1. Buoyancy Buoyant force vs. Weight Apparent weight
Apparent weight – Example 10-7
Measuring Density – Example 10-8
Partially submerged objects
Partially submerged Iceberg – Example
Hydrometer – Example 10-9
Helium Balloon – Example 10-10

2. Basic Buoyancy Force on top surface 𝐹 1 = 𝑃 1 𝐴=𝜌𝑔 ℎ 1 𝐴
𝐹 1 = 𝑃 1 𝐴=𝜌𝑔 ℎ 1 𝐴
Force on bottom surface
𝐹 2 = 𝑃 2 𝐴=𝜌𝑔 ℎ 2 𝐴
Force difference
( 𝐹 2 − 𝐹 1 )= (𝑃 2 − 𝑃 1 )𝐴
=𝜌𝑔 (ℎ 2 − ℎ 1 ) 𝐴
Since h2 > h1 upward
Side forces cancel
Buoyant force
𝐹 2 − 𝐹 1 = 𝐹 𝐵 = 𝜌𝑔 (ℎ 2 − ℎ 1 ) 𝐴 = 𝜌𝑔 𝑉 𝑑𝑖𝑠𝑝𝑙 = 𝑚 𝑑𝑖𝑠𝑝𝑙 𝑔

3. Archimedes' Principal Buoyant force up: 𝐹 𝐵 =𝜌 𝑉 𝑑𝑖𝑠𝑝 𝑔= 𝑚 𝑑𝑖𝑠𝑝 𝑔
Gravity force down: 𝐹 𝑊 =𝑚𝑔
Total (+ up) 𝐹= 𝐹 𝐵 − 𝐹 𝑊 = 𝑚 𝑑𝑖𝑠𝑝 𝑔−𝑚𝑔
Note: if volume filled with same fluid – Total force neutral

4. Example 1 – Apparent weight
Step 1 – empty statue volume of water
𝐹 𝐵𝑢𝑜𝑦 = 𝜌 𝑤𝑎𝑡𝑒𝑟 𝑉 𝑤𝑎𝑡𝑒𝑟 𝑔
=(1000 𝑘𝑔 𝑚 3 ) (3∙ 10 −2 𝑚 3 ) (9.8 𝑚 𝑠 2 )
=294 𝑁
Step 2 – refill statue volume with statue
𝐹 𝑠𝑡𝑎𝑡𝑢𝑒 = 𝑚 𝑠𝑡𝑎𝑡𝑢𝑒 𝑔=686 𝑁
Step 3 – apparent force down
𝐹 𝑎𝑝𝑝𝑡 = 𝐹 𝑠𝑡𝑎𝑡𝑢𝑒 − 𝐹 𝐵𝑢𝑜𝑦 =392 𝑁

5. Example 2 - Measure density
Apparent vs. real weight.
𝐹 𝑎𝑝𝑝𝑡 = 𝐹 𝑊 − 𝐹 𝐵
13.4 𝑘𝑔 𝑔=14.7 𝑘𝑔 𝑔 − 𝑚 𝑑𝑖𝑠𝑝 𝑔
Mass of water displaced.
𝑚 𝑑𝑖𝑠𝑝 =14.7 𝑘𝑔 −13.4 𝑘𝑔
=1.3 𝑘𝑔
Volume of water displaced.
𝑉= 𝑚 𝜌 = 1.3 𝑘𝑔 𝑘𝑔 𝑚 3 = 𝑚 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟 =1.3 𝐿= 𝑚 3

6. Example 2 – Density (cont)
Volume of crown equals volume of water displaced.
𝑉 𝑐𝑟𝑜𝑤𝑛 = 𝑉 𝑑𝑖𝑠𝑝 = 𝑚 3
Density of crown
𝜌= 𝑚 𝑐𝑟𝑜𝑤𝑛 𝑉 𝑐𝑟𝑜𝑤𝑛
𝜌= 14.7 𝑘𝑔 𝑚 3
=11,308 𝑘𝑔 𝑚 3

7. Partially submerged objects
Density less than water
1200 kg log with volume 2 m3
ρ = 600 kg/m3
(a) Totally submerged – buoyant force exceeds weight
(b) Partially submerged – buoyant force equals weight

8. Partially submerged objects (cont)
Buoyant force in fluid (partial volume)
𝐹 𝐵𝑢𝑜𝑦 = 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑔
Weight of object (total volume)
𝐹 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚 𝑤𝑜𝑜𝑑 𝑔
Equating
𝐹 𝐵𝑢𝑜𝑦 = 𝐹 𝑤𝑒𝑖𝑔ℎ𝑡
𝜌 𝑤𝑎𝑡𝑒𝑟 𝑉 𝑤𝑎𝑡𝑒𝑟 𝑔= 𝜌 𝑤𝑜𝑜𝑑 𝑉 𝑤𝑜𝑜𝑑 𝑔
𝑉 𝑤𝑎𝑡𝑒𝑟 𝑉 𝑤𝑜𝑜𝑑 = 𝜌 𝑤𝑜𝑜𝑑 𝜌 𝑤𝑎𝑡𝑒𝑟

9. Example 3 - Iceberg The old iceberg problem
𝑉 𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟 𝑉 𝑖𝑐𝑒 = 𝜌 𝑖𝑐𝑒 𝜌 𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟
Table 10-1
𝜌 𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟 =1025 𝑘𝑔/ 𝑚 3
𝜌 𝑖𝑐𝑒 =917 𝑘𝑔/ 𝑚 3
Equating
𝑉 seawater 𝑉 𝑖𝑐𝑒 = 917𝑘𝑔/ 𝑚 𝑘𝑔/ 𝑚 3 =0.89
only 0.11 above water (That’s just the tip of the iceberg!)

10. Example 4 – Hydrometer Winemaker’s tool
Effective density of hydrometer
𝜌 𝑒𝑓𝑓 = 45 𝑔 25 𝑐𝑚 2 𝑐𝑚 2
=0.9 𝑔 𝑐𝑚 3
OK to use g/cm3, since conversion will cancel

11. Example 4 – Hydrometer Submerged volume ratio
𝑉 𝑤𝑎𝑡𝑒𝑟 𝑉 𝑡𝑜𝑜𝑙 = 𝜌 𝑡𝑜𝑜𝑙 𝜌 𝑤𝑎𝑡𝑒𝑟
𝑉 𝑤𝑎𝑡𝑒𝑟 𝑉 𝑡𝑜𝑜𝑙 = 0.9 𝑔 𝑐𝑚 𝑔 𝑐𝑚 3
=0.9
Hydrometer should be 0.9 submerged in water
Mark at 0.9 * 25 or 22.5 cm

12. Example 5 – Balloon Buoyancy in a “pool” of air
Step 1 – empty balloon volume of air
𝐹 𝐵𝑢𝑜𝑦 = 𝜌 𝑎𝑖𝑟 𝑉 𝑎𝑖𝑟 𝑔
=(1.29 𝑘𝑔 𝑚 3 )𝑉𝑔
Step 2 – fill balloon with helium
add load weight
𝐹 𝑊𝑒𝑖𝑔ℎ𝑡 = 𝜌 𝐻𝑒 𝑉 𝐻𝑒 𝑔 + 𝑚 𝑜 𝑔
=( 𝑘𝑔 𝑚 3 )𝑉𝑔 +180 𝑘𝑔 𝑔

13. Example 6 – Balloon (cont)
Buoyancy in a “pool” of air
Equating up and down forces
(1.29 𝑘𝑔 𝑚 3 )𝑉𝑔
=( 𝑘𝑔 𝑚 3 )𝑉𝑔 +180 𝑘𝑔 𝑔
Solving for V
𝑉= 180 𝑘𝑔 𝑘𝑔 𝑚 3 − 𝑘𝑔 𝑚 3
𝑉=160 𝑚 3