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04-BS-7 · May 2015

Question 2 of 13: Force to Lift a Submerged Concrete-Block Gate

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-7 Mechanics of Fluids — May 2015 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs "do seven"; Section B (Analytical) offers 4 questions and instructs "do three." Every question is answered below (13 of 13), so students can use the full paper as a study resource.

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics (Ch. 2), the linear-momentum and energy equations (Ch. 3), pipe friction and the Moody chart (Ch. 6), and drag on immersed bodies (Ch. 7).

Question 2: Force to Lift a Submerged Concrete-Block Gate (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: the figure gives the plug diameter (2 m) and the two water depths (10 m fresh, 5 m sea) but no printed thickness for the concrete cylinder. This solution assumes a 1 m thick plug — the standard form of this classic textbook problem and consistent with the wall thickness drawn in the figure — and flags it explicitly rather than silently guessing.

Given.

QuantityValue
Fresh water depth above plug top10 m
Sea water depth above plug bottom5 m
Plug diameter2.0 m
Plug thickness (assumed, see the check note)1.0 m
Concrete density2400 kg/m³
Sea water density1025 kg/m³
Fresh water density1000 kg/m³

Find. The vertical force F needed to lift the plug clear of its seat.

Fresh waterSea waterF10 m5 m2.0 m diaF = 224.2 kN
Fig. Q2 — fresh water pushes down on the plug top, sea water pushes up on the plug bottom, and the plug's own weight acts down; F must overcome the net of all three.

Approach. Sum vertical forces on the plug as a free body: applied force up, sea-water pressure up on the bottom face, fresh-water pressure down on the top face, and weight down.

  1. Plug plan area. $$A = \frac{\pi}{4}D^2 = \frac{\pi}{4}(2.0)^2 = 3.1416\ \text{m}^2$$
  2. Weight of the concrete plug. $$W = \rho_{conc}\,g\,A\,t = 2400(9.81)(3.1416)(1.0) = 73\,966\ \text{N} = 73.97\ \text{kN}$$
  3. Pressure force pushing down on the top (fresh water, 10 m). $$F_{top} = \rho_{fresh}\,g\,h_f\,A = 1000(9.81)(10)(3.1416) = 308.19\ \text{kN}$$
  4. Pressure force pushing up on the bottom (sea water, 5 m). $$F_{bot} = \rho_{sea}\,g\,h_s\,A = 1025(9.81)(5)(3.1416) = 157.95\ \text{kN}$$
  5. Vertical equilibrium of the free plug (F acts up; weight and the top pressure force act down; the bottom pressure force acts up): $$F + F_{bot} = W + F_{top}$$ $$F = W + F_{top} - F_{bot} = 73.97 + 308.19 - 157.95 = \boxed{224.21\ \text{kN}}$$
QuantityResult
Plug weight73.97 kN
Fresh-water force (down)308.19 kN
Sea-water force (up)157.95 kN
Lift force F224.21 kN