CE Board Exam Randomizer

⬅ Back to Subject Topics

Archimedes' Principle

The buoyant force on a submerged or floating body is equal to the weight of the fluid displaced by the body.

$$BF = \gamma_f V_d$$ $$W = \gamma_b V_b$$

For floating equilibrium, the buoyant force equals the weight of the floating body.

$$BF = W$$ $$\gamma_f V_d = \gamma_b V_b$$

Floating, Sinking, and Draft

A body floats if buoyant force can balance its weight. It sinks if its weight is greater than the buoyant force available when fully submerged.

$$BF > W \Rightarrow \text{body tends to rise}$$ $$BF = W \Rightarrow \text{floating or neutral equilibrium}$$ $$BF < W \Rightarrow \text{body sinks}$$

For prismatic floating bodies, draft is found from the displaced volume.

$$V_d = A d$$ $$d = \frac{W}{\gamma_f A}$$

Apparent Weight and Submerged Force

When an object is submerged, its apparent weight is reduced by the buoyant force.

$$W_{app} = W - BF$$ $$BF = W - W_{app}$$ $$V = \frac{BF}{\gamma_f}$$

If a body must be held submerged, the required external force depends on whether the body tends to float up or sink down.

$$F + W = BF \quad \text{for downward holding force}$$ $$F + BF = W \quad \text{for upward lifting force}$$

Problem: Rectangular Scow Draft

A rectangular scow 5 m by 10 m with vertical sides weighs 40 tonnes. Determine its draft in fresh water and in sea water with $SG = 1.025$.

$$BF = W$$ $$5(10)d(1) = 40$$ $$d = 0.80 \text{ m}$$ $$(1.025)(5)(10)d = 40$$ $$d = 0.78 \text{ m}$$

Answer: Draft is 0.80 m in fresh water and 0.78 m in sea water.

Problem: Iceberg Volume

An iceberg with $SG = 0.90$ floats in salt water with $SG = 1.03$. If the volume above the surface is $500 \text{ m}^3$, find the total volume of the iceberg.

$$V_d = V - 500$$ $$BF = W$$ $$1.03(9.81)(V - 500) = 0.90(9.81)V$$ $$1.03V - 515 = 0.90V$$ $$V = 3962 \text{ m}^3$$

Answer: The iceberg volume is approximately $3962 \text{ m}^3$.

Problem: Irregular Object Volume

An irregular object weighs 300 N in air and 230 N when submerged in fresh water. Determine its volume.

$$BF = W - W_{app}$$ $$BF = 300 - 230 = 70 \text{ N}$$ $$BF = \gamma_w V$$ $$70 = 9810V$$ $$V = 0.00714 \text{ m}^3$$

Answer: The object volume is $0.00714 \text{ m}^3$.

Problem: Composite Floating Body — Wood with Steel Attachment

A wooden log has volume 0.50 m³ and specific gravity 0.60. A steel anchor piece (SG = 7.85) is attached to the bottom of the log so that the combined assembly just barely floats in fresh water (buoyant force exactly equals total weight). Determine the required volume of the steel anchor.

$$W_{wood} = \gamma_w (SG_{wood}) V_{wood} = 9.81(0.60)(0.50) = 2.943 \text{ kN}$$ $$W_{steel} = 9.81(7.85) V_s = 76.99 V_s$$

For just-floating equilibrium, buoyant force on the whole submerged system equals total weight. The assembly is fully submerged when just floating (barely):

$$BF = \gamma_w (V_{wood} + V_s) = 9.81(0.50 + V_s)$$ $$BF = W_{wood} + W_{steel}$$ $$9.81(0.50 + V_s) = 2.943 + 76.99 V_s$$ $$4.905 + 9.81 V_s = 2.943 + 76.99 V_s$$ $$1.962 = 67.18 V_s$$ $$V_s = 0.02920 \text{ m}^3$$

Answer: A steel piece of 0.0292 m³ (volume) is required. Any larger steel piece would cause the assembly to sink.

Problem: Concrete Block Anchored to the Bottom by a Chain

A concrete block (SG = 2.40) with volume 0.25 m³ rests on the bottom of a freshwater lake. A steel chain holds it to a pile. The block tends to float because a hollow chamber inside it gives it a net SG of only 0.85. Determine the tension in the chain required to hold the block submerged.

$$W = \gamma_w(SG_{net}) V = 9.81(0.85)(0.25) = 2.085 \text{ kN (downward)}$$ $$BF = \gamma_w V = 9.81(0.25) = 2.453 \text{ kN (upward)}$$

Since BF > W, the block tends to rise. The chain must pull it down:

$$\sum F = 0: \quad T + W = BF$$ $$T = BF - W = 2.453 - 2.085 = 0.368 \text{ kN} = 368 \text{ N}$$

Answer: The chain tension is 368 N. The chain must be attached at the bottom to hold the buoyant block down.

Problem: Barge — Maximum Cargo Load Before Sinking

A steel barge is 8.0 m long, 3.0 m wide, and has vertical sides and bottom. The barge itself weighs 50 kN. The freeboard (distance from waterline to top of barge walls) is 0.40 m when empty in fresh water. Determine: (a) the draft when empty, and (b) the maximum additional cargo load the barge can carry without sinking.

$$BF = W: \quad \gamma_w(8.0)(3.0)d = 50$$ $$9.81(24.0)d = 50 \Rightarrow d = \frac{50}{235.44} = 0.212 \text{ m (draft when empty)}$$ $$H_{total} = d + freeboard = 0.212 + 0.40 = 0.612 \text{ m (total barge depth)}$$

Maximum load: when the barge sinks to the gunwale (freeboard = 0), draft = total depth = 0.612 m.

$$BF_{max} = 9.81(8.0)(3.0)(0.612) = 144.43 \text{ kN}$$ $$W_{cargo,max} = BF_{max} - W_{barge} = 144.43 - 50 = 94.43 \text{ kN}$$

Answer: Empty draft is 0.212 m. Maximum additional cargo is 94.43 kN (≈9.63 tonnes) before the barge would take on water over the top of its sides.

Problem: Fraction of a Floating Log Submerged

A cylindrical log of length 4.0 m and diameter 0.30 m floats horizontally in a river (fresh water). The log has a specific gravity of 0.72. Determine: (a) the total weight of the log, (b) the fraction of the log volume that is submerged, and (c) whether the log would float in salt water (SG = 1.025) with more or less draft.

$$V_{log} = \frac{\pi(0.30)^2}{4}(4.0) = 0.2827 \text{ m}^3$$ $$W = \gamma_w(SG_{log})V = 9.81(0.72)(0.2827) = 1.995 \text{ kN}$$
$$BF = W: \quad \gamma_w V_d = 1.995 \text{ kN}$$ $$V_d = \frac{1.995}{9.81} = 0.2034 \text{ m}^3$$ $$\text{Fraction submerged} = \frac{V_d}{V_{log}} = \frac{0.2034}{0.2827} = 0.72 = SG_{log}$$

In salt water: $V_d = \frac{W}{\gamma_{sw}} = \frac{1.995}{9.81 \times 1.025} = 0.1984 \text{ m}^3$ (fraction = 0.70 — less submerged).

Answer: Log weighs 1.995 kN. 72% of its volume is submerged in fresh water (equal to its SG — this is always true for floating bodies). In salt water, only 70% is submerged — the log floats higher.