## College Physics (4th Edition)

(a) The aluminum has a greater buoyant force acting on it. (b) The wood has a greater buoyant force acting on it. (c) The buoyant force acting on the aluminum is $3.63~N$ while the buoyant force acting on the lead is $0.864~N$ The buoyant force acting on the wood is $9.8~N$ while the buoyant force acting on the steel is $1.22~N$
(a) The density of lead is $11,340~kg/m^3$ The density of aluminum is $2700~kg/m^3$ Since the density of aluminum is less than the density of lead, the volume of 1.0 kg of aluminum is greater than the volume of 1.0 kg of lead. The buoyant force is equal to the weight of water that is displaced. Since the aluminum displaces more water than the lead, the aluminum has a greater buoyant force acting on it. (b) The weight of the steel is equal to the weight of the wood. Since the wood is floating, the buoyant force acting on the wood is equal to the weight of the wood. Since the steel sinks, the buoyant force is less than the weight of the steel. Therefore, the wood has a greater buoyant force acting on it. (c) The density of lead is $11,340~kg/m^3$. We can find the volume of 1.0 kg of lead: $V_l = \frac{m}{\rho_l} = \frac{1~kg}{11,340~kg/m^3} = 8.82\times 10^{-5}~m^3$ The buoyant force is equal to the weight of water that is displaced. We can find the buoyant force acting on the lead: $F_b = V_l~\rho_w~g$ $F_b = (8.82\times 10^{-5}~m^3)(1000~kg/m^3)(9.80~m/s^2)$ $F_b = 0.864~N$ The density of aluminum is $2700~kg/m^3$. We can find the volume of 1.0 kg of aluminum: $V_a = \frac{m}{\rho_a} = \frac{1~kg}{2700~kg/m^3} = 3.70\times 10^{-4}~m^3$ The buoyant force is equal to the weight of water that is displaced. We can find the buoyant force acting on the aluminum: $F_b = V_a~\rho_w~g$ $F_b = (3.70\times 10^{-4}~m^3)(1000~kg/m^3)(9.80~m/s^2)$ $F_b = 3.63~N$ The buoyant force acting on the aluminum is $3.63~N$ while the buoyant force acting on the lead is $0.864~N$ Since the wood is floating, the buoyant force acting on the wood is equal to the weight of the wood. We can find the buoyant force acting on the wood: $F_b = m_w~g$ $F_b = (1.0~kg)(9.80~m/s^2)$ $F_b = 9.8~N$ The density of steel is $8050~kg/m^3$. We can find the volume of 1.0 kg of steel: $V_s = \frac{m}{\rho_s} = \frac{1~kg}{8050~kg/m^3} = 1.24\times 10^{-4}~m^3$ The buoyant force is equal to the weight of water that is displaced. We can find the buoyant force acting on the steel: $F_b = V_s~\rho_w~g$ $F_b = (1.24\times 10^{-4}~m^3)(1000~kg/m^3)(9.80~m/s^2)$ $F_b = 1.22~N$ The buoyant force acting on the wood is $9.8~N$ while the buoyant force acting on the steel is $1.22~N$