Sand on a Belt and a Falling Chain

The push of a stream

A hose, a stream of rice, a spray of hailstones and a line of bullets have one thing in common. Mass arrives at a surface continuously, at a steady rate. The mass flow rate \(\mu = dm/dt\) is the mass arriving each second, in kg/s. If each bit arrives at speed \(v\) and stops, the surface removes momentum \(\mu v\) every second. By the second law, it must push on the stream with a force \(\mu v\).

This is the thrust of the last lesson, seen from the surface. The stream pushes back on the surface with an equal force \(\mu v\). If the stream bounces straight back at the same speed, each bit's momentum reverses, and the change doubles to \(2\mu v\). For a jet of water of area \(A\) and density \(\rho\), a length \(v\) of jet arrives each second, so \(\mu = \rho A v\).

Sand on a moving belt

At a quarry, sand falls from a hopper onto a belt that runs at a steady speed \(v\). Each grain lands with no horizontal velocity. The belt must bring it to \(v\). So the belt hands horizontal momentum to the sand at the rate \(\mu v\). To keep the belt's speed steady, a motor must pull it with a force \(F = \mu v\). The power of that force is \(Fv = \mu v^2\).

Where does half the power go? Every second a mass \(\mu\) of sand reaches speed \(v\), and gains kinetic energy \(\tfrac12\mu v^2\). That is half of \(\mu v^2\). The other half becomes heat in the sliding. A grain does not reach the belt speed at once. The belt slides under it for a moment, as under the box on a belt in the chapter Friction, Part 2. Kinetic friction then makes heat equal to the kinetic energy the grain gains, whatever the value of \(\mu_k\).

The same thinking settles two puzzles about carts. Sand that leaks out through a hole in the floor of a moving cart leaves with the cart's own velocity. Its relative velocity is zero, so it gives no thrust, and the cart keeps its speed. Sand poured straight down into a moving cart, with no pull, is different. The sand arrives with no horizontal momentum, and the cart must share its own with it, so the cart slows down.

A chain falling onto a table

A uniform chain of mass \(M\) and length \(L\) hangs vertically, with its lower end just touching a table. It is let go. The links fall freely and pile up on the table as they arrive. How hard does the table push up when a length \(x\) has landed? The weight on the table is only \(Mgx/L\). The arriving links, though, bring momentum that the table must remove.

The figure above shows a chain of \(1.5\,\text{kg}\) and \(1.2\,\text{m}\). When \(0.8\,\text{m}\) has landed, the links arrive at \(\sqrt{2 \times 10 \times 0.8} = 4\,\text{m/s}\). The landed weight is \(10\,\text{N}\), and stopping the stream takes \(20\,\text{N}\) more, so \(N = 30\,\text{N}\). Just before the last link lands, \(N = 3Mg = 45\,\text{N}\). The moment the chain is at rest, \(N\) drops to \(15\,\text{N}\), its weight.

Now return to the rice on the kitchen scale. Suppose the cook pours \(0.1\,\text{kg}\) every second from \(20\,\text{cm}\) above the pan. The grains land at \(\sqrt{2 \times 10 \times 0.2} = 2\,\text{m/s}\). Stopping them takes \(\mu v = 0.1 \times 2 = 0.2\,\text{N}\), the weight of \(20\,\text{g}\). A grain takes \(0.2\,\text{s}\) to fall, so \(20\,\text{g}\) of rice is always in the air. The push of the stream is exactly the weight of the rice in the air. So the scale already counts the falling rice, and the cook can stop pouring when the display shows one kilogram.