Check: only the base width (15 m) and overall height (20 m) are dimensioned on the sketch, with a vertical upstream face and a sloped downstream face; the cross-section is therefore modelled as a right triangle (vertical face at the heel, sloped face down to the toe O), the standard shape for this class of textbook gravity-dam problem. No crest width is given, so this is the minimal shape consistent with the sketch.
Right-triangle dam section: vertical upstream (water) face, sloped downstream face to toe O. The resultant of weight and hydrostatic thrust crosses the base 7.30 m from O, inside the shaded middle-third band (5–10 m).
Find. The horizontal distance from the toe O at which the resultant force crosses the base, and whether it lies inside the middle third of the base.
Approach. Resolve the loading into the horizontal hydrostatic thrust on the vertical upstream face and the dam's own weight, take moments of both about the toe O, and divide the net moment by the total vertical force (the weight) to locate where the resultant crosses the base.
Hydrostatic thrust on the vertical face. Per metre length of dam,
$$F_H=\tfrac12\rho_w g h^2=\tfrac12(1000)(9.81)(18)^2=\boxed{1589.2\ \text{kN/m}}$$
acting at $h/3=6.00$ m above the base.
Weight of the dam. The triangular cross-section has area $\tfrac12 BH=\tfrac12(15)(20)=150\ \text{m}^2$, so per metre length
$$W=\rho_c g\left(\tfrac12BH\right)=(2400)(9.81)(150)=\boxed{3531.6\ \text{kN/m}}$$
with its centroid $\tfrac13B=5.00$ m from the heel, i.e. $x_W=\tfrac23B=10.00$ m from the toe O.
Moments about the toe O. The weight's resisting moment and the thrust's overturning moment (arm = height of application above the base) are
$$M_{resist}=Wx_W=(3531.6)(10.00)=35{,}316\ \text{kN}\cdot\text{m/m}\qquad M_{over}=F_H\!\left(\tfrac{h}{3}\right)=(1589.2)(6.00)=9535.3\ \text{kN}\cdot\text{m/m}$$
Location of the resultant on the base. The net moment divided by the total vertical force (weight only — the vertical face carries no vertical hydrostatic component) gives
$$\bar{x}=\frac{M_{resist}-M_{over}}{W}=\frac{35{,}316-9535.3}{3531.6}=\boxed{7.30\ \text{m from O}}$$
Middle-third check. The middle third of a 15 m base spans from $B/3=5.00$ m to $2B/3=10.00$ m from either edge. Since $5.00\lt7.30\lt10.00$, the resultant lies inside the middle third: the eccentricity from the base centre is only $e=|B/2-\bar{x}|=0.20\ \text{m}\ll B/6=2.50\ \text{m}$, so the base pressure stays compressive everywhere ($\sigma_{toe}=254.3$ kPa, $\sigma_{heel}=216.6$ kPa, both positive) and $\boxed{\text{the dam is SAFE}}$.
