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MODULE 5. BUOYANCY In addition to the force of gravity or weight, all objects submerged in a fluid are acted on by a force BF. The buoyant force acts upward and is equal to the weight of the fluid displaced by the object. This is known as Archimedes Principle. The upward buoyant force also acts through the center of gravity or centroid of the displaced volume which is known as the center of buoyancy. 𝐵𝐹 = 𝛾𝑊 𝑉𝐷 𝑤ℎ𝑒𝑟𝑒 𝐵𝐹 = 𝑏𝑢𝑜𝑦𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒 𝑖𝑛 𝑁𝑒𝑤𝑡𝑜𝑛 𝑜𝑟 𝑘𝑖𝑙𝑜 𝑁𝑒𝑤𝑡𝑜𝑛 𝛾𝑊 = 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑖𝑛 𝑁/𝑚3 𝑜𝑟 𝑘𝑁/𝑚3 𝑉𝐷 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑖𝑛 𝑚3 For a freely floating object (with no external forces) the weight of the object (acting downward) is equal to the buoyant force on the object (acting upward). 𝑊 = 𝐵𝐹 𝑊 = 𝛾𝑊 𝑉𝐷 This equation id useful in determining what part of an object will float below the surface of a liquid. For objects partially submerged in a liquid and a gas, the buoyant force of the gas is usually neglected. However, the buoyant force on a totally submerged body in a gas is very important in the study of balloons.

Solved Problems: 1. A stone weighs 468 N in air. When submerged in water if weighs 298 N. a) Find the volume of the metal. b) Find the specific weight of the metal. c) Find the specific gravity of the metal. Solution: a) Volume of stone: 𝑊 = 468 − 298 𝑊 = 170 𝑊=𝑉𝐷 170 = 𝑉 (9810) 𝑽 = 𝟎. 𝟎𝟏𝟕𝟑 𝒎𝟑 b) Specific weight: 𝑆𝑝. 𝑤𝑡. =

468 0.0173

𝑆𝑝. 𝑤𝑡. = 27052

𝑺𝒑. 𝒘𝒕. = 𝟐𝟕. 𝟎𝟓

𝒌𝑵 𝒎𝟑

c) Specific gravity: 𝑆𝑝. 𝑔𝑟. =

27.05 9.81

𝑺𝒑. 𝒈𝒓. = 𝟐. 𝟕𝟔 2. A.) What fraction of the volume of a solid object of sp.gr. 7.3 floats above the surface of a container of mercury? B.) If the volume of the object below the liquid surface is 0.014 m3, what is the wt. of the object. C.) What load applied vertically hat would cause the object to be fully submerged? Solution:

V2

V1

Mercury (13.6)

A. Fraction of volume of a solid object above the mercury: 𝑊 = 𝐵𝐹 (7.3)(𝑉)(9.81) = 𝑉1 (9.81)(13.6) 𝑉 = 1.863 𝑉1 𝑉1 = 0.536 𝑉 𝑉2 = 𝑉 − 0.536 𝑉 𝑉2 = 0.464 𝑉 𝑽𝟐 = 𝟎. 𝟒𝟔𝟒 𝑽

B. Wt. of object: 𝑉1 = 0.014 𝑉 = 1.863(0.014)

𝑉 = 0.026 𝑚3 𝑊 = 0.026(9.81)(7.3) 𝑾 = 𝟏. 𝟖𝟔 𝒌𝑵 C. Load to cause the object to submerged: 𝑃 = 𝑉2 (9.81)(13.6) 𝑃 = 0.464(0.026)(9.81)(13.6) 𝑷 = 𝟏. 𝟔𝟏 𝒌𝑵 3. A hollow cylinder 1.0 m. in diameter and 2 m. long weighs 3825 N. a. How many kN of lead weighing 110 kN/m3 must be fastened to the outside bottom to make the cylinder float vertically with 1.50 m. submerged in freshwater. b. How many kN of lead weighing 110 kN/m3 must be placed inside the cylinder to make the cylinder float vertically with 1.50 m. submerged in freshwater. c. What additional load must be placed inside the cylinder to make the top of the cylinder flush with the water surface. Solution: w.s.

W1

2m 1.5 m BF2

w W2

Lead BF2

a. Load placed outside the cylinder: 𝑊1 + 𝑊2 = 𝐵𝐹1 + 𝐵𝐹2 3.825 + 𝑊2 =

𝜋 (1)2 (1.5)(9.81) + 𝑉(9.81) 4

𝑉=

𝑊 𝐷

𝑊2 𝑉= 110

3.825 + 𝑊2 = 11.56 +

𝑊2 (9.81) 110

0.91 𝑊2 = 7.735 𝑾𝟐 = 𝟖. 𝟓 𝒌𝑵 b. Load placed inside the cylinder: 1m w.s.

W1 1.5m Lead

W3 w

BF2 𝑊1 + 𝑊3 = 𝐵𝐹 3.825 + 𝑊3 =

𝜋 (1)2 (1.5)(9.81) 4

𝑾𝟑 = 𝟕. 𝟕𝟑 𝒌𝑵

c. Additional load to make the top flush with the water surface: 1m w.s. 3.825

W4

2m 7.73 BF2

w

Additional load

Lead

3.825 + 7.73 + 𝑊4 =

𝜋 (1)2 (2)(9.81) 4

𝑾𝟒 = 𝟑. 𝟖𝟓 𝒌𝑵 4. Piece of metal weighs 350 N in air and when it is submerged completely in water it weighs 240 N. a. Find the volume of the metal b. Find the specific weight of the metal c. Find the specific gravity of the metal Solution: a. Volume of metal: W = 350 -240 W = 110 N 110 = V(9810) V = 0.0112 m3 b. Specific weight: 350 Specific weight = 0.0112 Specific weight = 31250 Specific weight = 31.25 kN/m3

c. Specific gravity: weight in air

SG = weight of equal volume of H2O 350

SG = (9810)0.0112 SG = 3.19 5. A piece of wood floats in water with 50 mm projecting above the water surface. When placed in glycerin with a specific gravity 1.35, the block projects 75 mm above the liquid surface. a. Find the height of the piece of wood b. Find the specific gravity of wood c. Find the weight of the wood if it has a cross sectional area of 200 mm x 200 mm

Solution: 0.05

0.075

h - 0.05

h - 0.075

water

a. Height of wood: W = 9.81(A)(h – 0.05) Sh (9.81)(A) = 9.81A(h – 0.05) Sh = h – 0.05 W= 9.81(1.35) A (h – 0.075) Sh (9.81) A = 9.81(1.35) S (h – 0.075) Sh = 1.35(h – 0.075) h – 0.05 = 1.35 (h-0.075) 0.35h = 0.5125 h = 0.146 m b. Specific gravity of wood: SG h = h – 0.05 SG (0.146) = 0.146 – 0.05 SG = 0.658 c. Weight of wood: W = 9.81(0.658)(0.146)(0.2)(0.2) W = 0.038 kN W = 38 N

glycerin