ARCHETYPE: Monkey Climbing a Rope

A monkey grabs a rope hanging from a branch and climbs. Simple enough — until an examiner asks the question that has been snapping imaginary ropes since it entered the NCERT exercise set: the rope can bear at most 600 N; in which of these four motions does it break? Climbing up fast, climbing up slowly, sliding down, accelerating down — most students rank them by effort, or by speed, and lose the marks. The rope does not care about effort, and it cannot feel speed. It feels exactly one thing: the force at the monkey's hands. And by Newton's second law, that force is dictated by the monkey's acceleration alone. That is the entire archetype. A climbing monkey is a lift problem in disguise — the same $m(g+a)$ and $m(g-a)$ readings as a passenger on a weighing scale, dressed in fur. Once you read every phrase in the problem ("climbs with acceleration", "uniform speed", "slides freely") as a statement about $a$ , the four-case menu collapses to one line of algebra. JEE then adds its favourite twist: hang a counterweight on the other side of a pulley and let the monkey climb relative to the rope — a setup with a genuinely beautiful punchline that Lewis Carroll used to torment Oxford dons. This article drills the setup to mastery: the tension ledger for a fixed rope, the breaking condition, the monkey-versus-counterweight machine, and the exam traps. Learn to translate verbs into accelerations, and every monkey problem ever set becomes the same three-line solution.

Key takeaways

A monkey of mass $m$ on a rope feels two forces: tension $T$ up, weight $mg$ down. Newton's second law gives $T = m(g + a)$ with $a$ the monkey's upward acceleration.
The rope reads acceleration, not speed and not effort: climbing up at any uniform speed gives $T = mg$ , exactly as if the monkey hung still.
Accelerating up (or decelerating while descending): $T = m(g+a) > mg$ . Accelerating down (or decelerating while rising): $T = m(g-a) < mg$ . Free slide: $T = 0$ .
The rope breaks when the required tension exceeds its rating: maximum climbing acceleration $a_{\max} = \dfrac{T_{\max}}{m} - g$ .
Over a pulley with an equal counterweight, monkey and weight always share the same acceleration — the monkey can never out-climb the bananas.
Key idea: translate every verb in the problem into the monkey's acceleration first; the tension is then always $T = m(g + a)$ with a signed $a$ .

The core insight

The tension in a rope carrying a climbing monkey is fixed by the monkey's acceleration through $T - mg = ma$ ; how fast the monkey moves, or how furiously it climbs, changes nothing unless it changes $a$ — so every monkey problem is solved by converting its verbs into a single signed acceleration.

Where this lives today

Swap the monkey for a human and the rope for a climbing rope, and equation (2) becomes safety engineering. Sport-climbing ropes are rated not by the climber's weight but by the fall arrest — the braking phase, where a falling climber decelerates over the rope's stretch and the tension spikes to several times body weight, exactly worked example 2 at scale. Dynamic ropes are designed to stretch precisely so that the stopping distance grows and $a = v^2/2s$ shrinks; the UIAA impact-force standard caps the peak tension a rope may transmit in a standardized fall. The same ledger governs elevator safety brakes, crane load ratings ("do not shock-load"), and the winch cables that lower rescue baskets from helicopters: steady lowering costs $mg$ , but every change of speed is billed at $m(g+a)$ , and the cable must be rated for the verbs, not the nouns.

The setup

Geometry first, because in an archetype the geometry is the physics. A light, inextensible rope hangs from a rigid support — a branch, a ceiling hook — and a monkey of mass $m$ grips it. The monkey may hang, climb up, climb down, or slide; the rope may also run over a pulley with a counterweight on the far side. In every version, the monkey touches nothing but the rope, so the free-body diagram of the monkey has exactly two entries: the rope's pull $T$ upward along the rope, and the weight $mg$ downward.

Figure

The monkey on a fixed rope — one free-body diagram, four exam cases

Two readings of this picture carry the whole archetype, and both earn explicit statements in a board answer:

The grip force and the tension are a third-law pair. The monkey pulls down on the rope with its hands; the rope pulls up on the monkey with the same magnitude $T$ . So "the force the monkey exerts on the rope" and "the tension at the monkey's hands" are one number. When a question asks either, compute the other. For a light rope, that tension is the same all along the rope and is also the force on the branch above.

The rope is a lift in disguise. Compare the monkey's FBD with a passenger standing on a lift's weighing scale: normal force $N$ up, $mg$ down, $N = m(g+a)$ . Replace $N$ by $T$ and the physics is identical — the archetype you drilled in Lifts & Apparent Weight transfers wholesale. A monkey accelerating upward is a "heavy" passenger; a monkey accelerating downward is a "light" one; a monkey in free slide is weightless with a rope running uselessly through its paws. The single most-tested sentence follows immediately: climbing at uniform speed changes nothing. Whether the monkey hangs at rest, climbs up at a steady $2$ m/s, or descends at a steady $5$ m/s, its acceleration is zero and $T = mg$ exactly. Examiners phrase this trap a dozen ways — "climbs up with uniform speed", "goes down at constant rate" — and every one of them is the same equation.

The master solution pattern

Every monkey problem begins by converting words into a signed acceleration, up positive. "Climbs up with acceleration $a$ " means $+a$ . "Climbs down with acceleration $a$ " means $-a$ . "Uniform speed" (any direction) means $0$ . "Slides down freely" means $-g$ . "Moving down and coming to rest" means the velocity is down but shrinking — acceleration up, so $+a$ . Only after this translation do you touch Newton's law. Newton's second law for the monkey, up positive:

T - mg = ma \qquad (1)

where: $T$ — the rope tension at the monkey's hands (equal, by the third law, to the force the monkey exerts on the rope); $m$ — the monkey's mass; $a$ — the signed vertical acceleration. Read it physically: the rope must first carry the weight ( $mg$ ) and then fund whatever acceleration the monkey demands ( $ma$ ). Solving for the tension,

T = m(g + a) \qquad (2)

where: $a > 0$ makes the rope work harder than the hanging weight, $a < 0$ lets it relax below $mg$ , and $a = -g$ (free slide, no grip) sends $T$ to zero. Equation (2) with a signed $a$ is the whole fixed-rope archetype — the four exam cases are four inputs to one formula. The rope survives only while the demanded tension stays inside its rating $T_{\max}$ :

m(g + a) \le T_{\max} \qquad\Longrightarrow\qquad a_{\max} = \frac{T_{\max}}{m} - g \qquad (3)

where: $a_{\max}$ — the largest upward acceleration the monkey can use without snapping the rope. For the NCERT classic ( $m = 40$ kg, $T_{\max} = 600$ N, $g = 10$ m/s²): $a_{\max} = 15 - 10 = 5$ m/s². Note the corollary examiners love: descending can never break a rope that hanging does not — while moving down with any downward acceleration, $T < mg$ ; the danger returns only when the monkey brakes its descent, because braking while descending is upward acceleration.

the problem says	acceleration (up +)	tension
hangs at rest / climbs either way at uniform speed	$0$	$T = mg$
climbs up with acceleration $a$	$+a$	$T = m(g+a)$
climbs / accelerates down with $a$	$-a$	$T = m(g-a)$
slides down freely (no grip)	$-g$	$T = 0$
descending, braking to rest with magnitude $a$	$+a$	$T = m(g+a)$

Derivation AThe tension ledger for a fixed rope — all four cases from one equation

We derive the full case table above from a single application of Newton's second law, then extract the breaking condition (3). The strategy is deliberate minimalism: draw the monkey's free-body diagram once, write the second law once with a signed acceleration, and let the four exam cases enter only as values of $a$ . Watch one subtlety — the direction of the velocity never appears anywhere in the derivation, which is exactly why "climbs at uniform speed" questions are free marks.

Step 1. Free-body diagram of the monkey: tension $T$ up (the rope can only pull along itself), weight $mg$ down. Nothing else touches the monkey. Why the monkey's climbing action does not add a third force: the hands' pull on the rope acts ON THE ROPE. Its third-law partner is the tension itself. Forces come in pairs acting on different bodies — the monkey's FBD keeps only the forces on the monkey.

Step 2. Newton's second law along the vertical, up positive:

T - mg = ma \;\;\Rightarrow\;\; T = m(g + a)

Step 3. Feed in the cases. Uniform speed (any direction): $a = 0 \Rightarrow T = mg$ . Accelerating up at $a$ : $T = m(g+a)$ . Accelerating down at $a$ : $T = m(g-a)$ . Free slide: the paws exert no grip, so $T = 0$ and indeed $a = -g$ consistently. Why "descending and stopping" lands in the heavy row: velocity down, speed shrinking, so the acceleration vector points up — the same translation ritual as lift problems. The rope's worst moment in any descent is the braking at the bottom.

Step 4. Breaking condition. The rope fails when $T > T_{\max}$ :

m(g + a) > T_{\max} \;\;\Rightarrow\;\; a > \frac{T_{\max}}{m} - g = a_{\max}

Only upward accelerations can threaten a rope rated above the hanging weight; downward accelerations reduce $T$ . Verified: NCERT values $m = 40$ kg, $T_{\max} = 600$ N, $g = 10$ : climbing up at $6$ m/s² demands $T = 40(16) = 640$ N — snaps; down at $4$ m/s² needs $240$ N — safe; uniform $5$ m/s needs $400$ N — safe; free fall needs $0$ — safe. Matches the NCERT answer key exactly.

The pulley variant — monkey versus counterweight

JEE's favourite escalation replaces the branch with a light, frictionless pulley and hangs a counterweight $M$ on the rope's other end. Now the rope can move, and the honest coordinate is the monkey's acceleration relative to the rope, $a_{\rm rel}$ (its climbing rate of change), because that is what the monkey controls. The rope's own acceleration is set by the counterweight's response. Two results carry every exam question. First, the general machine: if the counterweight accelerates upward at $A$ , the rope on the monkey's side runs downward at $A$ , so the monkey's absolute acceleration is $a_{\rm rel} - A$ upward. Newton's law on each body (one tension $T$ throughout the light rope over the light pulley — the Atwood idealizations):

T - Mg = MA, \qquad T - mg = m(a_{\rm rel} - A) \qquad (4)

where: two equations, two unknowns ( $T$ , $A$ ) once the monkey announces $a_{\rm rel}$ . Solve by elimination as in any Atwood problem. The special case examiners adore: keep the counterweight stationary ( $A = 0$ ) — then $T = Mg$ immediately, and the monkey must climb with $a_{\rm rel} = \dfrac{(M - m)g}{m}$ , which only works for $M > m$ . Second, the equal-mass theorem — Lewis Carroll's puzzle. Set $M = m$ in (4): subtracting the equations gives $MA = m(a_{\rm rel} - A)$ , so $A = \tfrac12 a_{\rm rel}$ and both bodies accelerate upward at the same rate $A$ . However the monkey climbs, rests, or lunges, the counterweight mirrors its ascent exactly — the monkey can never rise above the bananas on the far side.

Go deeper · Derivation B · 1 minMonkey versus counterweight — the general solution and the equal-mass mirror

Derivation BMonkey versus counterweight — the general solution and the equal-mass mirror

We derive the pulley-variant results (4) and the equal-mass theorem. The strategy: treat it as an Atwood machine in which one "block" (the monkey) has an extra, self-chosen acceleration relative to the rope. The crucial discipline is bookkeeping — three accelerations live here (monkey's absolute, counterweight's absolute, monkey's relative-to-rope) and exams are lost by conflating them. We take up as positive for both bodies.

Step 1. Idealizations: light rope, light frictionless pulley $\Rightarrow$ one tension $T$ throughout. Let the counterweight accelerate up at $A$ ; the rope being inextensible, the rope on the monkey's side accelerates down at $A$ .

Step 2. Kinematic constraint for the monkey. Climbing at $a_{\rm rel}$ relative to the rope, while its handhold accelerates down at $A$ , gives the monkey's absolute acceleration

a_{\rm monkey} = a_{\rm rel} - A

Why relative-to-rope is the natural variable: the monkey's muscles set how fast it moves along the rope; the ground-frame result depends on what the rope itself does. Exams state $a_{\rm rel}$ ("climbs with acceleration $a$ relative to the rope") — never assume it is the absolute acceleration.

Step 3. Newton's second law on each body:

\text{counterweight: } T - Mg = MA, \qquad \text{monkey: } T - mg = m(a_{\rm rel} - A)

Step 4. Eliminate $T$ : $Mg + MA = mg + m\,a_{\rm rel} - mA$ , so

A = \frac{m\,a_{\rm rel} - (m - M)g}{M + m}, \qquad T = M(g + A)

For the stationary-counterweight special case set $A = 0$ : $T = Mg$ and $a_{\rm rel} = \dfrac{(M-m)g}{m}$ .

Step 5. Equal masses ( $M = m$ ): the gravity terms cancel, leaving $A = \tfrac12 a_{\rm rel}$ — and the monkey's absolute acceleration is $a_{\rm rel} - A = \tfrac12 a_{\rm rel} = A$ . Monkey and counterweight rise identically. If the monkey climbs a length $\ell$ along the rope, each rises $\ell/2$ ; starting level with the bananas, it stays level with the bananas all the way to the pulley. Verified: limits — $a_{\rm rel} = 0$ recovers the ordinary Atwood machine $A = (m - M)g/(M+m)$ with sign conventions consistent (heavier monkey side descends); $M = m$ , $a_{\rm rel} = 0$ gives $A = 0$ , the balanced machine. Numbers: $m = 10$ , $M = 15$ kg, $g = 10$ : keeping the weight still needs $a_{\rm rel} = 5$ m/s² and $T = 150$ N — checks in worked example 3.

Figure

The pulley variant — the monkey climbs the rope, the counterweight answers

Limiting cases

case	result	meaning
$a = 0$ (any uniform speed)	$T = mg$	the rope cannot feel velocity — hanging and steady climbing are identical
$a = -g$ (free slide)	$T = 0$	no grip, no force: the rope runs through open paws
$a = a_{\max} = T_{\max}/m - g$	$T = T_{\max}$	the fastest survivable climb; any harder lunge snaps the rope
pulley, $M = m$ , any climbing	$A = \tfrac12 a_{\rm rel}$ both sides	the equal-mass mirror: monkey and bananas rise together
pulley, $A = 0$ demanded	$a_{\rm rel} = (M - m)g/m$	possible only for $M > m$ ; a light counterweight cannot be held still by climbing
pulley, $a_{\rm rel} = 0$	ordinary Atwood machine	the monkey riding the rope is just a block of mass $m$

Common trapA faster climb means a tighter rope

It does not — and this misconception is the archetype's number-one mark harvester. Speed never enters $T - mg = ma$ : a monkey climbing steadily at $8$ m/s loads the rope with exactly $mg$ , the same as one hanging asleep. What tightens the rope is changing speed — the lunge that begins a climb, the brake that ends a descent. The NCERT four-case classic is built entirely on this point: "climbs up with uniform speed of 5 m/s" is the safe case, while "climbs up with acceleration 6 m/s²" is the one that snaps the 600 N rope. Translate the verbs to accelerations before you compute anything, and note the symmetric trap on the way down: a monkey descending but slowing is accelerating upward, so its rope is tighter than $mg$ — the most dangerous instant of a descent is the stylish stop at the bottom.

Worked examples

1 · Board level — the NCERT four-case classic (CBSE/NEET pattern, 3 marks)

A monkey of mass $40$ kg climbs a rope that can withstand a maximum tension of $600$ N. In which of the following cases will the rope break: the monkey (a) climbs up with an acceleration of $6\ \mathrm{m/s^2}$ ; (b) climbs down with an acceleration of $4\ \mathrm{m/s^2}$ ; (c) climbs up with a uniform speed of $5\ \mathrm{m/s}$ ; (d) falls down the rope nearly freely? Take $g = 10\ \mathrm{m/s^2}$ and ignore the rope's mass.

Step 1. Translate each phrase into a signed acceleration (up positive): (a) $+6$ ; (b) $-4$ ; (c) $0$ ; (d) $-g = -10$ .

Step 2. Apply $T = m(g + a)$ case by case. (a) $T = 40(10 + 6) = 640$ N. (b) $T = 40(10 - 4) = 240$ N. (c) $T = 40(10) = 400$ N. (d) $T = 40(0) = 0$ N.

Step 3. Compare with $T_{\max} = 600$ N: only case (a) exceeds it. The rope breaks in case (a) alone.

Step 4. Sanity: the hanging weight is $400$ N, comfortably under $600$ N, so only an upward acceleration can threaten the rope — cases (b), (c), (d) never could. The maximum survivable climb is $a_{\max} = 600/40 - 10 = 5\ \mathrm{m/s^2}$ , and $6 > 5$ confirms the verdict.

2 · Board level — the braking descent (ISC/JEE-Main bridge)

A $10$ kg monkey slides down a long rope at a steady $6\ \mathrm{m/s}$ , then brings itself to rest uniformly over the final $1.5$ m. The rope is rated at $150$ N. Does it survive the stop? ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. During the steady slide, $a = 0$ and $T = mg = 100$ N — safe. All the drama is in the stop.

Step 2. Braking kinematics: $v^2 = u^2 + 2as$ along the motion gives $0 = 6^2 - 2a(1.5)$ , so $a = \dfrac{36}{3} = 12\ \mathrm{m/s^2}$ , directed upward (velocity down, speed shrinking).

Step 3. Tension while braking: $T = m(g + a) = 10(10 + 12) = \boxed{220\ \mathrm{N}}$ .

Step 4. $220 > 150$ : the rope snaps during the stop. The monkey descended safely the whole way and lost the rope in the last metre and a half — the braking phase, not the descent, is where ropes (and climbing anchors) fail. To survive, the stop must be gentler: $a \le 150/10 - 10 = 5\ \mathrm{m/s^2}$ , i.e. spread over at least $s = 36/(2 \times 5) = 3.6$ m.

3 · Competitive level — holding the counterweight still (JEE-Main pattern, ~2.5 minutes)

A monkey of mass $10$ kg holds a light rope that passes over a smooth pulley; a block of $15$ kg hangs on the other end. With what acceleration relative to the rope must the monkey climb so that the block stays stationary, and what is then the rope's tension? ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Block stationary $\Rightarrow$ its net force is zero $\Rightarrow T = Mg = 150$ N. The tension is fixed before the monkey is even considered — write this first, it is half the marks.

Step 2. Block still means the rope is still ( $A = 0$ ), so the monkey's absolute acceleration equals its relative one: $a_{\rm monkey} = a_{\rm rel}$ .

Step 3. Newton's law on the monkey: $T - mg = m\,a_{\rm rel}$ , so $150 - 100 = 10\,a_{\rm rel}$ giving $a_{\rm rel} = \boxed{5\ \mathrm{m/s^2}}$ upward, with $T = \boxed{150\ \mathrm{N}}$ .

Step 4. Read the physics: the monkey must pull the rope down with $150$ N — harder than its own weight — and the excess $50$ N drives its climb. A monkey heavier than the block could never hold it still this way: it would need a downward relative climb just to hang on. Timing note: ~2.5 min. The order of operations is the whole trick — tension from the stationary block first, monkey second. Students who start with the monkey's equation carry two unknowns and drown.

4 · Competitive level — Lewis Carroll's monkey and the bananas (JEE-Advanced flavour, ~4 minutes)

A monkey of mass $m$ and a bunch of bananas, also of mass $m$ , hang from opposite ends of a light rope over a light frictionless pulley, initially at rest and level with each other. The monkey climbs $2.0$ m along the rope. (a) How far do the bananas rise? (b) Does the monkey gain on the bananas? (c) The monkey then lets go with both hands for a moment — what happens?

Step 1. Equal masses: from derivation B, whatever climbing profile the monkey chooses, both sides share the same absolute acceleration $A = \tfrac12 a_{\rm rel}$ at every instant — so, starting together from rest, monkey and bananas always have identical velocities and displacements.

Step 2. Climbing $\ell = 2.0$ m along the rope splits evenly: the monkey rises $\ell/2 = 1.0$ m and the bananas rise $1.0$ m. Bananas rise 1.0 m — answer (a).

Step 3. (b) No. Level they began, level they remain: climbing hoists the bananas exactly as fast as it hoists the monkey. The monkey's only reward is that both approach the pulley; it can reach the bananas only there.

Step 4. (c) Hands off means no grip: the monkey's tension drops to zero, so the bananas' side also goes slack ( $T$ is one number), and both fall freely at $g$ , keeping their equal velocities. The mirror is unbreakable — grip or no grip, the symmetric machine never lets the monkey win by more than the pulley's geometry allows. Timing note: ~4 min as a reasoning problem, ~1 min once you own the equal-mass theorem. JEE-Advanced poses this as assertion–reason or multiple-correct; the killer distractor claims the bananas stay still "because the monkey's climbing is internal to the rope". It is not — the climb changes the tension, and the tension moves the bananas.

Exam traps

Reading speed as force. "Climbs up at a brisk uniform $8$ m/s" still means $T = mg$ . The rope carries acceleration, not velocity and not enthusiasm. If no rate-of-change is stated or derivable, the tension is the hanging weight.

The braking-descent sign flip. Down-and-slowing is upward acceleration: $T = m(g+a)$ , heavier than hanging. The two-readings collapse from lift problems transfers verbatim: {speeding up upward, slowing down downward} → tight rope; {speeding up downward, slowing up upward} → slack rope.

Adding a "climbing force" to the FBD. The monkey's pull acts on the rope; its third-law partner is the tension. The monkey's free-body diagram has exactly two forces — $T$ and $mg$ — in every case. A third arrow labelled "muscular force" double-books the interaction and wrecks the algebra.

Comparing accelerations instead of tensions with the rope rating. The rope breaks on tension, not on acceleration. Always compute $T$ and compare with $T_{\max}$ ; converting the rating to $a_{\max} = T_{\max}/m - g$ first is fine, but only for the upward-acceleration cases.

Relative versus absolute acceleration in the pulley variant. "Climbs with acceleration $a$ relative to the rope" is a constraint statement, not the ground-frame answer. The monkey's absolute acceleration is $a_{\rm rel} - A$ ; forgetting the rope's own motion is the standard JEE-Main error. Anchor: the stationary-counterweight case, where $A = 0$ makes them equal — the only case where sloppiness goes unpunished.

Force on the support. For a fixed light rope, the branch feels the same $T$ that the monkey feels — so a climbing lunge shakes the branch harder. Over a pulley, the axle feels $2T$ , exactly as in the Atwood machine.

Which $g$ . CBSE/ISC expect $9.8\ \mathrm{m/s^2}$ unless stated; the NCERT monkey problem itself declares $g = 10$ , as do JEE-Main/Advanced and NEET almost always. Use what the paper states and never mix values mid-solution.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1642–1727 · Cambridge & London

Newton's Principia (1687) supplied both laws this archetype runs on: the second law, which prices the monkey's acceleration in newtons of rope tension, and the third, which guarantees that the monkey's pull on the rope and the rope's pull on the monkey are one number. Newton was famously careful about the third law's bookkeeping — the pair acts on different bodies — which is precisely the discipline that keeps a "muscular force" arrow off the monkey's free-body diagram. He tested his laws on pendulums, planetary orbits and tides rather than monkeys; the primate version had to wait for the age of competitive examinations, which discovered that a climbing animal is the most vivid possible test of whether a student believes forces respond to acceleration rather than to effort.

History

1687

— Newton's Principia publishes the second and third laws — the complete toolkit for every rope-and-climber problem.

1893

— Lewis Carroll (the Oxford logician C. L. Dodgson) circulates his monkey-and-weight pulley puzzle; Victorian scientists argue publicly about whether the counterweight rises.

1905

— Textbook treatments settle Carroll's puzzle with momentum and force analysis: the equal-mass counterweight mirrors the monkey exactly.

1960

s — The 600-newton climbing-monkey problem enters Indian physics course materials, becoming one of the most-solved mechanics exercises on Earth.

2023

— The rationalized NCERT retains the monkey classic in Chapter 5's main exercises — four verbs, one equation, generations of students.

Going further

Texts. NCERT Physics Class 11, Ch. 5 (Laws of Motion), the monkey-and-rope exercise · HC Verma, Concepts of Physics Vol. 1, Ch. 5 (monkey-and-pulley problems) · Irodov, Problems in General Physics, §1.4 for the relative-acceleration machines.

Provenance. All results are standard. A verifier should recompute: the four NCERT cases ( $640/240/400/0$ N against $600$ N, only the first breaks); $a_{\max} = 5\ \mathrm{m/s^2}$ ; example 2's braking ( $a = 12\ \mathrm{m/s^2}$ up, $T = 220$ N, safe stop needs $\ge 3.6$ m); example 3 ( $T = 150$ N, $a_{\rm rel} = 5\ \mathrm{m/s^2}$ ); the equal-mass theorem $A = \tfrac12 a_{\rm rel}$ and the 1.0 m banana rise.

Prerequisites. Newton's Second Law · Tension & Ideal Strings · Lifts & Apparent Weight. Leads to. Limiting & Extreme Cases (technique) · Pulley With Mass (rotation) · Tension & Ideal Strings (UG deep dive).

Why doesn't the monkey's climbing effort appear as a force on its free-body diagram?

Because the effort acts on the rope, not on the monkey. The monkey's muscles make its hands pull the rope downward; the third-law partner of that pull is the rope's tension pulling the monkey up — and $T$ is already on the diagram. "Effort" changes the FBD only by changing the value of $T$ : lunge and $T$ rises, relax and it falls. Two forces, always: $T$ and $mg$ .

How can climbing up and hanging still give the same tension? Climbing feels so much harder!

The extra effort of steady climbing goes into hauling your body's weight through height — energy expenditure, not extra force. Force and energy are different currencies: the rope's tension is set by $T = m(g+a)$ , and steady climbing has $a = 0$ just like hanging. Your muscles burn because they do work $mgh$ against gravity, at whatever tension the acceleration demands. The rope bills for acceleration; your muscles bill for altitude.

Descending is supposed to slacken the rope. So how do ropes break during descents?

During the descent proper, they don't — any downward acceleration gives $T = m(g-a) < mg$ , and even a steady slide costs only $mg$ . The danger is the stop. Braking a downward velocity means accelerating upward, and $T = m(g+a)$ can spike far past the hanging weight if the stop is short: worked example 2's monkey needed $220$ N to stop in $1.5$ m though it weighs only $100$ N. Climbers call this shock-loading; the cure is always a longer, softer stop.

In the pulley problem, what exactly does "climbs with acceleration a relative to the rope" mean?

It describes hand-over-hand motion along the rope itself: if you painted marks on the rope, the monkey passes them at an increasing rate, gaining speed $a_{\rm rel}t$ along the rope. The ground-frame acceleration is different whenever the rope itself moves — absolute = $a_{\rm rel} - A$ if the monkey's side of rope accelerates down at $A$ . Exams give you $a_{\rm rel}$ because that is what the monkey chooses; the machine (through tension and the counterweight) decides $A$ .

Why can't the monkey ever get above the equal-mass bananas?

One rope, one tension. Whatever $T$ the monkey's climbing produces acts equally on the bananas' side, and with equal masses, equal net force means equal acceleration — at every instant. Identical accelerations from a level, resting start mean identical heights forever. The monkey's climbing does lift it — but it lifts the bananas just as fast. The only place they can meet is at the pulley, where the geometry ends the race in a draw.

If the monkey lets go completely, does the counterweight stay put?

No — everything falls. Letting go sets the grip force, and hence the tension, to zero along the whole rope. The counterweight then has only its weight acting: free fall at $g$ . The monkey too. Tension is a single number in a light rope; the monkey cannot zero its own side while leaving the other side taut.

Does the mass of the rope ever matter?

At this level, no — every exam version declares the rope light, which is what makes $T$ one number from the monkey's hands to the support. A heavy rope carries more tension near the top (each section holds the rope below it plus the monkey), so "where does the rope break?" becomes a real question with a real answer: at the top. That refinement lives in the UG tension article; for board and JEE work, "light rope" is your licence for a single $T$ .