Two friends shift a steel almirah across a room. A gentle push does nothing. A harder push does nothing either. They push harder still, and suddenly the almirah lurches forward. Once it is moving, the two find that they can keep it sliding with less effort than they needed to start it.
Static friction matches the push
Push a heavy almirah with \(50\,\text{N}\) and it stays at rest. By Newton's first law the net force on it is zero, so friction must be \(50\,\text{N}\) backward. Push with \(100\,\text{N}\) and it still stays at rest, so friction is now \(100\,\text{N}\). Friction on a body at rest is called static friction. Static friction is not a fixed number. It takes whatever value keeps the body at rest.
Static friction is a self-adjusting force, like the tension in a rope or the normal reaction from a floor. It is not decided by a formula alone. It is decided by the condition that the body stays at rest. So to find static friction, first find what the other forces along the surface add up to. Static friction is equal and opposite to that total.
A \(4\,\text{kg}\) block rests on a level floor. The coefficient of static friction is \(0.5\). You push the block horizontally with \(12\,\text{N}\), and it does not move. Find the friction on the block, in newtons. Take \(g = 10\,\text{m/s}^2\).
The limit
The welds at the contact points can hold only so much. Static friction therefore has a largest possible value, called limiting friction. Experiment shows that limiting friction is proportional to the normal reaction \(N\), the force pressing the two surfaces together. The constant of proportionality is the coefficient of static friction \(\mu_s\).
\[ f_s \le \mu_s N \]
The equals sign holds only when the body is on the point of sliding. At any smaller push, static friction is smaller than \(\mu_s N\). So \(\mu_s N\) is a ceiling, not the value of friction.
The coefficient \(\mu_s\) is a pure number. It depends on the two materials and on the state of their surfaces: dry or wet, clean or oily, polished or rough. It hardly depends on the area of contact or on the speed. For wood on wood it is about \(0.4\). For a rubber tyre on a dry road it is about \(1\).
The normal reaction \(N\) is not always equal to \(mg\). It equals \(mg\) only for a body on a level surface with no other vertical force on it. On a slope, or with a push that has a vertical part, find \(N\) from the forces at right angles to the surface first.
Kinetic friction
Once the push exceeds \(\mu_s N\), the block starts to slide. The friction on a sliding body is called kinetic friction. Kinetic friction is not self-adjusting. It has one value, proportional to the normal reaction, with its own constant, the coefficient of kinetic friction \(\mu_k\).
\[ f_k = \mu_k N \]
Kinetic friction hardly depends on the speed of sliding, over the speeds met in daily life. It points against the relative velocity, whatever the push is doing.
For every pair of surfaces, \(\mu_k\) is smaller than \(\mu_s\). At rest, the contact points have time to settle and bond firmly. While sliding, the surfaces ride over one another and each weld lasts only a moment. Fewer and weaker welds give less friction. So a body is harder to start than to keep moving.
Move the slider in the sim. Up to \(25\,\text{N}\), the friction arrow grows with the push, and the graph climbs along the line \(f = F\). Press play here and the block does not move. Just past \(25\,\text{N}\), the welds break. Friction drops at once to \(20\,\text{N}\), and the graph steps down. Now the push beats friction, and a net force acts forward, so the block accelerates.
Now return to the almirah. The gentle pushes were matched by static friction, so nothing moved. The push that finally exceeded \(\mu_s N\) broke the welds, and friction dropped at once to the smaller value \(\mu_k N\). The same push now exceeded friction by a margin, so the almirah lurched forward. After that, a push of only \(\mu_k N\) was enough to keep it sliding at a steady speed.
A steel almirah of mass \(80\,\text{kg}\) stands on a level floor. For this pair of surfaces \(\mu_s = 0.4\) and \(\mu_k = 0.3\). Take \(g = 10\,\text{m/s}^2\). (a) What horizontal push just starts the almirah moving? (b) What push keeps it sliding at a steady speed? (c) What is the friction when the push is \(200\,\text{N}\)?
(a) \(N = mg = 800\,\text{N}\). The push must reach the limit \(\mu_s N = 0.4 \times 800 = 320\,\text{N}\).
(b) At steady speed the net force is zero, so the push equals kinetic friction, \(\mu_k N = 0.3 \times 800 = 240\,\text{N}\).
(c) \(200\,\text{N}\) is less than \(320\,\text{N}\), so the almirah stays at rest. Static friction balances the push: \(f = 200\,\text{N}\), not \(320\,\text{N}\).
Why is it easier to keep a heavy box sliding than to start it sliding?
A block on a level floor has \(\mu_s N = 30\,\text{N}\). A push of \(18\,\text{N}\) acts on it, and the block stays at rest. What is the friction on the block?
The same block, with \(\mu_s N = 25\,\text{N}\) and \(\mu_k N = 20\,\text{N}\), is pushed in three ways. Rank the friction on it, largest first.
A push slowly increases on a heavy crate. At the instant the crate starts to move, it jumps forward suddenly. Explain the jump.
Up to the moment of sliding, static friction equals the push. At the point of sliding the push equals \(\mu_s N\). Once the crate slides, friction drops to \(\mu_k N\), which is smaller. The push has not dropped, so for an instant the push exceeds friction by \(\mu_s N - \mu_k N\). This net force gives the crate a sudden acceleration, which is the jump.
- Writing \(f = \mu_s N\) for every body at rest.\(\mu_s N\) is the largest possible static friction. For a body at rest, find friction from the balance of forces; it equals \(\mu_s N\) only on the point of sliding.
- Taking \(N = mg\) without checking.Resolve the forces at right angles to the surface first. On a slope, or with a push that has a vertical part, \(N\) differs from \(mg\).
- Using \(\mu_s\) for a body that is already sliding.A sliding body feels kinetic friction, \(\mu_k N\), whatever the push.