Buoyancy The relationship is → B = ρVg Any time an object is immersed in a fluid, it will experience an upward force (buoyant force) from the fluid. (you feel lighter in a swimming pool.) The relationship is → B = ρVg Where: B = buoyant force (N) ρ = density of the fluid (kg/m3) V = volume of displaced fluid (m3) g = 9.8 m/s2

Archimedes’ Principle the volume of the displaced fluid is equal to the volume of the submerged portion of the displacing object. Problem solving applications Use B as you would any other force in ΣF calculations. (B always is directed up)

Examples Calculate the weight of a rock of mass 32 kg and density 2800 kg/m3 when fully submerged in water of density 1000 kg/m3 ? GIVEN mrock = 32 kg (Fg = 313.6 N) ρrock = 2800 kg/m3 g = 9.8 m/s2 ρfluid = 1000 kg/m3

The weight of the rock in the water is First the volume of the rock can be determined. ρ = m / V therefore V = m/ρ V = (32 kg)  ( 2800 kg/m3) = 1.14 x 10-2 m3 this is also the volume of the displaced fluid (since the rock is fully submerged) the buoyant force calculated as follows: B = ρVg therefore B = (1000) ( 1.14 x 10-2) (9.8) B = 112 N The weight of the rock in the water is therefore 112 N less than it is in air or 201.6 N

the buoyant force calculated as follows: A helium balloon of volume 12 x 10-3 m3 has a total mass of 12 grams. Calculate the initial acceleration of the balloon when released in air (density = 1.3 kg/m3). GIVEN mballoon = 0.012 kg (Fg = 0.1176 N) Vballoon = 12 x 10-3 m3 since the balloon is fully submerged in air, this is also the volume of the displaced air g = 9.8 m/s2 ρfluid = 1.3 kg/m3 in this case the fluid the balloon is fully submerged in is air the buoyant force calculated as follows: B = ρVg therefore B = (1.3) ( 12 x 10-3) (9.8) B = 0.153 N

A free body diagram for the balloon is shown to the right. Σ F = ma Fg B - Fg = ma a Σ F = B - Fg 0.153 - 0.1176 = .012 (a) a = 2.95 m/s2

A wooden raft of density 600 kg/m3 is placed in sea water ( = 1030 kg/m3). If the mass of the raft is 300 kg, what additional mass can it hold and remain afloat? First the volume of the raft can be determined. ρ = m / V therefore V = m/ρ V = (300 kg)  ( 600 kg/m3) = 0.50 m3 this is also the volume of the displaced fluid (since the raft can remain afloat when fully submerged)

The weight the raft itself = (300)(9.8) = 2940 N the buoyant force calculated as follows: B = ρVg therefore B = (1030) ( 0.5) (9.8) B = 5047 N The weight the raft itself = (300)(9.8) = 2940 N The additional weight the fully submerged raft can hold is therefore 5047 – 2940 = 2107N The additional mass is 2107 ÷ 9.8 = 215 kg

A gold ( = 19,300 kg/m3) bar is suspected to have a hollow center. It has a mass of 38.25 g in air and an apparent mass of 36.22 g in water. How large is the hole in the center? First the volume of the bar, if it were solid gold, can be determined. ρ = m / V therefore V = m/ρ V = (0.03825 kg)  ( 19,300 kg/m3) = 1.98 x 10-6 m3 The buoyant force can be determined as the difference in weight of the bar in air and in water. B = (0.03825)(9.8) – ( .03622)(9.8) = 0.0199 N B = ρVg therefore 0.02 = (1000) (V) (9.8) V= 2.03 x 10-6 m3

Therefore the hole is 5 x 10-8 m3 The calculated volume is bigger than the volume of a solid gold bar by 2.03 x 10-6 m3 – 1.98 x 10-6 m3 Therefore the hole is 5 x 10-8 m3