Question 3 of 13: Hinged Gate — Open or Remain Closed?
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2013-May. 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.
Reference texts: Crowe, C.T., Elger, D.F. & Roberson, J.A., Engineering Fluid Mechanics (the exam's own Moody chart and drag-coefficient chart are reproduced from this text); Douglas, J.F., Gasiorek, J.M., Swaffield, J.A. & Jack, L.B., Fluid Mechanics; White, F.M., Fluid Mechanics.
Check — assumptions used across this paper:
Where a question does not restate an ambient temperature, air is taken at the ISA sea-level standard of 15°C, giving ρair = 1.21 kg/m³ (the Constants table's 15°C value) — used in Q5, Q6, Q7 and Q9.
Q8's Moody chart and Q9's sphere drag-coefficient chart are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 (the Moody chart is a graphical solution of Colebrook–White) and the Morrison (2013) curve-fit for sphere drag versus Reynolds number for Q9 (which reproduces the published "Sphere" curve to within a few percent over this Re range).
Q1, Q3 and Q11 depend on their figures; the relevant crop of the exam page is reproduced beside each, and every reading used is taken from it.
Question 3: Hinged Gate — Open or Remain Closed? (5 marks)
Given. Water depth above pivot $x=2\ \text{m}$; gate width (into the page) $b=1.0\ \text{m}$; gate mass $m=400\ \text{kg}$; horizontal arm length $L=1.2\ \text{m}$; centre of gravity 0.3 m to the right of and 0.3 m above pivot O. From the figure, the reservoir is on the left of the vertical leaf. The horizontal arm runs to the right from O, and its tip bears up against a lip at the crest of a curved sill of radius 1.2 m about O. The space under the arm is open to the reservoir below O, so the underside of the arm is loaded by water at the pivot depth $x$. The top of the arm and the right face of the leaf are dry.
[Figure not reproduced: Source figure, Question 3 (exam page 3): L-shaped gate pivoted at O. See the official exam paper or the cited reference text.]
Exam figure (page 3). Water to the left of the leaf and under the arm. Clockwise rotation about O (leaf tipping downstream, arm tip swinging down along the curved sill) opens the gate. The lip at the sill crest stops it turning the other way.
Find. Whether the net moment about O rotates the gate clockwise, i.e. opens it.
Approach. Take moments about O for all three loads. The water thrust on the leaf acts clockwise (opening). The gate weight also acts clockwise, since the CG is to the right of O. The uplift of the water under the arm acts anticlockwise and holds the gate shut. The gate opens if the clockwise total wins.
Hydrostatic force on the leaf. Centroid depth $h_c=x/2=1.0\ \text{m}$, area $A=xb=2.0\ \text{m}^2$:
$$F_L = \rho_w g h_c A = 1000\times9.81\times1.0\times2.0 = 19{,}620\ \text{N}$$
Centre of pressure and leaf moment (clockwise, opening). $I_c = bx^3/12 = 0.667\ \text{m}^4$, so $y_{cp}=h_c+I_c/(h_cA) = 1.0+0.667/2.0 = 1.333\ \text{m}$ below the surface, i.e. $x/3 = 0.667\ \text{m}$ above O:
$$M_L = F_L\times\frac{x}{3} = \rho_w g b\frac{x^3}{6} = 19{,}620\times0.6667 = 13{,}080\ \text{N}\cdot\text{m}$$
Weight moment (clockwise, opening). Only the horizontal CG offset matters for a vertical force: $M_W = mg\times0.3 = 400\times9.81\times0.3 = 1177.2\ \text{N}\cdot\text{m}$.
Uplift under the arm (anticlockwise, closing). The arm is horizontal at the pivot depth, so its underside carries a uniform pressure $\rho_w g x$:
$$F_U = \rho_w g x\,bL = 1000\times9.81\times2.0\times1.0\times1.2 = 23{,}544\ \text{N},\qquad M_U = F_U\frac{L}{2} = 23{,}544\times0.6 = 14{,}126.4\ \text{N}\cdot\text{m}$$
Net moment. $M_{net} = M_L + M_W - M_U = 13{,}080 + 1177.2 - 14{,}126.4 = \boxed{+130.8\ \text{N}\cdot\text{m}\ \text{(clockwise)}}$. This is positive, so $\boxed{\text{the gate opens}}$, but only just.
How marginal. Setting $M_{net}=0$ gives $\rho_w g\,x^3/6 + 1177.2 = \rho_w g\,x L^2/2$, which solves to $x_{crit}\approx1.99\ \text{m}$. At 2 m the water is barely 1 cm above the opening depth. This is how an automatic flip gate works: the leaf moment grows as $x^3$ and the arm uplift only as $x$, so above the critical depth the leaf moment wins and the gate tips open.
