Put a pile of books on a tea tray and pull the tray gently across a table. The books ride along. Jerk the tray hard and the tray shoots out from under the books, which are left almost where they were. The same books and the same tray behave in two different ways, depending only on how hard you pull.
Friction is the only link
Place a block of mass \(m\) on a block of mass \(M\), which rests on a smooth floor. A force \(F\) pulls the lower block. Nothing touches the top block except the lower block. So the only horizontal force on the top block is the friction from the lower block. If the top block is to move at all, that friction must point forward, along \(F\). By Newton's third law, the top block pushes the lower block with an equal friction backward. The two frictions form a third-law pair.
Two bodies in contact exert friction on each other. The friction on the first and the friction on the second are equal in size and opposite in direction. On a stack of blocks, friction forward on one block always comes with the same friction backward on the other.
Suppose the blocks move together with a common acceleration \(a\). Treat them as one body of mass \(m + M\). The friction pair is internal to this body and cancels, so \(a = F/(m + M)\). Now look at the top block alone. Friction is the only horizontal force on it, so it must be \(f = ma = mF/(m + M)\). This is the friction the top block needs to keep up.
The test for slipping
The friction the top block needs grows with \(F\). The friction the lower block can give is at most \(\mu_s N = \mu_s mg\), because the normal force on the top block is its weight. As long as the need is within the limit, the blocks move together. When the need passes the limit, the lower block cannot drag the top block fast enough, and the top block slips back relative to it.
(1) Assume the bodies move together and find the common acceleration from the total mass. (2) Find the friction needed by the body that is not pushed directly. (3) Compare it with the limit \(\mu_s N\). If it is within the limit, they move together. If not, they slip, and kinetic friction \(\mu_k N\) acts on each body, opposing their relative motion.
The formula has a simple reading. The top block can have at most the acceleration \(\mu_s g\), the most that friction can give it. The pair moves together only while the common acceleration \(F/(m + M)\) is no more than \(\mu_s g\). This is the trolley and the box of Lesson 1 again, with the trolley replaced by a block.
A \(3\,\text{kg}\) block rests on a \(5\,\text{kg}\) block on a smooth floor, with \(\mu_s = 0.25\) between them. Take \(g = 10\,\text{m/s}^2\). What is the largest force on the lower block, in newtons, for which they move together?
In the sim, \(F = 12\,\text{N}\) acts on the lower block and the blocks move together. What is the friction on the lower block?
After the slip: two bodies, two accelerations
Once the top block slips, the blocks are two separate bodies with different accelerations. The friction between them is now kinetic, \(\mu_k mg\). It still points forward on the top block and backward on the lower block, because the top block slides back relative to the lower one. Write Newton's second law for each block on its own.
\[ a_{\text{top}} = \dfrac{\mu_k mg}{m} = \mu_k g, \qquad a_{\text{lower}} = \dfrac{F - \mu_k mg}{M} \]
The top block's acceleration is now fixed at \(\mu_k g\), however hard you pull. All the extra force goes into the lower block. In the sim, with \(\mu = 0.3\) for both kinds of friction, the top block stays at \(3\,\text{m/s}^2\) while the lower block reaches \(6\,\text{m/s}^2\) at \(F = 30\,\text{N}\).
If \(\mu_s\) and \(\mu_k\) are different, use each in its place. The test for slipping uses \(\mu_s\), because it asks whether static friction can hold. The accelerations after the slip use \(\mu_k\), because the surfaces are sliding. Most problems at this level give one \(\mu\) for both.
A \(2\,\text{kg}\) pile of books lies on a \(4\,\text{kg}\) tray on a smooth table, with \(\mu = 0.3\) between books and tray for both kinds of friction. Take \(g = 10\,\text{m/s}^2\). The tray is pulled with \(24\,\text{N}\). Find the acceleration of each.
Test first. Together: \(a = 24/6 = 4\,\text{m/s}^2\). The books would need \(f = 2 \times 4 = 8\,\text{N}\). The limit is \(\mu mg = 0.3 \times 20 = 6\,\text{N}\).
\(8 > 6\), so the books slip. Books: \(2\,a_{\text{top}} = 6\). Tray: \(4\,a_{\text{lower}} = 24 - 6\).
\(a_{\text{top}} = 3\,\text{m/s}^2\). \(a_{\text{lower}} = 18/4 = 4.5\,\text{m/s}^2\).
The books accelerate at \(3\,\text{m/s}^2\) and the tray at \(4.5\,\text{m/s}^2\). Relative to the tray, the books slide backward at \(1.5\,\text{m/s}^2\). From rest, they slide \(0.75\,\text{m}\) back along the tray in the first second.
In the sim, \(F = 30\,\text{N}\) acts on the \(4\,\text{kg}\) lower block, and the top \(2\,\text{kg}\) block slips with \(\mu_k = 0.3\). Take \(g = 10\,\text{m/s}^2\). What is the acceleration of the lower block, in m/s²?
The force on the top block
Now pull the top block instead. The same test works, but the roles swap. The lower block is the one that is not pushed directly. Friction from the top block is the only horizontal force on it, forward. So the lower block needs \(f = Ma = MF/(m + M)\), and the limit is still \(\mu_s mg\), set by the weight of the top block. The stack moves together only while \(F \le \mu_s mg\,(m + M)/M\).
Compare the two limits for the same blocks: \(18\,\text{N}\) on the lower block, \(9\,\text{N}\) on the top block. Pulling the top block, the heavy lower block must be dragged by friction alone, and that needs more friction for the same acceleration. Friction on the lower block now points forward, the direction of motion, which is the trolley lesson once more.
A \(2\,\text{kg}\) block lies on a \(4\,\text{kg}\) block on a smooth floor, \(\mu_s = 0.3\) between them. Which statements are true?
A truck carrying steel sheets that are not tied down depends on friction between the sheets and the bed. If the truck brakes harder than \(\mu_s g\), the friction the sheets need exceeds the limit, and they slide forward, towards the cabin. With \(\mu_s = 0.3\), that is only \(3\,\text{m/s}^2\), an ordinary firm stop. Straps add a force that friction alone cannot supply. This is why transport rules require loads to be secured, not just stacked.
- Taking the friction between the blocks as \(\mu_s mg\) before testing.While they move together, friction is only what the unpushed block needs, \((\text{its mass}) \times a\). The limit is reached only at the slipping point.
- Using the total mass after the blocks slip.After the slip each block has its own acceleration. Write one equation for each block.
- Drawing friction in the same direction on both blocks.Friction between two bodies is a third-law pair: forward on one, backward on the other.