Question 3 of 13: Sliding Friction Required to Hold a Concrete Dam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2013-Dec. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.
Air density is taken from the paper's own Constants table at the temperature each question states: 1.19 kg/m³ at 20°C (Q5's wind, Q7's inlet air).
Q1's touching-rod array is modelled as a repeating square unit cell of four mutually tangent rods (pitch = rod diameter, per the question's own "closely packed" wording), giving a curvilinear-square pore whose perimeter/area ratio drives the capillary rise.
Q8's Moody diagram and Q9's drag-coefficient diagram are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 and the Morrison (2013) curve-fit for sphere drag vs. Reynolds number for Q9.
Question 3: Sliding Friction Required to Hold a Concrete Dam (5 marks)
Trapezoidal dam cross-section, unit length into the page; vertical upstream face carries the full hydrostatic thrust horizontally.
Find. Minimum coefficient of friction μ between the dam base and its foundation.
Approach. Because the upstream face is vertical, the hydrostatic resultant is purely horizontal; compute it and compare against the friction resistance available from the dam's own self-weight (no uplift, so the full weight bears on the base).
Hydrostatic thrust. Per unit length, $F=\tfrac12\rho g h^{2}=\tfrac12(1000)(9.81)(4)^{2}=\boxed{78{,}480\ \text{N} = 78.48\ \text{kN}}$.
Dam weight. Trapezoidal cross-sectional area $A=\tfrac12(2+6)(5)=20\ \text{m}^2$, so
$$W = \rho_c g A = 2400\times9.81\times20 = \boxed{470{,}880\ \text{N} = 470.9\ \text{kN}}$$
Minimum friction coefficient. With no uplift the base normal force equals the full weight, so sliding equilibrium ($\mu W = F$) requires
$$\mu_{\min} = \frac{F}{W} = \frac{78{,}480}{470{,}880} = \boxed{0.167}$$