At a village well, a steel bucket goes down on a rope and fills with water. While it is under the water, a gentle pull brings it up. The moment its rim breaks the surface, the rope drags hard, as if the bucket had suddenly become several times heavier. Nothing has been added to it on the way up.
The pushes on a cube
Recall from the chapter on Pressure in Fluids that the pressure at a depth \(h\) in a liquid of density \(\rho\) is \(P = P_0 + \rho g h\). Here \(P_0\) is the pressure at the surface. The pressure pushes on every surface the liquid touches, at right angles to that surface. A body under water is therefore pushed from every side at once.
Hold a cube of side \(a\) under water, with its top face at a depth \(h\). Look first at two opposite side faces. They span the same range of depths, so the water pushes on them equally, in opposite directions. The side pushes cancel in pairs. The top and the bottom faces are different. The bottom face is deeper by \(a\), so the water pushes up on it harder than it pushes down on the top face.
\[ F_1 = (P_0 + \rho g h)\,a^2 \]
Why. the push down on the top face is the pressure at depth \(h\) times the area \(a^2\).
\[ F_2 = \big(P_0 + \rho g (h + a)\big)\,a^2 \]
Why. the push up on the bottom face, which is at depth \(h + a\).
\[ F_B = F_2 - F_1 = \rho g a \cdot a^2 \]
Why. the side pushes cancel. The terms \(P_0 a^2\) and \(\rho g h a^2\) appear in both pushes and cancel too.
\[ F_B = \rho g V \]
Read it. \(a^3\) is the volume \(V\) of the cube. The depth \(h\) has dropped out.
The net push of the water, \(F_2 - F_1\), points up. It is called the buoyant force, or buoyancy, and written \(F_B\). The result does not contain \(h\). A cube held at \(1\,\text{m}\) and at \(10\,\text{m}\) feels the same buoyant force, although every single push is much larger at \(10\,\text{m}\). The atmosphere drops out as well, because \(P_0\) adds the same push to the top face and the bottom face.
Try the sim above. With the top face \(0.2\,\text{m}\) down, the top face carries \(320\,\text{N}\) and the bottom face \(960\,\text{N}\). At \(0.8\,\text{m}\) they carry \(1280\,\text{N}\) and \(1920\,\text{N}\). The difference is \(640\,\text{N}\) both times, which is \(\rho g V = 1000 \times 10 \times 0.064\). Now raise the cube until its top is out of the water. The top face has no water on it, and the buoyant force shrinks with the part still under.
The buoyant force on a body in a fluid is the resultant of all the pushes of the fluid on the surface of the body. In a fluid at rest it points vertically upward.
A body of any shape can be cut into thin vertical columns. On each column, the push up at the bottom exceeds the push down at the top. The excess is \(\rho g\) times the column's height times its area, the weight of an equal column of fluid. Add up the columns, and the buoyant force is \(\rho_f V g\) for any shape.
The buoyant force depends on the fluid and on the volume under the surface. It does not depend on what the body is made of. An iron ball and a wooden ball of the same size, both held under water, feel the same buoyant force. Use the density of the fluid in \(F_B = \rho_f V g\), never the density of the body.
A stone hangs on a string, fully under water in a deep well. It is lowered from \(2\,\text{m}\) below the surface to \(8\,\text{m}\) below. Take water to be incompressible. What happens to the buoyant force on the stone?
A block \(0.5\,\text{m}\) by \(0.4\,\text{m}\) by \(0.2\,\text{m}\) is held fully under water. Take \(g = 10\,\text{m/s}^2\) and water as \(1000\,\text{kg/m}^3\). Find the buoyant force on it, in newtons.
The replacement argument
There is an argument that gives the buoyant force for any shape without any algebra. Picture a parcel of water, of any shape, inside a tank of still water. Mark its boundary by an imaginary surface. The parcel is at rest, so the forces on it balance. Its weight acts down, and the pushes of the water around it act on its surface. So those pushes add up to a force equal to the parcel's weight, acting up.
Now take the parcel away, and put a stone of exactly the same shape in its place. The water around it is the same water at the same depths. It pushes on the stone's surface exactly as it pushed on the parcel. So the stone feels the same upward force: the weight of the water it has pushed out of its place. That water is called the displaced fluid.
A body wholly or partly immersed in a fluid is pushed up by a force equal to the weight of the fluid it displaces.
In Fig 1.1 only one arrow changes from chip to chip: the weight. The water is the same, so the pushes and \(F_B\) are the same. Whether the body sinks or rises depends only on how its weight compares with \(F_B\).
Sink or rise
Hold a body of density \(\rho_b\) and volume \(V\) fully under a fluid of density \(\rho_f\), and let it go. Its weight is \(W = \rho_b V g\), and the buoyant force is \(F_B = \rho_f V g\). The volume and \(g\) are common to both, so the densities decide.
- If \(\rho_b > \rho_f\), then \(W > F_B\) and the body sinks, like a coin dropped in water.
- If \(\rho_b < \rho_f\), then \(W < F_B\) and the body rises, like wood let go under water.
- If \(\rho_b = \rho_f\), the forces balance and the body stays wherever it is put.
The buoyant force also has a line of action. The parcel of water was in balance, so the pushes on it acted up through its centre of gravity. For the stone in its place, the pushes are the same. So the buoyant force acts up through the centre of gravity of the displaced fluid, a point called the centre of buoyancy. For a uniform block fully under water, it is the centre of the block.
No fluid underneath, no push up. The principle assumes the fluid touches the body all round. A flat block pressed tight on the floor of a tank, with no water under it, gets no push from below at all. The water above then presses it down onto the floor. This is part of the reason an object sunk in soft mud can be very hard to pull free.
Three balls of the same size, of aluminium (\(2700\,\text{kg/m}^3\)), wax (\(900\,\text{kg/m}^3\)) and steel (\(7800\,\text{kg/m}^3\)), are held fully under water and let go. Which statements are true?
Archimedes lived in Syracuse, a Greek city in Sicily. His book On Floating Bodies starts from a single assumption about a fluid. A part of the fluid that is pushed less hard gives way to a part pushed harder. From it he proved that a solid lighter than the fluid sinks until it displaces its own weight. A heavier solid, he showed, is lighter in the fluid by the weight of the fluid it displaces. He went on to work out when a floating body turns itself upright, the question of the last lesson of this chapter.
Weighing in a fluid
Hang a stone from a spring balance, and the balance reads its weight \(W\). Now lower the stone fully into water. Three forces act on it: \(W\) down, the pull \(T\) of the balance up, and \(F_B\) up. The stone is at rest, so \(T = W - F_B\). The balance reads \(T\), which is called the apparent weight of the stone in water.
Write the weight as \(W = \rho_b V g\). Then \(F_B/W = \rho_f/\rho_b\), and the apparent weight takes a second form. A stone of density \(2500\,\text{kg/m}^3\) loses \(1000/2500 = 0.4\) of its weight in water. A steel ball of \(7800\,\text{kg/m}^3\) loses only about \(0.13\) of it.
Gravity has not changed. The stone's weight is still \(mg\) under water. The balance reads less because the water holds up part of the load, not because the stone has become lighter.
A stone weighs \(50\,\text{N}\) in air and \(30\,\text{N}\) when fully under water. Take \(g = 10\,\text{m/s}^2\). Find the buoyant force on it, its volume and its density.
The loss of weight is the buoyant force: \(F_B = 50 - 30 = 20\,\text{N}\).
\(F_B = \rho_w V g\) gives the volume. The mass is \(W/g\), and the density is the mass over the volume.
\(V = 20/(1000 \times 10) = 2 \times 10^{-3}\,\text{m}^3\). The mass is \(50/10 = 5\,\text{kg}\), so \(\rho = 5/(2 \times 10^{-3}) = 2500\,\text{kg/m}^3\).
The buoyant force is \(20\,\text{N}\), the volume \(2\) litres and the density \(2500\,\text{kg/m}^3\), a typical rock.
A \(2.7\,\text{kg}\) block of aluminium, of density \(2700\,\text{kg/m}^3\), hangs from a spring balance fully under water. Take \(g = 10\,\text{m/s}^2\). What does the balance read, in newtons?
Return now to the bucket at the well. Say the bucket is made of \(1.6\,\text{kg}\) of steel, of density \(8000\,\text{kg/m}^3\), and holds \(10\) litres of water. Under the surface, the water inside the bucket is surrounded by water. It displaces its own weight, so its apparent weight is zero. The steel has a volume of \(2 \times 10^{-4}\,\text{m}^3\), so it loses \(2\,\text{N}\) of its \(16\,\text{N}\). The rope pulls with only \(14\,\text{N}\).
Out of the well, the rope must hold \(16\,\text{N}\) of steel and \(100\,\text{N}\) of water, \(116\,\text{N}\) in all. As the rim leaves the surface, the pull on the rope grows from \(14\,\text{N}\) to \(116\,\text{N}\), more than eight times. The water in the bucket gained no weight. It simply lost the support of the water around it.
A person standing waist-deep in a swimming pool feels lighter on the feet. Explain why, and say whether the person's weight has changed.
The water pushes up on the part of the body under the surface with a buoyant force equal to the weight of the water displaced. So the floor needs to push up with only the weight minus the buoyant force, and that smaller push is what the feet feel. The weight, mg, has not changed. Only the share of it carried by the floor has become smaller.
- Thinking the buoyant force grows as a fully submerged body goes deeper.Both pushes grow by the same amount, so their difference, \(\rho_f V g\), does not change with depth.
- Using the density of the body in \(F_B = \rho V g\).The buoyant force is the weight of the displaced fluid, so \(\rho\) is the density of the fluid.
- Believing a body weighs less in water because gravity is weaker there.The weight is unchanged. The balance reads less because the water carries part of the load.