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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.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q336

HGE - Hydraulics / Buoyancy / Engr. Janclyde Espinosa (Clidez)

A barge delivers 25mm reinforcing steel bars, 6m long. It travels from the Manila Bay (salt water) and unload its cargo at a point in Pasig River (fresh water). From salt to fresh water, the barge sinks 50mm, and it rises 250mm after unloading the cargo. The volume submerged in salt water is 346m3.

What is the draft in sea water in meters?

  1. 1.67
  2. 1.33
  3. 2.67
  4. 2.33

What is the weight of the barge in kN?

  1. 2993.74
  2. 3876.23
  3. 3496.09
  4. 2874.56

How many pieces of 25mm rebars are unloaded if the specfic gravity is 7.8?

  1. 2229
  2. 2260
  3. 2248
  4. 2237
For the same load, displacement in salt water equals displacement in fresh water after the barge sinks 0.05 m:
$\gamma_sAT_s=\gamma_wA(T_s+0.05)$
Using seawater $s.g.=1.03$:
$1.03T_s=T_s+0.05$
$0.03T_s=0.05$
$\boxed{T_s=1.67\text{ m}}$

Question Bank: q343

HGE - Hydraulics / Buoyancy / Engr. Janclyde Espinosa (Clidez)

A ship with vertical sides near the waterline weighs 40MN including its cargo and has a draft of 6.7 meters in seawater (s.g. = 1.026). Unloading 2 MN of its cargo, the draft decreases to 6.4m. With its cargo reduced, the ship enters a harbor of fresh water. Evaluate the draft of the ship in fresh water, in meters.

Answer:

  1. 6.56m
  2. 7.34m
  3. 4.63m
  4. 5.78m
After unloading, the ship draft in seawater is 6.4 m. For vertical sides, displaced volume is proportional to draft. For the same reduced load in fresh water:
$\gamma_wAT_f=\gamma_sA(6.4)$
$T_f=1.026(6.4)=6.566$ m
$\boxed{T_f\approx 6.56\text{ m}}$

Question Bank: q351

HGE - Hydraulics / Buoyancy / Engr. Janclyde Espinosa (Clidez)

A sphere of radius 400mm is immersed in seawater (s.g. = 1.026) by anchoring it to the bottom of the seabed. The mooring line was observed to have a tension of 800N. Evaluate the specific weight of the sphere, in kN/m3.

q351

Answer:

  1. 7.08
  2. 7.56
  3. 6.39
  4. 6.04
The anchored sphere is fully submerged. Buoyancy acts upward, while weight and cable tension act downward:
$B=W+T$ so $W=B-T$
$V=\frac{4}{3}\pi(0.4)^3=0.2681$ m3
$B=1.026(9.81)(0.2681)(1000)=2698.27$ N
$W=2698.27-800=1898.27$ N
$\gamma_s=\frac{W}{V}=7080.9$ N/m3
$\boxed{\gamma_s=7.08\text{ kN/m}^3}$

Question Bank: q352

HGE - Hydraulics / Buoyancy / Engr. Janclyde Espinosa (Clidez)

A cylinder having a diameter of 1.20m and weighing 800N is held in position in sea water by a wire tied to an anchor block resting at the bottom of the sea such that 0.30m of the cylinder is below the surface of the water with its axis vertical. The anchor block has a volume of 0.50 cubic meter and weighs 24 kN per cubic meter in air. Assume sea water to have a specific gravity = 1.03. Neglecting the weight and volume of the cable,

Evaluate the buoyant force on the cylinder for the position described, in kN.

  1. 3.43
  2. 4.21
  3. 3.56
  4. 4.68

Evaluate the tensile force in the wire when the top of the cylinder is 0.60m above the water surface, in kN.

  1. 2.628
  2. 2.876
  3. 2.475
  4. 2.583

Evaluate the rise in the tide that will lift the anchor from the bottom of the sea, in meters.

  1. 0.378
  2. 0.426
  3. 0.353
  4. 0.417
The buoyant force on the cylinder equals the weight of displaced seawater:
$B=\gamma_{sw}V_{sub}$
$V_{sub}=\frac{\pi(1.20)^2}{4}(0.30)=0.3393$ m3
$B=1.03(9.81)(0.3393)=3.428$ kN
$\boxed{B\approx3.43\text{ kN}}$

Question Bank: q358

HGE - Hydraulics / Buoyancy / Engr. Janclyde Espinosa (Clidez)

A container holds two layers of different liquids, one having specific gravity of 1.2 and the other having a specific gravity of 1.5. A solid spherical metal having a diameter of 200mm and specific gravity of 7.4 is submerged in such a manner that half of the sphere is on the top layer and the other half in the bottom layer of liquid.

Evaluate the buoyant force acting on the sphere, in kN.

  1. 0.055
  2. 0.249
  3. 0.667
  4. 0.286

Determine the tension in the cable attached to the sphere to normal position, in kN.

  1. 0.249
  2. 0.055
  3. 0.667
  4. 0.286

If both liquids are water, evaluate the buoyant force acting on the sphere, in kN.

  1. 0.0411
  2. 0.0511
  3. 0.0611
  4. 0.0311

Part 1.

Half of the sphere is in liquid with s.g. 1.2 and half is in liquid with s.g. 1.5. The buoyant force is the sum of the displaced liquid weights:
$B=\gamma_wV\left(\frac{1.2+1.5}{2}\right)$
$V=\frac{4}{3}\pi(0.10)^3=0.004189$ m3
$B=9.81(0.004189)(1.35)=0.0555$ kN
$\boxed{B\approx0.055\text{ kN}}$

Part 2.

The sphere weight is based on its specific gravity:
$W=7.4\gamma_wV=7.4(9.81)\left[\frac{4}{3}\pi(0.10)^3\right]=0.3041$ kN
From the two-liquid displacement, $B=0.0555$ kN. The cable tension is:
$T=W-B=0.3041-0.0555=0.2486$ kN
$\boxed{T\approx0.249\text{ kN}}$

Part 3.

If both liquids are water, the buoyant force is simply the weight of displaced water:
$B=\gamma_wV$
$B=9.81\left[\frac{4}{3}\pi(0.10)^3\right]=0.0411$ kN
$\boxed{B=0.0411\text{ kN}}$

Question Bank: q362

HGE - Hydraulics / Buoyancy / Engr. Janclyde Espinosa (Clidez)

A ship having a displacement of 20000 metric tones enters a harbor of fresh water. The ship captain recorded a draft of 8.4 m. while the ship was still in seawater (sp.gr. = 1.03). Obtain the draft in meters of the ship in fresh water if the horizontal section of the ship below the water line is 3000 m3 in both instances.

Answer:

  1. 8.594
  2. 8.206
  3. 8.608
  4. 8.624
When the ship moves from seawater to fresh water, the extra displacement volume needed is:
$\Delta V=20{,}000-\frac{20{,}000}{1.03}=582.52$ m3
With waterplane area $A=3000$ m2:
$\Delta T=\frac{582.52}{3000}=0.1942$ m
$T_f=8.4+0.1942=8.594$ m
$\boxed{T_f=8.594}$

Question Bank: q363

HGE - Hydraulics / Buoyancy / Engr. Janclyde Espinosa (Clidez)

An iceberg having a specific gravity of 0.90 floats in salt water having a specific gravity of 1.03. If the volume of the ice above the surface is 500 m3, what is the volume of the iceberg in m3?

Answer:

  1. 3962
  2. 3926
  3. 3629
  4. 3296
For floating equilibrium:
$\gamma_iV=\gamma_sV_{sub}$
$V_{sub}=\frac{0.90}{1.03}V$
The volume above water is:
$V_{above}=V-V_{sub}=V\left(1-\frac{0.90}{1.03}\right)=500$
$V=3961.54$ m3
$\boxed{V\approx3962}$

Question Bank: v18

HGE - Hydraulics / Buoyancy / HGE May 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A vertical cylinder of diameter 1.1 m, height 1 m, and weight 0.9 kN is held in sea water of specific gravity 1.03 by a cable. Use $\gamma_w=9.81$ kN/m$^3$.

If 0.45 m of the cylinder is submerged, evaluate the buoyant force on the cylinder, in kN.

  1. 4.88 kN
  2. 5.36 kN
  3. 3.20 kN
  4. 4.32 kN

Evaluate the cable tension when the top of the cylinder is 0.7 m above the water surface, in kN.

  1. 1.98 kN
  2. 3.07 kN
  3. 2.54 kN
  4. 1.70 kN

The cable anchors to a block of volume 0.5 m3 and unit weight in air 24 kN/m3 on the sea bed, with the cylinder initially 0.3 m submerged. Evaluate the rise in tide that will just lift the anchor, in meters.

  1. 0.567 m
  2. 0.517 m
  3. 0.617 m
  4. 0.667 m

The unit weight of sea water is $\gamma_{sw}=9.81\,SG$.

Part 1 — Buoyant force. It equals the weight of displaced sea water:

$$B=\gamma_{sw}\left(\frac{\pi D^{2}}{4}\,h_{sub}\right).$$

Part 2 — Cable tension. The submerged length is $h=H_{cyl}-\text{above}$, so by vertical equilibrium

$$T=B-W_c=\gamma_{sw}\frac{\pi D^{2}}{4}\,h-W_c.$$

Part 3 — Tide to lift the anchor. The anchor lifts when the cable tension equals its submerged weight, $T_{max}=W_a-B_a$ with $W_a=\gamma_a V_a$ and $B_a=\gamma_{sw}V_a$. The cylinder must then supply buoyancy $B_c=T_{max}+W_c$, i.$e$. submergence

$$h_{req}=\frac{B_c}{\gamma_{sw}\,\pi D^{2}/4},\qquad \text{rise}=\left|h_{req}-h_0\right|.$$

Computed answers:
1. 4.32 kN
2. 1.98 kN
3. 0.517 m

Question Bank: v53

HGE - Hydraulics / Buoyancy / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A cube 0.3 m on each side is held in equilibrium under water by attaching lightweight foam. The unit weights of the cube and foam are 20 kN/m3 and 1.1 kN/m3. Determine the minimum foam volume.

  1. 0.0262 m³
  2. 0.0316 m³
  3. 0.0553 m³
  4. 0.0193 m³
Minimum foam occurs when both bodies are fully submerged. Vertical equilibrium gives $$\gamma_fV_f+\gamma_cV_c=\gamma_w(V_f+V_c),$$ so $$V_f=\frac{(\gamma_c-\gamma_w)V_c}{\gamma_w-\gamma_f}=\frac{(20-9.81)(0.027)}{9.81-1.1}=0.0315878300804\text{ m}^3.$$
Computed answer: 0.0316 m³

Question Bank: v54

HGE - Hydraulics / Buoyancy / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A square timber 0.6 m on each side and 1.9 m high floats vertically in water with 0.3 m outstanding. Evaluate its weight in kN.

  1. 7.23 kN
  2. 2.99 kN
  3. 5.65 kN
  4. 4.52 kN
A floating body weighs the same as the displaced water: $$W=\gamma_wV_D=\gamma_wb^2(H-h_f)=9.81(0.6)^2(1.9-0.3)=5.65056\text{ kN}.$$
Computed answer: 5.65 kN

Question Bank: v55

HGE - Hydraulics / Buoyancy / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A wood block of specific gravity 0.84 and volume 0.16 m3 floats across an upper liquid of specific gravity 0.45 and water below. Determine the volume submerged in the upper liquid.

  1. 0.047 m³
  2. 0.065 m³
  3. 0.084 m³
  4. 0.028 m³
Let $V_u$ be in the upper liquid and $V-V_u$ in water. Buoyancy equals weight: $$(V-V_u)+SG_uV_u=SG_wV.$$ Hence $$V_u=V\frac{1-SG_w}{1-SG_u}=0.16\frac{1-0.84}{1-0.45}=0.0465454545455\text{ m}^3.$$
Computed answer: 0.047 m³

Question Bank: v56

HGE - Hydraulics / Buoyancy / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A sunken boat has an effective submerged weight of 13 kN. Find the minimum volume of an attached balloon needed to raise it in seawater of specific gravity 1.03. Neglect balloon and air weight.

  1. 1.29 m³
  2. 0.63 m³
  3. 1.71 m³
  4. 2.84 m³
At impending lift, balloon buoyancy equals the effective boat weight: $$\gamma_wSG_{sw}V=W,$$ so $$V=\frac{13}{9.81(1.03)}=1.28658096058\text{ m}^3.$$
Computed answer: 1.29 m³

Question Bank: v57

HGE - Hydraulics / Buoyancy / HGE November 2019

Formula-mode item rendered with fixed values for lecture/PDF export.

A ship displacing 28000 metric tons has draft 12 m in seawater of specific gravity 1.04. Its waterline area is 2700 m2. Find its draft in fresh water.

  1. 12.27 m
  2. 12.56 m
  3. 12.40 m
  4. 11.84 m
The ship must displace more fresh water: $$\Delta V=W\left(\frac1{\rho_{fw}}-\frac1{\rho_{sw}}\right)=1076.92307692\text{ m}^3.$$ The draft increase is $\Delta d=\Delta V/A_w=0.39886039886$ m, hence $$d_{fw}=12+0.39886039886=12.3988603989\text{ m}.$$
Computed answer: 12.40 m