A rectangular block of wood floats in a beaker of water. A block of lead, of exactly the same dimensions as the block of wood, is placed in the same beaker and sinks to the bottom.

Which block experiences the largest buoyant force?

1. The wooden block
2. The lead block
3. The forces are equal
4. Can't tell

Which block displaces more water? The lead block. The buoyant force equals the weight of the displaced fluid, so the lead block experiences the largest buoyant force.

A beaker of water sits on a scale. If you dip your little finger into the water, what happens to the scale reading? Assume that no water spills from the beaker in this process.

1. The scale reading goes up
2. The scale reading goes down
3. The scale reading stays the same

The scale reading goes up. When your finger is in the water the water applies an upward buoyant force on your finger. By Newton's third law, your finger applies an equal and opposite force down on the water. The scale reading equals the normal force between the scale and the beaker - the normal force equals the force of gravity on the beaker and water plus the extra force associated with your finger.

Another way to look at this is that putting your finger in the water raises the water level in the beaker. This raises the pressure at the bottom of the beaker, as well as the downward force exerted on the bottom of the beaker by the water. The upward force from the scale must increase by the same amount to compensate.

A large beaker of water is placed on a scale, and the scale reads 40 N. A block of wood with a weight of 10 N is then placed in the beaker. It floats with exactly half of its volume submerged. Assuming that none of the water spilled out of the beaker when the block was added, what does the scale read now?

1. 40 N
2. 45 N
3. 50 N
4. None of the above.

The scale, which is at the bottom of the pile, has to support the entire weight of everything on top of it. The scale reads 50 N.

The key to many buoyancy problems is to treat the buoyant force like all the other forces we've dealt with so far. What's the first step? Draw a free-body diagram.

A basketball floats in a bathtub of water. The ball has a mass of 0.5 kg and a diameter of 22 cm.

(a) What is the buoyant force?
(b) What is the volume of water displaced by the ball?
(c) What is the average density of the basketball?

(a) To find the buoyant force, simply draw a free-body diagram. The force of gravity is balanced by the buoyant force:

ΣF = ma

F_b - mg = 0

F_b = mg = 4.9 N

(b) By Archimedes' principle, the buoyant force is equal to the weight of fluid displaced.

F_b = ρV_dispg

V_disp= F_b/ρg = 4.9/(1000*9.8) = 5 x 10^-4 m³

(c) To find the density of the ball, we need to determine its volume. The volume of a sphere is:

V = (4/3)πr³

With r = 0.11 m, we get:

volume of basketball = V = 5.58 x 10^-3 m³

The density is mass divided by volume:

ρ = m/V = 0.5 / 5.58 x 10^-3 = 90 kg/m³

Another way to find density is to use the volume of displaced fluid. For a floating object, the weight of the object equals the buoyant force, which equals the weight of the displaced fluid.

mg = F_b = ρ_fluid V_disp g

m = ρ_object V, so:

ρ_object V = ρ_fluid V_disp

Factors of g cancelled. Re-arranging this gives, for a floating object:

ρ_object / ρ_fluid = V_disp/V