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

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

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

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

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 1000 𝑘𝑔 𝑚 3 =0.0013 𝑚 3 1.3 𝑘𝑔 𝑤𝑎𝑡𝑒𝑟 =1.3 𝐿=0.0013 𝑚 3

Example 2 – Density (cont) Volume of crown equals volume of water displaced. 𝑉 𝑐𝑟𝑜𝑤𝑛 = 𝑉 𝑑𝑖𝑠𝑝 =0.0013 𝑚 3 Density of crown 𝜌= 𝑚 𝑐𝑟𝑜𝑤𝑛 𝑉 𝑐𝑟𝑜𝑤𝑛 𝜌= 14.7 𝑘𝑔 0.0013 𝑚 3 =11,308 𝑘𝑔 𝑚 3

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

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

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 3 1025𝑘𝑔/ 𝑚 3 =0.89 only 0.11 above water (That’s just the tip of the iceberg!)

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

Example 4 – Hydrometer Submerged volume ratio 𝑉 𝑤𝑎𝑡𝑒𝑟 𝑉 𝑡𝑜𝑜𝑙 = 𝜌 𝑡𝑜𝑜𝑙 𝜌 𝑤𝑎𝑡𝑒𝑟 𝑉 𝑤𝑎𝑡𝑒𝑟 𝑉 𝑡𝑜𝑜𝑙 = 0.9 𝑔 𝑐𝑚 3 1.0 𝑔 𝑐𝑚 3 =0.9 Hydrometer should be 0.9 submerged in water Mark at 0.9 * 25 or 22.5 cm

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 =(0.179 𝑘𝑔 𝑚 3 )𝑉𝑔 +180 𝑘𝑔 𝑔

Example 6 – Balloon (cont) Buoyancy in a “pool” of air Equating up and down forces (1.29 𝑘𝑔 𝑚 3 )𝑉𝑔 =(0.179 𝑘𝑔 𝑚 3 )𝑉𝑔 +180 𝑘𝑔 𝑔 Solving for V 𝑉= 180 𝑘𝑔 1.29 𝑘𝑔 𝑚 3 − 0.179 𝑘𝑔 𝑚 3 𝑉=160 𝑚 3