A chain of mass m = 0.80 kg and length l = 1.5 m rests on a rough-surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the overhanging part equals η = 1/3 of the chain length. What will be the total work performed by the friction forces acting on the chain by the moment it slides completely off the table?
The total work of friction is A = −η(1 − η)mgl/2. With the numbers on screen, A = ( J). It is negative because friction opposes the motion.
Problem statement
Problem 1.85. A chain of mass m = 0.80 kg and length l = 1.5 m rests on a rough-surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the overhanging part equals η = 1/3 of the chain length. What will be the total work performed by the friction forces acting on the chain by the moment it slides completely off the table?
We assume the chain is uniform, stays straight and taut, and turns the table edge smoothly with no friction at the edge. We take g = 9.8 m/s².
We assume the friction coefficient k is not given. It is found from the start condition: the chain is just on the point of sliding when η = 1/3.
We assume the values m = 0.80 kg, l = 1.5 m and η = 1/3 from the problem as the starting values. All three are sliders. Changing η also changes k, because k = η/(1 − η).
Symbols: λ = m/l is the mass of one metre of chain, x the length still on the table, y the length hanging, f the friction.
The big idea in plain words
Friction on the chain is not constant. It depends on how much chain still lies on the table, because the table pushes up only on that part. As the chain slides off, friction shrinks to zero. So we cannot just multiply a force by a distance. We add the work in small pieces, or use the triangle shape of the force against distance.
The solution in each frame
Only the table frame is needed. The table is at rest in it, and friction is defined in it.
The solution step by step
Step 1, the friction coefficient. At the start the chain is on the point of sliding. The weight of the overhang equals the largest friction on the part lying on the table. The overhang weighs ληlg. The part on the table has length (1 − η)l, so its friction limit is kλ(1 − η)lg. Setting them equal: k = η/(1 − η) = .
Step 2, friction at one moment. When a length x lies on the table, the table pushes up with N = λxg. The friction is f = kN = kλxg. At the start, x = (1 − η)l = m, and f = ηmg = N, which is just the weight of the overhang.
Step 3, the work in a small piece. The chain slides a small distance, so a length dx leaves the table and x falls by dx. Friction opposes the motion, so dA = −f dx = −kλgx dx.
Step 4, add the pieces. The length x goes from (1 − η)l down to zero. The sum of x dx is half of x₀². So A = −kλg(1 − η)²l²/2. Put in k = η/(1 − η) and λ = m/l: A = −η(1 − η)mgl/2 = .
Step 5, the cases. If η is very small there is almost no overhang, k is small and the work is small. If η is large, little chain lies on the table, and the work is small again. The work is largest at η = 1/2, where it is J for these values of m and l.
A separate calculation does not use the formula for the work. It follows the real motion of the chain in 1 millisecond steps. The chain gets a one-in-a-million nudge to start, since it begins exactly at the limit. At each step it adds up friction times the small distance moved. It gets A = J, against J from the formula. It also gets the speed at which the chain leaves the table as m/s, against m/s from energy. The largest relative difference is .
Five methods
1. Force method
Find the friction at each moment and add the work. This is steps 1 to 4.
2. Energy method
Work done by gravity on the chain as it slides off is mgl(1 − η²)/2. Friction takes away |A|. What is left is the kinetic energy at the end, mgl(1 − η)/2, so the chain leaves at v = √(gl(1 − η)) = m/s. This uses the answer to check it.
3. Graph method
Friction against the length that is still on the table is a straight line from ηmg at the start down to zero. The area under it is a triangle: |A| = ½ × ηmg × (1 − η)l. That gives the same result without calculus. The second graph shows this work growing as the chain slides off.
4. Calculus method
A = −∫f dx = −kλg ∫x dx from 0 to (1 − η)l. This is step 4.
5. Check method
Units: mgl is joules. The numbers η and (1 − η) have no unit. Limits: η = 0 and η = 1 both give zero work. The largest work is at η = 1/2.
Fastest: the graph method, with the triangle. Most reliable: the force method with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Mass m | Yes | The weights, the friction and the work. Not k |
| Length l | Yes | The picture, the distances and the work. Not k |
| Overhang η | Yes | The friction coefficient k, the starting positions and the work |
| Given-answer mode | Sets η | The page finds η for a chosen work. It gives the smaller of two values, and warns if the work is above the largest possible |
| Play and overhang bar | No | Play lets the overhang grow from the start value to the full chain, holds 1.2 s, then repeats |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Mass | m | kg |
| Length | l | m |
| Start overhang | η | |
| Mass per metre | λ | kg/m |
| Friction coefficient | k | |
| Chain on the table at the start | x₀ | m |
| Friction at the start | ηmg | N |
| Friction work | A |
Exam traps
- Using the friction at the start for the whole slide. It falls to zero, so the work is half of that.
- Looking for k in the problem. It is not given. It comes from the start condition.
- Using N = mg for the chain. Only the part on the table presses on it.
- Counting the hanging part as if it feels friction. It does not touch the table.
- Giving a positive answer. The work of friction is negative.