One graph holds the whole story
Take a block of mass \(5\,\text{kg}\) on a level floor, with \(\mu_s = 0.4\) and \(\mu_k = 0.3\). The normal reaction is \(N = mg = 50\,\text{N}\). So the limit of static friction is \(\mu_s N = 20\,\text{N}\), and kinetic friction is \(\mu_k N = 15\,\text{N}\). Now push the block horizontally with a force \(F\) that grows slowly from zero, and plot the friction \(f\) against \(F\).
The push on the block grows steadily from \(0\) to \(40\,\text{N}\). How does the friction on the block change?
Read the graph from left to right. On the first part the block is at rest, and friction equals the push, so the line rises at \(45°\). The line ends at the limit, \(f = \mu_s N\). This point is the corner of the graph. Here the block is on the point of sliding. Just past the corner, the block slides, and friction falls to \(\mu_k N\). After that the graph is flat, because kinetic friction does not depend on the push.
Think of the contact points as a row of small hooks. A gentle push bends the hooks only a little, and the hooks pull back exactly as hard as you push. At the limit the hooks snap. After that they catch and slip, catch and slip, and pull back with a smaller, steady force.
On the graph of friction against applied force, what does the flat part on the right tell you?
The method
Every problem about a block pushed along a level floor is solved the same way. First find \(N\) from the vertical forces. Then compare the push with the limit \(\mu_s N\). The comparison decides which kind of friction acts, and only then can you write the equation of motion.
If the push \(F \le \mu_s N\), the block stays at rest, friction \(f = F\), and \(a = 0\). If \(F > \mu_s N\), the block slides, friction \(f = \mu_k N\), and the acceleration is \(a = (F - \mu_k N)/m\).
The \(5\,\text{kg}\) block with \(\mu_s = 0.4\) and \(\mu_k = 0.3\) starts at rest and is pushed horizontally with (a) \(10\,\text{N}\), (b) \(20\,\text{N}\), (c) \(30\,\text{N}\). Find the friction and the acceleration in each case. Take \(g = 10\,\text{m/s}^2\).
\(N = mg = 50\,\text{N}\). Limit \(\mu_s N = 20\,\text{N}\). Kinetic friction \(\mu_k N = 15\,\text{N}\).
At rest: \(f = F\), \(a = 0\). Sliding: \(f = \mu_k N\), \(a = (F - \mu_k N)/m\).
(a) \(10 < 20\): at rest, \(f = 10\,\text{N}\), \(a = 0\). (b) \(20 = 20\): on the point of sliding, \(f = 20\,\text{N}\), \(a = 0\). (c) \(30 > 20\): sliding, \(f = 15\,\text{N}\), \(a = (30 - 15)/5 = 3\,\text{m/s}^2\).
Friction is \(10\,\text{N}\), \(20\,\text{N}\) and \(15\,\text{N}\). Only the largest push moves the block, at \(3\,\text{m/s}^2\). The largest friction came with the middle push, exactly at the corner of the graph.
A \(2\,\text{kg}\) block rests on a level floor with \(\mu_s = 0.5\) and \(\mu_k = 0.3\). A horizontal push of \(15\,\text{N}\) acts on it. Find the acceleration of the block, in \(\text{m/s}^2\). Take \(g = 10\,\text{m/s}^2\).
The same \(2\,\text{kg}\) block (\(\mu_s = 0.5\), \(\mu_k = 0.3\)) starts at rest and is pushed with \(8\,\text{N}\). Which statement is correct?
The \(5\,\text{kg}\) block of the example (\(\mu_s N = 20\,\text{N}\), \(\mu_k N = 15\,\text{N}\)) is already sliding. The push is now reduced to \(17\,\text{N}\). A friend says the block must stop, because \(17\,\text{N}\) is less than the \(20\,\text{N}\) needed to start it. Is the friend right?
No. The block is sliding, so the friction on it is kinetic, 15 N, not the static limit of 20 N. The push of 17 N exceeds 15 N, so a net force of 2 N acts forward. The block keeps speeding up at 2/5 = 0.4 m/s². The limit of 20 N matters only for starting a block at rest. A sliding block slows down only if the push falls below 15 N.
A wooden study table of mass about \(30\,\text{kg}\) stands on a tiled floor, with \(\mu_s\) about \(0.4\). Estimate the horizontal push needed to start sliding it, in newtons.
Before he became famous for the law of electric force, Coulomb was a military engineer who had built fortifications in the French colony of Martinique. Ship builders needed to know the force to drag heavy hulls down a slipway. Coulomb loaded wooden sledges, dragged them with weights over pulleys, and timed them. He found the two regimes of this lesson: a larger force to start sliding, and a smaller, nearly steady force once sliding. He also found that the starting force grew when the surfaces were left in contact for longer. His memoir on simple machines won the prize of the Paris Academy in 1781.
- Using \(\mu_k N\) before checking whether the block slides.Compare the push with \(\mu_s N\) first. A block that never starts to slide feels only static friction.
- Thinking that the largest friction comes with the largest push.The largest friction is at the corner of the graph, on the point of sliding. A sliding block has less friction than that.
- Thinking a sliding block needs the starting push to keep moving.Once sliding, friction is \(\mu_k N\). Any push above \(\mu_k N\) keeps the block speeding up.