

1. A standard basketball (mass = 624 grams; 24.3 cm in diameter) is held fully under water. Calculate the buoyant force and weight. When released, does the ball sink to the bottom or float to the surface? If it floats, what percentage of it is sticking out of the water? If it sinks, what is the normal force, F_{N} with which it sits on the bottom of the pool? SOLUTION The weight of the ball is To calculate the buoyancy, we need the volume of displaced water, which is the volume of the ball because it is being held completely submerged. The buoyant force is equal to the weight of that volume of water. That's a lot stronger than the 6.1N downward pull of gravity, so the ball will rise to the surface when released. The density of the ball is


2. Six objects (AF) are in a liquid, as shown. None of them are moving.
Arrange them in order of density, from lowest to highest.
SOLUTION: The more of an object's volume is above the water surface, the less dense it is. Object B must therefore be the least dense, followed by D, A, and F. Object E is next, because it is neutrally buoyant and equal in density to the liquid. Object C is negatively buoyant because it is more dense than the fluid. Full answer to the question: B, D, A, F, E, C.


3. Water ice has a density of 0.91 g/cm³, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. (a) What is the cube's weight? (b) What volume of liquid water must be displaced in order to support the floating cube? (c) How much of the cube is under the surface of the water? SOLUTION: (a) The cube's weight is (b) The buoyant force must equal the cube's weight. Take the equation for buoyant force, solve it for V_{df}, and plug in the numbers. (c) The volume of the cube itself is 0.001m³, so the percentage under the surface is...
4. You have a block of a mystery material, 12 cm long, 11 cm wide and 3.5 cm thick. Its mass is 1155 grams. (a) What is its density? 2.5 g/cm³ or 2500 kg/m³ (d) Repeat parts b and c, only instead of water, the tank is full of mercury. The object is less dense than mercury (13.6 g/cm³), so the object will float in mercury. The ratio of their densities, is 2.5/13.6 = 0.18. So 18% of the object is below the surface of the mercury, meaning that 82% must be sticking up above the surface. 

