A piece of wood is floating at the surface of some water as illustrated. The wood has a circular cross section and a height h = 3.0 cm. The density of the wood is 0.41 g/cm³. The distance y from the surface of the water to the bottom of the wood is A) 1.8 cm. B) 0.81 cm. C) 3.2 cm. D) 1.2 cm. E) 0.41 cm.

Question

Quizplus · Accepted Answer

The buoyant force on the wood is equal to the weight of the water displaced by the wood. Since the wood is floating at the surface of the water, the volume of water displaced is equal to the volume of the part of the wood submerged in the water. The volume of the wood is V = π(r^2)(h) and the weight of the wood is W = (density of wood)(V)(g) = (0.41 g/cm^3)(π(r^2)(h))(9.81 m/s^2). The buoyant force is equal to the weight of the water displaced, which is Fb = (density of water)(V)(g) = (1000 kg/m^3)(π(r^2)(y))(9.81 m/s^2). Setting the buoyant force equal to the weight of the wood, we have:

1000πr^2y = 0.41πr^2h
y = (0.41/1000)(h) = 1.23 cm

Therefore, the distance y from the surface of the water to the bottom of the wood is D) 1.2 cm.

A Piece of Wood Is Floating at the Surface of Some