^2 + {R_y}^2}$

$R = \sqrt{4803.85^2 + 27\,000^2}$

$R = 27\,424.02 \, \text{ lb}$

$\tan \theta_x = \dfrac{R_y}{R_x}$

$\tan \theta_x = \dfrac{27\,000}{4803.85}$

$\theta_x = 79.91^\circ$

Righting moment

$RM = 24\,000(18 - 7) + 6000(4)$

$RM = 288\,000 \, \text{ lb}\cdot\text{ft}$

Overturning moment

$OM = 10\,000(6)$

$OM = 60\,000 \, \text{ lb}\cdot\text{ft}$

Moment at the toe (downstream side - point B)

$M_B = RM - OM$

$M_B = 288\,000 - 60\,000$

$M_B = 228\,000 \, \text{ lb}\cdot\text{ft counterclockwise}$

Location of R_{y} as measured from the toe

$R_yx = M_B$

$27\,000x = 228\,000$

$x = 8.44 \, \text{ ft to the left of B}$ (within the middle third)

Thus, R = 27 424.02 lb downward to the right at θ_{x} = 79.91° and passes through the base at 8.44 ft to the left of B which is within the middle third.

Related post: Foundation (Soil) pressure of gravity dam.