## College Physics (4th Edition)

(a) The scale reading is $0.60~W$ (b) The scale reading is $0.64~W$
(a) Let $\rho_w$ be the density of water. Then the stone has a density of $2.50~\rho_w$. Let $V$ be the volume of the stone. We can write an expression for the weight of the stone: $W = 2.50~\rho_w~V~g$ According to Archimedes' principle, the buoyant force is equal to the weight of the water that is displaced. We can find the buoyant force: $F_b = \rho_w~V~g = \frac{W}{2.50} = 0.40~W$ We can consider the forces on the stone to find the scale reading $F_s$: $\sum F = 0$ $F_s+F_b-W = 0$ $F_s = W - F_b$ $F_s = W - 0.40~W$ $F_s = 0.60~W$ The scale reading is $0.60~W$ (b) According to Archimedes' principle, the buoyant force is equal to the weight of the oil that is displaced. We can find the buoyant force: $F_b = \rho_o~V~g$ $F_b = 0.90~\rho_w~V~g$ $F_b = 0.90~(\frac{W}{2.50})$ $F_b = 0.36~W$ We can consider the forces on the stone to find the scale reading $F_s$: $\sum F = 0$ $F_s+F_b-W = 0$ $F_s = W - F_b$ $F_s = W - 0.36~W$ $F_s = 0.64~W$ The scale reading is $0.64~W$.