ARCHETYPE: Spring Balances & Weighing Machines

Hang a 1 kg mass from a spring balance and it reads 1 kg. Reassuring — and misleading, because the balance has no idea what a kilogram is. Inside the case there is only a spring, and a spring can report exactly one thing: how hard it is being stretched. The dial converts that force into "kg" by quietly assuming your weight equals $mg$ with the factory's value of $g$ . The moment anything disturbs that assumption — an accelerating lift, a second balance in the chain, a pulley bolted to the hook, a beaker of water under the load — the reading and the mass part company, and an examiner is waiting at exactly that fork. Every board that tests this archetype builds the same short list of ambushes. Two balances in series: do they share the load or both read the full weight? A balance pulled by 10 N from both ends: 20 N or 10 N? An Atwood machine hanging from a balance: does it read the total weight of the blocks even while they accelerate? A stone lowered into water without touching the beaker: what happens to the balance above and the weighing machine below? Each has a one-line answer once you hold the master key: a spring balance is a string that tells you its tension; a weighing machine is a floor that tells you its normal force. This article drills that key through every standard variant — series and tug-of-war readings, lifts, pulleys, and the submerged-block classic — plus the traps that separate an instrument's reading from the truth it seems to promise.

Key takeaways

A spring balance reads the tension in its own spring — the force at its hook — calibrated in kg by dividing by the factory's $g$ . A weighing machine reads the normal force on its platform.
Golden rule: replace a light spring balance by a string and ask for the tension; replace a weighing machine by a floor and ask for $N$ .
Two light balances in series both read the full weight — tension is one number along the line; nothing is "shared".
A balance pulled by $F$ from each end reads $F$ , not $2F$ : one pull is the load, the other is the anchor.
In a lift, readings become apparent weight: $m(g+a)$ accelerating up, $m(g-a)$ accelerating down, zero in free fall.
An accelerating Atwood machine hung from a balance reads $2T = \dfrac{4m_1m_2g}{m_1+m_2}$ — less than the blocks' total weight.
Key idea: instruments report forces, not masses — find the tension (balance) or the normal force (machine) with an ordinary FBD, and the "reading" falls out.

The core insight

A spring balance reads the tension at its hook and a weighing machine reads the normal force on its platform; every reading problem reduces to an ordinary free-body-diagram calculation of that one force — the kilogram markings are just newtons divided by the calibrator's $g$ .

Where this lives today

Every supermarket hanging scale, luggage scale and fishing scale is this archetype in a plastic case — and so is the accelerometer in your phone. An accelerometer is a tiny mass on a silicon spring whose "reading" is exactly the apparent-weight tension of equation (2): it reports $g$ at rest, zero in free fall, and the extra $m a$ every time you brake or corner. Load cells under highway weighbridges are weighing machines reading $N$ ; astronauts "weigh" themselves on the ISS with a vibrating chair (measuring mass through oscillation frequency) precisely because a spring balance in orbital free fall reads zero forever — the ultimate lift problem. And the submerged-block bookkeeping of worked example 4 is how marine engineers compute mooring-line loads: what buoyancy carries, the crane's balance no longer must.

The setup

Two instruments, two forces. A spring balance is a spring in a case: the load stretches the spring, and the pointer reports the stretch, which Hooke's law converts to the tension in the spring. A weighing machine (bathroom scale, platform scale) is a stiff spring you stand on: it compresses, and the dial reports the normal force pressing on the platform. Neither device contains any mass-measuring magic — the kg markings are newtons divided by $9.8$ (or $10$ ), decided at the factory.

Figure

Two instruments, two forces — the balance reads tension, the machine reads normal force

Two idealizations make every exam version solvable in one line, and both deserve explicit mention in a board answer:

The balance is light. A massless spring (and case) must have zero net force, so the pulls on its two ends are equal — the balance transmits tension unchanged, exactly like an ideal string. That is the licence for the golden substitution: mentally delete the balance, replace it with a string, and compute the tension there. The reading is that tension. (In newtons; divide by the paper's $g$ if the dial is marked in kg.)

The hook and the anchor both pull. A balance in use is always pulled from both ends — the load on the hook, and whatever holds the balance up on the other end. When it hangs from a ceiling with a 10 N weight below, the ceiling pulls up with 10 N too. So "pulled by 10 N at each end" is not a new situation; it is every situation, and the reading is 10 N, not 20 N. The second pull is not a second load — it is the anchor doing its job. For the weighing machine the same logic runs through the normal force: the platform reads $N$ , whatever $N$ turns out to be — $mg$ when everything is static, $m(g \pm a)$ in an accelerating lift, less than $mg$ if something else helps hold you up.

The master solution pattern

Every reading problem is the same three moves. Replace the instrument by the force it measures (balance → string tension at that point; machine → normal force). Draw the FBD of a conveniently chosen body. Solve for that one force with Newton's second law — including $a$ if anything accelerates. The static hang gives the calibration case:

R = T = mg \qquad (1)

where: $R$ — the reading in newtons; the dial shows $R/g$ in kilograms. For a 5 kg mass at $g = 10$ : $R = 50$ N, dial shows "5 kg". This is the only case where the reading equals the true weight — every variant below breaks it deliberately. Put the balance (or machine) in a lift with acceleration $a$ and the reading becomes apparent weight:

R = m(g + a) \quad \text{(up positive)} \qquad (2)

where: the signed- $a$ ritual from the lift archetype applies verbatim — up-speeding or down-braking reads heavy, down-speeding or up-braking reads light, free fall reads zero. A balance is just a lift passenger holding the load by a string. Bolt an Atwood machine (masses $m_1 \ne m_2$ over a light pulley) to the hook and the reading drops below the total weight:

R = 2T = \frac{4\,m_1 m_2\, g}{m_1 + m_2} \;<\; (m_1 + m_2)g \qquad (3)

where: the balance supports the pulley, which feels the two string tensions, not the blocks' weights directly. While the blocks accelerate, their centre of mass accelerates downward, so the support carries less than the full weight — the shortfall is exactly $(m_2 - m_1)a$ . Equality holds only for a balanced machine.

situation	what to compute	reading
mass hangs statically	tension	$mg$
two light balances in series	one tension throughout	each reads $mg$
pulled by $F$ from both ends	tension	$F$ , never $2F$
in a lift, acceleration $a$ (up +)	tension / normal force	$m(g+a)$
Atwood machine on the hook	$2T$ at the pulley	$4m_1m_2g/(m_1+m_2)$
block on frictionless incline, balance along slope	tension	$mg\sin\theta$
block submerged in water (hangs from balance)	tension	$mg - F_{\rm buoyancy}$

Derivation AWhy a light balance reads the same at both ends — series chains and the tug-of-war

We derive the transmission rule that powers the series-balances and both-ends-pulled classics: a light spring balance carries one tension, reads it, and passes it on unchanged. The strategy is the same massless-body argument that gives ideal strings a single tension, applied to the balance as a whole and then to a chain of them. Watch the punchline: "10 N on each end" turns out to be the definition of a 10 N reading, not a doubling of it.

Step 1. Model the balance as a light spring with hooks. Newton's second law for a massless object: $F_{\rm net} = 0 \cdot a = 0$ always. So the pull on the top hook and the pull on the bottom hook are equal in magnitude — call it $T$ . The pointer, geared to the spring's stretch, displays exactly this $T$ . Why massless matters: a balance with real weight $w$ hung vertically reads differently at each end (top anchor carries $T + w$ ). Exams say "light balance" to hand you the one-tension idealization — state it back to them for the method mark.

Step 2. Series chain: balance A hangs from the ceiling, balance B hangs from A's hook, mass $m$ hangs from B. Replace both balances by string segments: it is ONE line from ceiling to mass, so one tension $mg$ throughout. Put the balances back: A reads $mg$ and B reads $mg$ . Nothing halves; a tension is not a cargo to be split but a state of the whole line.

Step 3. Tug-of-war: a balance held horizontally, pulled with $10$ N leftward at one end and $10$ N rightward at the other. By Step 1 this is just a taut line with tension $10$ N — the reading is 10 N. Now compare the "ordinary" case: weight of $10$ N below, ceiling above. The ceiling pulls up with $10$ N (the balance is light and unaccelerated). Identical force diagram — the ordinary case is the tug-of-war. Why 20 N is wrong: the two pulls are the two ends of ONE state of stress. Counting both is like weighing yourself once on each foot and adding the readings.

Step 4. General rule extracted: to predict any balance's reading, cut the line at the balance and compute the tension there from either side's FBD — whichever side is simpler. Verified: series chain with a heavy balance B of weight $w$ : A reads $mg + w$ , B still reads $mg$ — matches the massless limit as $w \to 0$ . The 10 N tug-of-war matches the hanging case by symmetry. Standard treatments: HC Verma Ch. 5, NCERT Ch. 5 exemplar problems.

Go deeper · Derivation B · 1 minThe Atwood machine on a spring balance — why the reading drops while the blocks run

Derivation BThe Atwood machine on a spring balance — why the reading drops while the blocks run

We derive result (3): a light pulley carrying masses $m_1 < m_2$ hangs from a spring balance, the blocks are released, and the balance reads less than $(m_1+m_2)g$ . The strategy: the balance supports the pulley, so we need the force the string exerts on the pulley — two tension pulls — with $T$ imported from the Atwood solution. Then we check the answer against the centre-of-mass shortcut, which is the fastest exam route.

Step 1. The Atwood results (from the archetype): $a = \dfrac{(m_2 - m_1)g}{m_1 + m_2}$ , $\;T = \dfrac{2m_1m_2g}{m_1+m_2}$ , one tension throughout the light string.

Step 2. FBD of the light pulley: balance pulls up with $R$ ; the string, passing over it, pulls down with $T$ on each side. Massless pulley $\Rightarrow$ zero net force:

R = 2T = \frac{4m_1m_2g}{m_1+m_2}

Step 3. Compare with the static total weight $(m_1+m_2)g$ . Their difference:

(m_1+m_2)g - \frac{4m_1m_2g}{m_1+m_2} = \frac{(m_1 - m_2)^2 g}{m_1+m_2} \ge 0

so the running machine always reads less, with equality only when $m_1 = m_2$ (no acceleration). Why the reading drops: the system's centre of mass accelerates downward — the heavy block gains speed downward faster than the light one gains it upward. External forces must sum to (total mass) × (COM acceleration): $R - (m_1+m_2)g = -(m_1+m_2)a_{\rm com}$ , and a downward $a_{\rm com}$ means $R$ falls short of the weight.

Step 4. The COM shortcut for exams: $a_{\rm com} = \dfrac{(m_2-m_1)}{(m_1+m_2)}a$ downward, giving $R = (m_1+m_2)g - \dfrac{(m_2-m_1)^2 g}{m_1+m_2}$ — algebraically identical to $2T$ , but computable in one line if you already trust the Atwood $a$ . Verified: $m_1 = 3$ , $m_2 = 5$ kg, $g = 10$ : $T = 37.5$ N, $R = 75$ N versus static $80$ N — the balance "loses" 5 N, which equals $(m_2-m_1)^2g/(m_1+m_2) = 4 \times 10/8$ ✓. Limits: $m_1 = m_2 = m$ gives $R = 4m^2g/2m = 2mg$ , the full weight ✓; $m_1 \to 0$ gives $R \to 0$ , a lone free-falling block leaves the string slack ✓.

Figure

The Atwood machine on a spring balance — the reading is 2T, not the total weight

Limiting cases

case	result	meaning
static hang	$R = mg$	the calibration case — the only honest "kg" reading
free fall ( $a = -g$ )	$R = 0$	balance, load and lift fall together; the spring relaxes
lift accelerating up at $g$	$R = 2mg$	apparent weight doubles — dial shows twice the mass
balanced Atwood on hook ( $m_1 = m_2$ )	$R = (m_1+m_2)g$	no acceleration, no shortfall
extreme Atwood ( $m_1 \to 0$ )	$R \to 0$	slack string: the balance holds a pulley and nothing else
incline angle $\theta \to 90°$	$mg\sin\theta \to mg$	the slope becomes a cliff; the balance holds the full weight
fully submerged, density ratio $\rho_b/\rho_w \to 1$	$R \to 0$	neutral buoyancy: water carries the whole load

Common trapTwo balances in series share the weight between them

They do not — both read the full weight, and this single trap fills option (b) of a hundred MCQs. A light balance is a tension-transmitting device: hang a 1 kg mass from balance B, hang B from balance A, and the line from ceiling to mass carries one tension, $mg \approx 10$ N, everywhere along it. A reads 1 kg, B reads 1 kg. The "sharing" intuition secretly imagines the weight as a fluid poured into containers; tension is not a cargo, it is a state of the line. The same confusion powers the tug-of-war distractor: a balance pulled by 10 N from each end reads 10 N (that is just a taut 10 N line), never 20 N — and never 0 N either, since the spring is stretched, not slack. If you remember one picture, remember this: every working balance is already being pulled from both ends; the second pull is the anchor, not a second load.

Worked examples

1 · Board level — the series chain and the tug-of-war (ICSE/CBSE pattern, 3 marks)

Two light spring balances A and B are joined in series: A hangs from the ceiling, B hangs from A, and a $2.0$ kg mass hangs from B. (a) What does each balance read? (b) The pair is then unhooked and two students pull the ends horizontally with $30$ N each; what does each balance read now? ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. State the idealization (earns its mark): light balances transmit tension unchanged — replace both by string segments.

Step 2. (a) One line, one tension: $T = mg = 2.0 \times 10 = 20$ N. A reads 20 N (2.0 kg) and B reads 20 N (2.0 kg). Nothing is shared.

Step 3. (b) A straight line pulled with $30$ N at each end is a line under tension $30$ N. Each balance reads 30 N (3.0 kg). Not $60$ N — the two pulls are the two ends of one stress state.

Step 4. Sanity: hang a 3 kg mass from a ceiling-mounted balance and you reproduce case (b) exactly — load pulls down 30 N, ceiling pulls up 30 N, reading 30 N. The tug-of-war is not a special case; it is the ordinary case rotated sideways.

2 · Board level — the balance in a lift (NEET/CBSE pattern)

A $5.0$ kg fish hangs from a spring balance in a lift. Find the reading when the lift (a) accelerates upward at $2.0\ \mathrm{m/s^2}$ , (b) moves upward at a steady $4.0$ m/s, (c) accelerates downward at $3.0\ \mathrm{m/s^2}$ , (d) falls freely. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. The balance reads the string tension holding the fish: $R = T = m(g + a)$ , $a$ signed up.

Step 2. (a) $R = 5(10 + 2) = 60$ N — dial shows 6.0 kg, heavier than truth. (b) Uniform velocity: $a = 0$ , $R = 50$ N — 5.0 kg, speed is invisible. (c) $R = 5(10 - 3) = 35$ N — 3.5 kg. (d) $R = 5(0) = 0$ — the dial reads zero while the fish, balance and lift fall together.

Step 3. Sanity check with the two-readings collapse: only the acceleration direction matters; (b) reads true at any speed, and a lift braking while descending would match (a), not (c).

Step 4. Note the vocabulary trap: the fish's mass never changes — only the reading does. "Apparent weight" is the balance's honest report of the tension it feels.

3 · Competitive level — the Atwood machine on the hook (JEE-Main pattern, ~2.5 minutes)

Masses of $3.0$ kg and $5.0$ kg hang from a light string over a light frictionless pulley. The pulley itself hangs from a spring balance fixed to the ceiling. Find the balance reading (in newtons) (a) while the masses are held at rest, (b) after they are released. ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. (a) Held at rest, the whole assembly is static: the balance carries the full weight, $R = (3+5) \times 10 = \boxed{80\ \mathrm{N}}$ .

Step 2. (b) Released, the machine runs: $T = \dfrac{2m_1m_2g}{m_1+m_2} = \dfrac{2(3)(5)(10)}{8} = 37.5$ N.

Step 3. The balance supports the pulley, which feels $2T$ : $R = 2(37.5) = \boxed{75\ \mathrm{N}}$ — five newtons less than the static answer.

Step 4. Cross-check via the centre of mass: $a = \dfrac{(5-3)10}{8} = 2.5\ \mathrm{m/s^2}$ ; shortfall $= (m_2 - m_1)a = 2 \times 2.5 = 5$ N ✓. The COM accelerates downward, so the support carries less than the weight. Timing note: ~2.5 min. The entire question is knowing that the balance reads $2T$ , not $(m_1+m_2)g$ — memorize $R = 4m_1m_2g/(m_1+m_2)$ and parts (a), (b) are two substitutions. The static option 80 N sits among the answers as bait.

4 · Competitive level — the submerged block, two instruments at once (JEE-Main/NEET stretch, ~4 minutes)

A metal block of mass $2.0$ kg and density $8000\ \mathrm{kg/m^3}$ hangs from a spring balance. A beaker of water sits on a weighing machine, reading $10$ N. The block is now lowered until fully submerged, without touching the beaker. Find the new readings of (a) the spring balance, (b) the weighing machine. ( $g = 10\ \mathrm{m/s^2}$ , $\rho_{\rm water} = 1000\ \mathrm{kg/m^3}$ .)

Step 1. Block's volume: $V = m/\rho = 2.0/8000 = 2.5 \times 10^{-4}\ \mathrm{m^3}$ . Upthrust (Archimedes — the buoyancy module derives it): $F_B = \rho_{\rm w} V g = 1000 \times 2.5\times10^{-4} \times 10 = 2.5$ N, pushing the block up.

Step 2. (a) FBD of the block: $T + F_B - mg = 0$ , so $T = 20 - 2.5 = \boxed{17.5\ \mathrm{N}}$ . The balance drops from 20 N to 17.5 N — the water now carries an eighth of the load (density ratio $1000/8000$ ).

Step 3. (b) Newton's third law: the block pushes down on the water with $2.5$ N. The weighing machine must support the beaker plus this extra push: $R = 10 + 2.5 = \boxed{12.5\ \mathrm{N}}$ .

Step 4. Global sanity: total upward support $= 17.5 + 12.5 = 30$ N $=$ total weight $20 + 10$ N ✓. The 2.5 N did not vanish — it moved from the string to the platform. Examiners set exactly this bookkeeping as the final mark. Timing note: ~4 min. The reflex to train: buoyancy always shows up TWICE, once subtracted upstairs, once added downstairs, and the two readings must still sum to the total weight. Checking that sum takes five seconds and catches every sign slip.

Exam traps

Reading = mass. The dial's kilograms are newtons in disguise: $R/g$ with the factory's $g$ . Any acceleration, buoyancy, or extra support breaks the equality. Say "the balance reads the tension" in the first line of your answer and the marker knows you know.

Series balances "share" the load. Both read the full weight — one line, one tension. The heavier-instruments refinement (top balance reads load + weight of lower balance) appears only when the paper gives the balances mass.

Doubling the two-ended pull. $F$ at each end reads $F$ . A balance cannot even exist with a pull at only one end — it would accelerate off. The anchor's pull is built into every reading you have ever taken.

The Atwood static reflex. A running Atwood on a hook reads $2T < (m_1+m_2)g$ . Writing the total weight is the single most common error; the fix is automatic once you support the pulley in your FBD rather than the blocks.

Lift sign slips. Down-and-braking reads heavy; up-and-braking reads light. Translate the verbs to a signed $a$ first (the lift archetype's ritual), then $R = m(g+a)$ .

Submerged bookkeeping. The spring balance loses exactly what the weighing machine gains ( $F_B$ ), and the two readings still sum to the total weight. Forgetting the third-law push on the water — or double-counting it — breaks the sum; always run the five-second total-weight check.

Which $g$ . CBSE/ISC expect $9.8\ \mathrm{m/s^2}$ unless stated; JEE-Main and NEET almost always declare $g = 10$ . A dial marked in kg silently used one of them — on numerical papers, use the stated value throughout and never mix.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · Royal Society, London

Robert Hooke published the spring law that makes force-measuring instruments possible — ut tensio, sic vis, "as the extension, so the force" (1678) — after first hiding it in an anagram to protect his priority. As the Royal Society's Curator of Experiments he built and calibrated instruments obsessively: spring watches, barometers, the compound microscope of Micrographia. A linear spring is a force-to-length converter, and that is the entire secret of the spring balance: mark equal force steps on a scale of equal length steps and the instrument reads honestly forever — about force. The kilogram markings came later, courtesy of manufacturers who assumed $g$ would sit still; every lift question in this archetype is a small revenge on that assumption.

History

1678

— Hooke publishes ut tensio, sic vis: spring extension is proportional to force — the operating principle of every spring balance.

1770

— Richard Salter of Bilston begins manufacturing spring balances commercially; the pocket "Salter scale" spreads through British trade.

1849

— Postal and market scales standardize on spring mechanisms — cheap, portable, and honest about force if not about mass.

1901

— The 3rd CGPM formally separates mass (kilogram) from force (weight), making "a balance reads force, not mass" official metrology.

2000

s — Digital scales replace springs with strain gauges, but the physics is unchanged: they read the normal force and divide by a stored g.

Going further

Texts. NCERT Physics Class 11, Ch. 5 (Laws of Motion) · HC Verma, Concepts of Physics Vol. 1, Ch. 5 (spring-balance problems) · DC Pandey, Mechanics 1, the reading-of-instruments drill sets.

Provenance. All results are standard. A verifier should recompute: series chain and tug-of-war readings (20 N/20 N; 30 N/30 N); the lift cases ( $60/50/35/0$ N for 5 kg); the Atwood reading $R = 2T = 75$ N vs static 80 N with the 5 N COM shortfall; the submerged pair ( $17.5$ N and $12.5$ N, summing with $30$ N total) via $F_B = 2.5$ N.

Prerequisites. Spring Force & Hooke's Law · Weight, Normal Force & Apparent Weight · Lifts & Apparent Weight. Leads to. Hanging Masses & Three-Force Equilibrium · Floating-Body Problems (fluids) · Limiting & Extreme Cases (technique).

If a spring balance doesn't measure mass, why is it marked in kilograms?

Convenience and an assumption. The spring measures force; the manufacturer divides by a standard $g$ (9.8 or 10 N/kg) and prints "kg" on the dial. On the ground floor, holding still, the assumption holds and the dial tells the truth. In a lift, on the Moon ( $g_{\rm moon} \approx g/6$ : a 6 kg mass reads "1 kg"), or under water, the assumption fails and the dial reports $T/g_{\rm factory}$ — a number, not a mass. A beam balance, comparing weights on both arms, cancels $g$ and does measure mass anywhere.

Why do both balances in a series chain read the full weight? Surely the top one carries more?

The top one would carry more only if the lower balance had weight of its own. For light balances the whole line from ceiling to load is one massless chain, and a massless chain must have equal pulls at every cross-section — one tension, $mg$ , everywhere. Give balance B a real weight $w$ and indeed A reads $mg + w$ while B still reads $mg$ : the exam's "light" is what collapses the chain to one number.

In the tug-of-war, why isn't the answer 20 N when two people each pull with 10 N?

Because a reading of "10 N" already means "pulled by 10 N from each end". A balance pulled from one end only would fly off, accelerating — no steady reading at all. Cut any taut rope mentally: each half pulls the other with the tension, so both end-pulls belong to a single stress state. Adding them counts one thing twice. The hanging case makes it vivid: load 10 N below, ceiling 10 N above, reading 10 N.

Why does the balance under an accelerating Atwood machine read less than the total weight?

Because the system's centre of mass is accelerating downward — the heavy side speeds down while the light side speeds up, and mass-weighted, down wins. The external forces (reading up, weight down) must produce that downward COM acceleration, so the reading falls short of the weight by exactly $(m_2 - m_1)a = (m_2-m_1)^2 g/(m_1+m_2)$ . Lock the pulley, or hang equal masses, and the reading returns to the full weight.

Does the reading change the instant the lift starts moving, or only while it accelerates?

Only while it accelerates. The reading tracks $m(g+a)$ , and $a$ is nonzero just during the speed-up and slow-down phases. A lift cruising between floors at constant velocity — any velocity — gives the same reading as standing still. This is worth internalizing as a motion-diary: spike at departure, honest cruise, dip at arrival (for an upward ride), with the mirror pattern coming down.

For the submerged block, why does the weighing machine's reading go UP? The block never touches the beaker.

The block touches the water, and that is enough. The water pushes the block up with the buoyant force (2.5 N in the example), so by Newton's third law the block pushes the water down with 2.5 N. The water passes that push to the beaker, the beaker to the platform. No contact with the beaker is needed — the water is a middleman, not a bystander. The five-second audit: balance's loss = machine's gain, and the two readings still sum to the total weight.

What does a weighing machine read if I pull up on a rope hanging from the ceiling while standing on it?

Less than your weight — by exactly your pull. FBD of you: $N$ up, rope tension $T$ up, $mg$ down, so $N = mg - T$ . Pull with 200 N and a 600 N you reads 400 N. Pull harder than you weigh and your feet leave the platform ( $N$ cannot go negative): the machine reads zero and you are climbing, which is where this archetype hands over to the monkey on a rope.