Tension in inextensible strings: Worked examples

The scenario matrix

Before solving anything, let us map the territory. Each row below is a case class — a genuinely different situation this topic can produce. The examples that follow are tagged with the cell they cover, and together they fill every row.

#	Case class	What is new / what can go wrong	Example
A	Both masses hang (Atwood) — unequal	which side goes down? sign of $a$	Ex 1
B	Balanced case $m_{1} = m_{2}$ — degenerate	$a = 0$ , static, $T = m g$	Ex 2
C	One mass zero — degenerate limit	free-fall check $a \to g$ , $T \to 0$	Ex 2
D	Horizontal + vertical (table + hanger)	weight cancels on the table body	Ex 3
E	Force pulls a chain of blocks	system-first, then isolate for $T$	Ex 4
F	Limiting behaviour ( $m \to \infty$ )	does $T$ approach a static value?	Ex 4
G	Two hanging masses lifted by an external upward force	tension can exceed $m g$	Ex 5
H	Incline + hanger — sign of $sin θ$	which way does the system slide?	Ex 6
I	Real-world word problem (elevator + person)	apparent weight = tension in cable	Ex 7
J	Exam twist — three masses, find middle tension	tension differs between segments	Ex 8

Example 1 — Cell A · Atwood, unequal masses

Forecast: The heavier side ( $m_{1}$ ) must fall, the lighter side must rise. Do you expect $T$ to be more than $m_{2} g = 20 N$ or less? (Guess before reading — think: the string has to lift $m_{2}$ and speed it up, so...)

Reading the figure below: the picture shows both masses hanging from the lavender pulley at the top. Follow the coral arrows — those are the weights $m_{1} g$ and $m_{2} g$ , both pointing straight down. The lavender arrows are the tension $T$ , pulling each block up toward the string. The small slate arrows on the outside show the actual motion: $m_{1}$ moves down, $m_{2}$ moves up, with the same acceleration magnitude $a$ .

Figure — Tension in inextensible strings

Step 1 — Draw the free body diagram for each mass (as in the figure). Why this step? Newton's law applies to one body at a time; each body feels only its own weight (down) and the tension (up).

For $m_{1}$ , choose down as positive (it falls): $m_{1} g - T = m_{1} a$ For $m_{2}$ , choose up as positive (it rises): $T - m_{2} g = m_{2} a$ Why different "positive" for each body? Because the inextensible string makes them accelerate the same amount $a$ but in opposite directions. Letting each body's own motion be positive lets us use one number $a$ for both.

Step 2 — Add the two equations to cancel $T$ : $m_{1} g - m_{2} g = (m_{1} + m_{2}) a \Rightarrow a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( 6 - 2 ) \cdot 10}{8} = 5 m/s^{2}$ Why add? The $- T$ and $+ T$ are equal and opposite; adding annihilates the unknown we don't want yet.

Step 3 — Derive the general tension, then plug in. Back-substitute into $T - m_{2} g = m_{2} a$ : $T = m_{2} (g + a) = m_{2} (g + \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}) = m_{2} \cdot \frac{( m _{1} + m _{2} ) g + ( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ Why do the algebra here? This is where the famous Atwood tension formula comes from — it is not magic, it is just $T = m_{2} (g + a)$ with $a$ substituted and simplified. Plugging in numbers: $T = \frac{2 \cdot 6 \cdot 2}{8} \cdot 10 = 30 N$

Verify: $T = 30 N > m_{2} g = 20 N$ ✓ — matching the forecast (the string lifts and accelerates $m_{2}$ ). Direct route: $T = m_{2} (g + a) = 2 (10 + 5) = 30 N$ ✓ — same answer, confirming the derived formula.

Example 2 — Cells B & C · Degenerate Atwood

Forecast: In (a) the two sides are perfectly matched — a tug-of-war between equals. What must $a$ be? In (b) there is nothing on one side; what is the other side doing?

Step 1 — Use the general formulas we derived in Example 1: $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}, T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$ Why reuse them? Degenerate cases are best checked by plugging into the general result — that is exactly how you sanity-check any formula.

Step 2 — Case (a), balanced: $a = \frac{( 4 - 4 ) \cdot 10}{8} = 0, T = \frac{2 \cdot 4 \cdot 4}{8} \cdot 10 = 40 N$ Why $a = 0$ ? No net driving difference, so nothing accelerates — the system is static. Each side simply holds its own weight: $T = m g = 4 \cdot 10 = 40 N$ ✓.

Step 3 — Case (b), one side empty ( $m_{2} = 0$ ): $a = \frac{( 4 - 0 ) \cdot 10}{4} = 10 = g, T = \frac{2 \cdot 4 \cdot 0}{4} \cdot 10 = 0$ Why $a = g$ and $T = 0$ ? With nothing to pull against, $m_{1}$ is in free fall; a slack, force-free string carries zero tension.

Verify: (a) $a = 0$ , $T = m g = 40 N$ ✓ (static balance). (b) $a = g = 10$ , $T = 0$ ✓ (free fall). Both degenerate limits behave physically.

Example 3 — Cell D · Block on table + hanging block

Reading the figure below: the block $m_{1}$ (lavender) sits on the slate table line; the string runs flat to the butter-coloured pulley at the edge, then bends down to the hanging $m_{2}$ (coral). Notice the mint arrow pushing up on $m_{1}$ — that is the normal force $N$ , exactly cancelling $m_{1}$ 's weight so nothing happens vertically on the table. The only un-cancelled force on $m_{1}$ is the lavender tension arrow pointing horizontally toward the pulley. On $m_{2}$ , the coral weight arrow (down) beats the lavender tension arrow (up), so $m_{2}$ accelerates down — the slate arrow labelled $a$ .

Forecast: The hanging mass pulls the whole system. Will the tension be equal to, more than, or less than $m_{2} g = 30 N$ ? (Hint: $m_{2}$ is accelerating downward, so the upward $T$ can't fully match its weight.)

Step 1 — FBD of $m_{2}$ (falls, down positive): $m_{2} g - T = m_{2} a$ FBD of $m_{1}$ (slides horizontally, direction-of-pull positive): $T = m_{1} a$ Why does weight not appear for $m_{1}$ ? On the table its weight is cancelled by the normal force (the mint arrow in the figure); the only horizontal force is $T$ .

Step 2 — Substitute $T = m_{1} a$ into the first equation: $m_{2} g - m_{1} a = m_{2} a \Rightarrow a = \frac{m _{2} g}{m _{1} + m _{2}} = \frac{3 \cdot 10}{5} = 6 m/s^{2}$ Why substitute? It eliminates $T$ so we can solve for $a$ first.

Step 3 — Get $T$ : $T = m_{1} a = 2 \cdot 6 = 12 N$

Verify: $T = 12 N < m_{2} g = 30 N$ ✓ (matches forecast: a falling mass has $T < m g$ ). Cross-check with the boxed formula $T = \frac{m _{1} m _{2} g}{m _{1} + m _{2}} = \frac{2 \cdot 3 \cdot 10}{5} = 12 N$ ✓.

Example 4 — Cells E & F · Chain of blocks pulled by a force, plus a limit

Forecast: The string only has to accelerate $m_{1}$ (the block behind). So should $T$ be the whole $12 N$ , or only a fraction of it?

Step 1 — Treat both blocks as one system: $F = (m_{1} + m_{2}) a \Rightarrow a = \frac{F}{m _{1} + m _{2}} = \frac{12}{6} = 2 m/s^{2}$ Why the system view? Internal tension cancels for the combined body (equal and opposite on the two ends), leaving only the external $F$ .

Step 2 — Isolate $m_{1}$ (the string is its only horizontal force): $T = m_{1} a = 4 \cdot 2 = 8 N$ Why $m_{1}$ and not $m_{2}$ ? For $m_{1}$ , $T$ is the sole accelerating force, so $T = m_{1} a$ directly. For $m_{2}$ you'd have $F - T = m_{2} a$ (an extra force), which is messier.

Step 3 — The limit (Cell F): with $F$ fixed, $T = m_{1} a = m_{1} \cdot \frac{F}{m _{1} + m _{2}} = \frac{m _{1} F}{m _{1} + m _{2}} m_{2} \to \infty 0$ Why zero? A hugely heavy front block barely accelerates ( $a \to 0$ ), so the string need pull $m_{1}$ with almost no force.

Verify: $a = 2$ , $T = 8 N$ . Check $m_{2}$ 's own equation: $F - T = 12 - 8 = 4 = m_{2} a = 2 \cdot 2 = 4$ ✓. Limit $T \to 0$ as $m_{2} \to \infty$ ✓.

Example 5 — Cell G · Two masses lifted by an upward force

Forecast: The whole assembly accelerates upward. Will the lower tension $T_{1}$ (holding $m_{1}$ ) be more or less than $m_{1} g = 10 N$ ?

Step 1 — System (both masses) to find $a$ : $F - (m_{1} + m_{2}) g = (m_{1} + m_{2}) a \Rightarrow a = \frac{F}{m _{1} + m _{2}} - g = \frac{30}{2} - 10 = 5 m/s^{2} (up)$ Why system first? The internal string tension cancels; only $F$ and total weight remain.

Step 2 — Isolate $m_{1}$ (feels its weight down and lower tension $T_{1}$ up): $T_{1} - m_{1} g = m_{1} a \Rightarrow T_{1} = m_{1} (g + a) = 1 (10 + 5) = 15 N$ Why up-positive for $m_{1}$ ? It accelerates upward, so upward is its natural positive direction.

Verify: $T_{1} = 15 N > m_{1} g = 10 N$ ✓ (accelerating up needs extra upward pull). Check $m_{2}$ : forces are $F$ up, $T_{1}$ down, weight down: $F - T_{1} - m_{2} g = 30 - 15 - 10 = 5 = m_{2} a = 1 \cdot 5$ ✓.

Example 6 — Cell H · Incline + hanging mass (sign of $sin θ$ )

Reading the figure below: the lavender wedge is the incline; its base angle is the slate-labelled $3 0^{\circ}$ . The mint block $m_{1}$ rests on the slope, its string running up along the slope to the butter pulley at the apex, then down to the coral hanging block $m_{2}$ . The key coral arrow on the slope is not the full weight — it is only the along-slope component $m_{1} g sin θ$ , the part of gravity that actually tries to slide $m_{1}$ downhill. The lavender arrows are the tension $T$ : one pulling $m_{1}$ up the slope, one holding $m_{2}$ up. Compare the coral down-slope pull against $m_{2}$ 's full weight to decide which way the system moves.

Forecast: The incline pulls $m_{1}$ down the slope with only the component $m_{1} g sin θ = 2 \cdot 10 \cdot 0.5 = 10 N$ , while $m_{2}$ pulls with its full weight $30 N$ . Which way does the system move — does $m_{2}$ descend, or does $m_{1}$ slide down and haul $m_{2}$ up?

Step 1 — Resolve $m_{1}$ 's weight along the slope. Why resolve? On an incline gravity has two parts: along the slope ( $m_{1} g sin θ$ , which can pull $m_{1}$ down the ramp) and into the slope ( $m_{1} g cos θ$ , balanced by the normal force). Only the along-slope part competes with the string.

Assume $m_{2}$ descends (so $m_{1}$ climbs the ramp), each with acceleration $a$ :

$m_{2}$ (down positive): $m_{2} g - T = m_{2} a$
$m_{1}$ (up-the-slope positive): $T - m_{1} g sin θ = m_{1} a$

Step 2 — Add to cancel $T$ : $m_{2} g - m_{1} g sin θ = (m_{1} + m_{2}) a \Rightarrow a = \frac{( m _{2} - m _{1} s i n θ ) g}{m _{1} + m _{2}} = \frac{( 3 - 2 \cdot 0.5 ) \cdot 10}{5} = 4 m/s^{2}$ Why check the sign of the numerator? If $m_{2} - m_{1} sin θ$ came out negative, our assumed direction was wrong and the block would slide down the incline instead. Here it is $+ 2 > 0$ , so $m_{2}$ really does descend. ✓

Step 3 — Tension from $m_{2} g - T = m_{2} a$ : $T = m_{2} (g - a) = 3 (10 - 4) = 18 N$

Verify: $a = 4$ , $T = 18 N$ . Check $m_{1}$ : $T - m_{1} g sin θ = 18 - 10 = 8 = m_{1} a = 2 \cdot 4 = 8$ ✓. And $T = 18 < m_{2} g = 30$ ✓ (descending mass).

Worked example Degenerate edge check — the perfectly balanced incline (

$m_{2} = m_{1} sin θ$ ) Statement: Keep $m_{1} = 2 kg$ , $θ = 3 0^{\circ}$ (so $m_{1} sin θ = 1$ ), but now set $m_{2} = 1 kg$ exactly. What happens? Step 1 — Plug into the general acceleration: $a = \frac{( m _{2} - m _{1} s i n θ ) g}{m _{1} + m _{2}} = \frac{( 1 - 2 \cdot 0.5 ) \cdot 10}{3} = \frac{0}{3} = 0$ Why does the numerator vanish? The hanging weight $m_{2} g$ exactly equals the down-slope pull $m_{1} g sin θ$ ; the two tugs cancel, so nothing accelerates. Step 2 — Tension in this static case: with $a = 0$ , $T = m_{2} (g - a) = m_{2} g = 1 \cdot 10 = 10 N$ . And indeed on the slope $T = m_{1} g sin θ = 10 N$ — both sides agree. Verify: $a = 0$ , $T = 10 N = m_{2} g = m_{1} g sin θ$ ✓. This is the incline's version of the balanced Atwood (Cell B): the numerator $m_{2} - m_{1} sin θ$ passing through zero is exactly the boundary between "system slides one way" and "the other".

Example 7 — Cell I · Real-world: person in an elevator

Forecast: Accelerating upward, will the person feel heavier or lighter than their usual $600 N$ ? And will the cable carry more than the total static weight $5000 N$ ?

Step 1 — FBD of the person alone. Two forces act: the normal force $N$ from the scale (up) and the weight $m g$ (down). The person accelerates up with $a$ , so take up as positive: $N - m g = ma \Rightarrow N = m (g + a) = 60 (10 + 2) = 720 N$ Why is the scale reading $N$ ? The scale reads the force the person presses down on it; by Newton's third law that equals the upward $N$ the scale pushes back. So the "reading" is $N$ .

Step 2 — FBD of the whole elevator system. Treat elevator + person as one body of mass $M = 500 kg$ ; the cable tension $T$ pulls up, total weight $M g$ pulls down, and the whole thing accelerates up with $a$ : $T - M g = M a \Rightarrow T = M (g + a) = 500 (10 + 2) = 6000 N$ Why bundle person + elevator for the cable? The cable holds everything inside; the internal floor-on-person and person-on-floor forces cancel for the combined body, leaving only $T$ and the total weight.

Verify: $N = 720 N > 600 N$ ✓ (feels heavier when accelerating up — this is "apparent weight"). $T = 6000 N > 5000 N$ ✓ (cable works harder than in a static hang). Units: $kg \cdot (m/s^{2}) = N$ ✓.

Example 8 — Cell J · Exam twist: three masses, find the middle tension

Forecast: There are two different strings, so the two tensions need not be equal. Which do you expect to be larger — the string closer to the pull, or the one at the back?

Step 1 — System first for $a$ : $F = (m_{1} + m_{2} + m_{3}) a \Rightarrow a = \frac{12}{6} = 2 m/s^{2}$ Why the system view? All three share one acceleration; internal tensions cancel for the whole train, leaving only $F$ .

Step 2 — Isolate the back block $m_{3}$ (only $T_{23}$ pulls it): $T_{23} = m_{3} a = 3 \cdot 2 = 6 N$ Why $m_{3}$ alone? The last string must accelerate only $m_{3}$ , so $T_{23} = m_{3} a$ directly.

Step 3 — Isolate the group ${m_{2} + m_{3}}$ together (string $T_{12}$ accelerates both of them): $T_{12} = (m_{2} + m_{3}) a = (2 + 3) \cdot 2 = 10 N$ Why group $m_{2}$ and $m_{3}$ ? Everything behind the middle string is pulled by that string, so $T_{12}$ carries the load of both trailing blocks.

Verify: $T_{12} = 10 N > T_{23} = 6 N$ ✓ — the front string carries more, as forecast. Check $m_{1}$ : $F - T_{12} = 12 - 10 = 2 = m_{1} a = 1 \cdot 2$ ✓. Check $m_{2}$ : $T_{12} - T_{23} = 10 - 6 = 4 = m_{2} a = 2 \cdot 2$ ✓. This confirms tension differs between strings — a direct link to when tensions are and aren't equal.

Active Recall

Recall Which cells produce

$T > m g$ vs $T < m g$ ? Falling / descending mass → $T < m g$ (Ex 1, 3, 6). Accelerating upward → $T > m g$ (Ex 5, 7). Static → $T = m g$ (Ex 2a, incline edge check).

Recall What tells you your assumed direction was wrong?

A negative acceleration numerator (Ex 6: $m_{2} - m_{1} sin θ$ ). If it is exactly zero, the system is balanced and static ( $a = 0$ ).

Recall Are two different strings guaranteed equal tension?

No. Within one massless string tension is uniform, but two separate strings (Ex 8) carry whatever their own trailing loads demand.

Connections

Newton's Second Law — every step above is $F = ma$ on one chosen body.
Free Body Diagrams — the drawing that makes each equation obvious.
Constraint Relations — why one $a$ serves all connected bodies.
Atwood Machine — Cells A, B, C.
Frictionless Pulleys vs Pulleys with Inertia — the limit where Cell J's "different tensions" also invades a single pulley.
Normal Force — the vertical partner used in Cells D and I.