Worked examples · Sheet 3

Floating bodies and immersed fraction

Theory used: Condition for flotation Archimedes' principle Density

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A homogeneous wooden cube, of side $20.0 cm$ and density $600 kg/ m^{3}$ , floats in water with two faces horizontal. Compute the immersed fraction of the volume, the height of the immersed part and the buoyant force.

S = ρ_{a} V_{imm} g, P = ρ_{c} V g, S = P

\frac{V _{imm}}{V} = \frac{ρ _{c}}{ρ _{a}}, h_{imm} = \frac{ρ _{c}}{ρ _{a}} L, S = ρ_{c} L^{3} g

\frac{V _{imm}}{V} = \frac{600}{1000} = 0.600, h_{imm} = 0.600 \cdot 20.0 cm = 12.0 cm

S = 600 \cdot (0.200)^{3} \cdot 9.81 N = 47.1 N

The density of the wood is less than that of water, so by the condition for flotation the cube floats. At equilibrium the only vertical forces are the weight, applied at the center of mass, and the buoyant force, applied at the center of buoyancy, therefore they have the same magnitude. From the equality $ρ_{a} V_{imm} g = ρ_{c} V g$ it follows that the immersed fraction is the ratio of the densities, and it equals $0.600$ .

The cube has horizontal faces, so the immersed part is a rectangular box with the same base as the cube; the immersed volume is proportional to the immersed height, which is therefore $60 %$ of the side, that is, $12.0 cm$ . The buoyant force equals the weight of the cube, which is computed from the density and the volume $L^{3}$ expressed in cubic meters; the buoyant force is $47.1 N$ , and it is equal to the weight of the cube.

Exercise 2

An iceberg floats in the sea with $50.0 m^{3}$ of its volume out of the water. The densities of ice and seawater are $917 kg/ m^{3}$ and $1025 kg/ m^{3}$ . Compute the total volume and the mass of the iceberg.

\frac{V _{imm}}{V} = \frac{ρ _{g}}{ρ _{m}} ⟹ \frac{V _{em}}{V} = 1 - \frac{ρ _{g}}{ρ _{m}} = \frac{ρ _{m} - ρ _{g}}{ρ _{m}}

V = \frac{ρ _{m}}{ρ _{m} - ρ _{g}} V_{em}, m = ρ_{g} V

V = \frac{1025}{1025 - 917} \cdot 50.0 m^{3} = \frac{1025}{108} \cdot 50.0 m^{3} = 474.5 m^{3} \approx 475 m^{3}

m = 917 kg/ m^{3} \cdot 474.5 m^{3} = 4.35 \cdot 1 0^{5} kg

Ice is less dense than seawater, so the iceberg floats and by the condition for flotation the immersed fraction is the ratio of the densities. The emerged fraction is the complement to one, $\frac{108}{1025} \approx 0.105$ . The emerged volume is known, therefore the total volume is obtained by dividing the emerged volume by the emerged fraction, and it turns out to be about nine and a half times the visible volume. The mass is the product of the density of ice and the total volume, computed with the unrounded value $474.5 m^{3}$ .

Exercise 3

A raft has mass $120 kg$ and total volume $0.400 m^{3}$ . Compute the maximum load that the raft can carry in fresh water without going below the water surface, and how many people of $75 kg$ it can carry.

S_{m a x} = ρ_{a} V g, (m + m_{c}) g \leq ρ_{a} V g

m_{c} \leq ρ_{a} V - m

m_{c, m a x} = 1000 \cdot 0.400 kg - 120 kg = 280 kg, \frac{280}{75} \approx 3.7

The buoyant force grows with the immersed volume, so it reaches its maximum value $ρ_{a} V g$ when the raft is entirely immersed. The raft with its load floats as long as the total weight does not exceed this maximum buoyant force; the factor $g$ cancels and the condition becomes an inequality between masses. The maximum load is therefore $280 kg$ .

The number of people must be an integer and its product with $75 kg$ must not exceed $280 kg$ ; since $3 \cdot 75 = 225$ and $4 \cdot 75 = 300$ , the raft carries at most three people.

Exercise 4

In a container a layer of oil of density $800 kg/ m^{3}$ and thickness $10 cm$ floats on water. A homogeneous cube of side $10.0 cm$ and density $900 kg/ m^{3}$ is placed in the container and settles in equilibrium with two faces horizontal. Compute what fraction of the volume of the cube is in the water and what fraction is in the oil.

The cube straddles the interface: half of its volume is in the water and half in the oil

S = ρ_{a} V_{a} g + ρ_{o} V_{o} g, V_{a} + V_{o} = V, S = ρ_{c} V g

ρ_{a} x + ρ_{o} (1 - x) = ρ_{c}, x = \frac{V _{a}}{V} ⟹ x = \frac{ρ _{c} - ρ _{o}}{ρ _{a} - ρ _{o}}

x = \frac{900 - 800}{1000 - 800} = \frac{100}{200} = 0.50, 1 - x = 0.50

The density of the cube is greater than that of the oil and less than that of the water. In the oil alone the cube would sink, in the water alone it would float, so it stops straddling the interface, with its lower base in the water and its upper face in the oil. The oil layer is $10 cm$ thick, enough to cover the upper part of the cube, therefore the cube is completely immersed in the two liquids.

Archimedes’ principle applies to each part: the part in the water receives the buoyant force $ρ_{a} V_{a} g$ and the part in the oil the buoyant force $ρ_{o} V_{o} g$ , and the total buoyant force is the weight of the two displaced liquids. At equilibrium the total buoyant force equals the weight of the cube; dividing by $V g$ we obtain a first-degree equation in the fraction $x$ immersed in the water. The result is $x = 0.50$ , so the cube is half in the water and half in the oil, that is, with $5.0 cm$ in each liquid.