NEET Physics — Friction

Maximum Length of Hung Chain – Complete Notes, Revision, Important Questions & Downloads

Maximum Length of Hung Chain covers the TOC subtopic Chain Hanging Over Edge. A uniform chain of total length l lies on a table. A portion l' hangs over the edge. The chain is on the verge of sliding when friction between chain and table exactly balances the weight of the hanging part. The condition gives μ = l'/(l−l'), where l−l' is the length on the table. Solving for the maximum hanging length: l'_max = μl/(1+μ). NEET tests this as both a formula-application and a conceptual question about friction and distributed mass.

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Chain Hanging Over EdgeFriction Ch.5Distributed Mass Friction

Expected QuestionsQ

0–1

NEET asks: given μ and total length l, find maximum hanging length l'. Or: given l and l', what is μ? Occasionally asks for the fraction (l'/l) that can hang = μ/(1+μ). The uniformity of the chain is key — mass per unit length is constant, so weight of any portion is proportional to its length.

Time Required⏱

15 min

5 min to understand the setup: mass of hanging part ∝ l'; mass on table ∝ (l−l'); limiting friction = μ × normal force from table = μ × mass on table × g. 5 min to derive: μ=(hanged weight)/(friction) → μl'=(l−l') → l'=μl/(1+μ). 5 min to practise 3 numerical examples with different μ and l.

Difficulty⚡

Easy

This is one of the more straightforward friction applications. The key simplification is that for a uniform chain, mass ∝ length. So mass of hanging part = m×(l'/l) where m is total mass, and mass on table = m×(l−l')/l. Normal force = mass-on-table × g = mg(l−l')/l. Friction = μ × normal = μmg(l−l')/l. Weight of hanging part = mg(l'/l). At limit: mg(l'/l) = μmg(l−l')/l → l' = μ(l−l') → l'=μl/(1+μ).

NRI USA Curriculum GapUS

Low

AP Physics 1 does not include the chain-hanging-over-edge problem as a named result. However, students who understand distributed-mass systems can derive it quickly. The key insight (mass proportional to length for uniform chain) is a simple generalisation of point-mass friction that NRI students can grasp.

1Subtopics

4+Practice Questions

4Free Downloads

15 minPrep Time

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NEET Weightage — Maximum Length of Hung Chain

Friction (Chapter 5)

NEET Year	Questions from this Topic	Bar	Marks
2024	1	1 Q	4
2023	0	0 Q	0
2022	1	1 Q	4
2021	0	0 Q	0
2020	0	0 Q	0
2019	0	0 Q	0
6-Year Total (2019–2024)	0–2		0–8

SETUP: Uniform chain of total length l, mass m. Mass per unit length λ = m/l. Portion l' hangs over the table edge; portion (l−l') remains on table. Normal force on table portion: N = λ(l−l')g = mg(l−l')/l. Maximum static friction: f_max = μN = μmg(l−l')/l. Weight of hanging portion pulling it down: W_hang = λl'g = mgl'/l.
LIMITING CONDITION: At the verge of sliding: W_hang = f_max → mgl'/l = μmg(l−l')/l → l' = μ(l−l') → l' = μl − μl' → l'(1+μ) = μl → l' = μl/(1+μ). This is the maximum hanging length. Coefficient formula: μ = l'/(l−l') [ratio of hanging to table-resting lengths].
KEY RESULTS: Maximum hanging length l'_max = μl/(1+μ). Maximum fraction hanging = l'/l = μ/(1+μ). Length on table at limit = l−l' = l/(1+μ). Check: μ=1 → l'=l/2 (exactly half hangs); μ=0 → l'=0 (frictionless, any hanging part slides); μ→∞ → l'→l (entire chain can hang). Quick memory: l'/(l−l')=μ — the ratio of lengths equals μ.

📊

0.3

Avg Questions / Year

🎯

Total Marks (6 yrs)

📈

Direct

Pattern

⚠️

Easy

Difficulty

How to Solve the Chain Hanging Over Edge Problem

Step 1 — Write forces for hanging and table portions Weight pulling chain off the edge: W = mgl'/l. Normal force from table supporting the on-table portion: N = mg(l−l')/l. Maximum static friction: f = μN = μmg(l−l')/l.

Step 2 — Apply limiting equilibrium At the maximum hanging length, friction just balances the hanging weight: mgl'/l = μmg(l−l')/l. Cancel common factors: l' = μ(l−l').

Step 3 — Solve for l' l' = μl − μl' → l'(1+μ) = μl → l'_max = μl/(1+μ). Ratio form: μ = l'/(l−l'). Fraction of chain hanging: l'/l = μ/(1+μ).

Study Materials — Maximum Length of Hung Chain

PDF · Cheat Sheet · MCQ Set · PYQ

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Full Notes

Uniform chain length l, mass m. Hanging length l'. Table length (l−l'). Limiting condition: l'=μ(l−l') → l'_max=μl/(1+μ). μ=l'/(l−l'). Fraction: l'/l=μ/(1+μ). Special: μ=1→half chain can hang.

1 subtopicDistributed massLimiting equilibrium

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Formula Sheet

l'_max=μl/(1+μ); μ=l'/(l−l'); fraction=μ/(1+μ); table length at limit=l/(1+μ).

4 formulas1 pageQuick reference

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MCQ Practice

8 MCQs: find l', find μ, find fractions, compare two chains with different μ.

8 MCQs1 formulaSolved

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PYQ

NEET previous year questions on hanging chain friction with complete solutions.

2+ year-tagged Qs2015–2024Solved

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Subtopics — Maximum Length of Hung Chain

2-Column Table

Column A	Column B
Chain Hanging Over Edge↗

Rapid Revision — Maximum Length of Hung Chain

Concept → Trap → Example

1) Chain Hanging Over Edge — Complete Derivation

Chain Hanging Over Edge

PROBLEM: Uniform chain of total length l, mass m, coefficient of friction μ between chain and table. A length l' hangs over the edge. Find the maximum l' for which the chain remains stationary (does not slide off). MASS DISTRIBUTION: For a uniform chain, mass per unit length = m/l. Mass of hanging portion = m × (l'/l). Mass of table portion = m × (l−l')/l. FORCES: Weight pulling the chain off the table = [m(l'/l)] × g = mgl'/l (due to hanging mass). Normal force from table on the table portion = [m(l−l')/l] × g = mg(l−l')/l (table supports only the portion lying on it — the hanging part is not in contact with the table). Maximum static friction = μ × Normal = μmg(l−l')/l. LIMITING EQUILIBRIUM: Chain is on the verge of sliding when gravity on hanging part exactly equals maximum friction: mgl'/l = μmg(l−l')/l. Cancel mg/l from both sides: l' = μ(l−l'). Expand: l' = μl − μl'. Collect l' terms: l' + μl' = μl → l'(1+μ) = μl → l' = μl/(1+μ). FORMULA: μ=l'/(l−l'); l'_max=μl/(1+μ). FRACTION OF CHAIN THAT CAN HANG: l'/l = μ/(1+μ).

NUMERIC CHECK: μ=1/3, l=120 cm. l'=μl/(1+μ)=(1/3×120)/(1+1/3)=40/(4/3)=40×3/4=30 cm. So 30 cm hangs at limit. Verify: l'/(l−l')=30/90=1/3=μ ✓. Chain on table = 90 cm. Maximum fraction hanging = (1/3)/(4/3) = 1/4 of total length.
COMMON NEET TRAP: Students often write μ=hanging/(total length) instead of μ=hanging/(table length). The correct ratio is l'/(l−l'), not l'/l. If 1/4 of the chain hangs at the limit, then l'/l=1/4 and l'/(l−l')=1/4/(3/4)=1/3=μ. These two ratios (1/4 and 1/3) are often confused. The formula μ=l'/(l−l') is unambiguous — denominator is the table-resting length.
ALTERNATE FORM: Maximum fraction that can hang = μ/(1+μ). Example: μ=0.5 → fraction=0.5/1.5=1/3. μ=1 → fraction=1/2. μ=2 → fraction=2/3. As μ→∞ → fraction→1 (entire chain hangs). For μ=3/7: fraction=3/(7+3)=3/10 (30% can hang). This fraction form is useful when NEET asks for percentages.

Example (NEET-style)A uniform chain of length 1 m is placed on a rough table (μ = 0.25). Maximum length that can hang over the edge: l'_max = μl/(1+μ) = 0.25×1/(1.25) = 0.25/1.25 = 0.2 m = 20 cm. Fraction = 1/5 of total length. Table-resting length at limit = 1−0.2 = 0.8 m. Verify: μ = l'/(l−l') = 0.2/0.8 = 0.25 ✓.

US Curriculum Gaps — Maximum Length of Hung Chain

Topics in this section are in NEET but may be framed differently in US physics courses.

Distributed Mass Systems in AP Physics 1

AP Physics 1 covers friction as a point-force on rigid bodies. Problems involving distributed mass (like a chain where normal force varies with the portion on the table) are not in the standard AP Physics 1 curriculum. NEET extends the friction concept to flexible, distributed-mass systems where the normal force itself depends on the variable l'. NRI students need to understand that for a uniform chain, weight and normal force scale linearly with length.

AP Physics 1: friction on point masses and rigid bodies; hanging chain not tested
NEET: chain problem uses distributed mass — normal force = μ × (table-resting length portion weight)
Key insight: for uniform chain, mass proportional to length — so all force ratios become length ratios

Static Equilibrium with Variable Load in University Physics

University Physics (Halliday & Resnick) covers static equilibrium but typically not the hanging chain with friction as a named problem. The chain problem is a simplified version of a flexible rope-on-surface problem. NRI students in US university physics courses learn to treat distributed loads in statics but need to specifically apply limiting friction conditions to solve this NEET type.

US university: distributed loads in statics; hanging chain with friction not a standard named problem
NEET: memorise μ = l'/(l−l') — ratio of hanging length to table length equals μ
Quick check: μ=1 means equal lengths on table and hanging — easy verification with special-case reasoning

NEET-Style Practice — Maximum Length of Hung Chain

4 Questions

1A uniform chain of length 2 m is on a rough table (μ = 0.4). What is the maximum length that can hang over the edge without the chain sliding?Chain Hanging Over Edge

0.4 m

0.571 m

0.8 m

1.0 m

l'_max = μl/(1+μ) = 0.4×2/(1+0.4) = 0.8/1.4 = 4/7 ≈ 0.571 m. Answer: Option B. Verify: table portion = 2−0.571 = 1.429 m. μ = 0.571/1.429 = 4/10 = 0.4 ✓. Trap: Option A (0.4 m) uses μ = l'/l instead of l'/(l−l'). If l'=0.4m: μ=0.4/(2−0.4)=0.4/1.6=0.25 ≠ 0.4. Option C (0.8 m): μ=0.8/1.2=0.667 ≠ 0.4.

2A uniform chain of length 1 m is placed on a table. 1/4 m hangs off the edge without sliding. What is the minimum coefficient of friction?Finding μ

μ = 1/4

μ = 1/3

μ = 1/2

μ = 3/4

l' = 0.25 m, l = 1 m, l−l' = 0.75 m. μ = l'/(l−l') = 0.25/0.75 = 1/3. Answer: Option B. Note: μ ≠ l'/l = 0.25/1 = 1/4 (Option A, wrong). The formula uses the table-resting length (0.75 m) in the denominator, not the total length. The minimum μ means just-not-sliding: any μ < 1/3 would allow the chain to slide with 1/4 m hanging.

3If the coefficient of friction between a uniform chain and a table is μ, what fraction of the chain can hang over the edge without sliding?Fraction Formula

μ/(1+μ)

1/(1+μ)

μ/(1−μ)

l'/l = μ/(1+μ). Derived from l'_max = μl/(1+μ) → l'_max/l = μ/(1+μ). For μ=1: fraction=1/2 (half hangs). For μ=0: fraction=0 (nothing hangs on frictionless). For μ=2: fraction=2/3. Option C (1/(1+μ)) is the fraction ON the table, not hanging. Option D (μ/(1−μ)) diverges at μ=1 — clearly wrong.

4A uniform chain of length l and mass M lies on a rough table (μ = 1/3). Another uniform chain of same length l but mass 2M lies on the same table. For each chain, what is the ratio of maximum hanging lengths (chain-1/chain-2)?Mass Independence

1:2 (proportional to mass)

1:1 (independent of mass)

2:1

Cannot be determined without knowing g

l'_max = μl/(1+μ) depends only on l and μ, NOT on the chain's mass or linear density. Both chains have the same l and same μ=1/3, so l'_max = (1/3)l/(1+1/3) = (1/3)l/(4/3) = l/4 for BOTH chains. Ratio = 1:1. This is because the g and mass cancel in the limiting condition: mgl'/l = μmg(l−l')/l → l' = μ(l−l'). The mass m cancels completely. Answer: Option B. This is an important conceptual point: the maximum hanging fraction depends only on μ, not on how heavy the chain is.

Practice Problems — Maximum Length of Hung Chain

Click "Reveal Answer" after attempting

1A uniform chain of length 60 cm has μ = 0.5 with the table. Find: (a) maximum hanging length, (b) fraction of chain on table at limiting condition.

(a) 20 cm; (b) 2/3 of total chain on table

(a) 30 cm; (b) 1/2 on table

(a) 20 cm; (b) 1/3 on table

(a) 12 cm; (b) 3/4 on table

👁 Reveal Answer

l'_max=μl/(1+μ)=0.5×60/1.5=30/1.5=20 cm. Table length=60−20=40 cm. Fraction on table=40/60=2/3. Answer: Option A. Verify: μ=l'/(l−l')=20/40=0.5 ✓. Fraction hanging=20/60=1/3=μ/(1+μ)=0.5/1.5=1/3 ✓.

2For a uniform chain on a table, if μ is doubled, the maximum fraction of chain that can hang over the edge:

Also doubles

More than doubles

Less than doubles

Stays the same

👁 Reveal Answer

Fraction = μ/(1+μ). Let's check with μ and 2μ. If μ=0.5: fraction=0.5/1.5=1/3. If 2μ=1: fraction=1/2. 1/2 > 2×(1/3)=2/3? No: 1/2 < 2/3. So the fraction less than doubles. Answer: Option C. The function f(μ)=μ/(1+μ) is concave in μ (f''<0), so doubling μ gives less than double the fraction. This is a conceptual NEET trap.

3A chain (total length l) hangs with 1/3 of its length over the table edge in limiting equilibrium. Find μ and the ratio of friction force to weight of hanging part at the limiting position.

μ=0.5; ratio=1 (friction equals hanging weight at limit)

μ=1/2 but ratio is 2 (friction is double the weight)

μ=1/3; ratio=1

μ=2/3; ratio=1

👁 Reveal Answer

l'=l/3 (1/3 hangs), l−l'=2l/3 (2/3 on table). μ=l'/(l−l')=(l/3)/(2l/3)=1/2. Friction force at limit=μN=μ×mg(l−l')/l=(1/2)×mg(2/3)=mg/3. Weight of hanging part=mg×(1/3)=mg/3. Ratio=friction/hanging weight=1. This is always 1 at the limiting condition — friction exactly equals hanging weight (that's the definition of the limit). Answer: Option A.

4Two chains A and B have the same length l but different μ: μ_A=0.2, μ_B=0.3. Each is placed on its own table. What are their maximum hanging fractions? Which chain allows more to hang?

A: 1/6, B: 3/13; B allows more

A: 1/6, B: 3/13 ≈ 0.231; B allows more

A: 0.167, B: 0.231; B allows more (same numerical answer)

All of the above are consistent

👁 Reveal Answer

For A: f_A=μ_A/(1+μ_A)=0.2/1.2=1/6≈0.167. For B: f_B=0.3/1.3=3/13≈0.231. B allows more chain to hang. Answer: all options agree numerically. The key NEET fact: higher μ → larger fraction can hang. μ_B > μ_A → f_B > f_A. Also, l'_A=0.167l and l'_B=0.231l — the absolute hanging lengths (in cm) if l=100 cm would be 16.7 cm and 23.1 cm respectively.

Physics — Friction Revision Checklist

Check off chapters as you revise

Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

Chain Hanging Over Edge: Write the limiting condition from memory: l' = μ(l−l'), give the final formula l'_max = μl/(1+μ), and state what each variable represents.

Chain Hanging Over Edge: Derive from forces: hanging weight = mgl'/l; normal force = mg(l−l')/l; friction = μN; equate for limiting case. Cancel mg/l to get l' = μ(l−l').

Chain Hanging Over Edge: Practise: l=2m, μ=0.5 → l'=1m (half hangs when μ=1). l=2m, μ=0.25 → l'=0.4m. l=3m, μ=1/3 → l'=3/(4)=0.75m.

Chain Hanging Over Edge: Memorise the fraction form: l'/l = μ/(1+μ). And table-fraction = 1/(1+μ). Their sum = μ/(1+μ)+1/(1+μ) = 1 ✓.

Chain Hanging Over Edge: Remember: the formula is INDEPENDENT of chain mass. Two chains of same length and same μ but different masses have the same maximum hanging length. Mass cancels in the limiting condition.

FAQ — Maximum Length of Hung Chain

Notes · Downloads · Revision · Important Questions

What is the formula for maximum hanging length of a chain on a rough table?

l'_max = μl/(1+μ), where l is the total chain length and μ is the coefficient of static friction between chain and table. Equivalently: μ = l'/(l−l'), where l' is the hanging length and (l−l') is the table-resting length. The fraction of chain that can hang without sliding = μ/(1+μ).

Why does the mass of the chain not appear in the formula?

The limiting condition is: weight of hanging part = maximum friction. Weight of hanging part = m(l'/l)g. Maximum friction = μ × N = μ × m(l−l')/l × g. Setting them equal: m(l'/l)g = μm(l−l')/l × g. The mass m and g cancel, leaving only lengths: l' = μ(l−l'). So the maximum hanging fraction depends only on μ and the length ratio, not on how heavy the chain is.

What does μ = l'/(l−l') represent intuitively?

μ equals the ratio of the hanging chain length to the table chain length. Intuitively: the hanging part pulls the chain off, and the table part resists via friction. The heavier the hanging part (longer l'), the larger the pull. The heavier the table part (longer l−l'), the greater the friction. For equilibrium, their ratio equals μ. If μ=1, equal lengths must be on each side; if μ=2, the hanging part can be twice the table length.

What happens if the chain is not uniform?

If the chain is non-uniform (mass per unit length varies along the length), the problem becomes more complex. The uniform-chain formula l'=μ(l−l') assumes that mass is proportional to length. For a non-uniform chain, you'd need to integrate the mass distribution to find the weight of the hanging portion and the normal force of the table portion separately. NEET always specifies 'uniform chain' to allow the simple length-ratio approach.

How do I find μ if I know how much of the chain hangs?

Use μ = l'/(l−l'), where l' = hanging length and l−l' = table-resting length. Example: 30 cm hangs, 90 cm on table → μ = 30/90 = 1/3. Note: do NOT use l'/l (hanging fraction); the denominator must be the TABLE length, not the total length.

Can more than half the chain hang if μ < 1?

Only if μ > 1. From l'/l = μ/(1+μ): for l'/l > 1/2, we need μ/(1+μ) > 1/2 → 2μ > 1+μ → μ > 1. So if μ < 1, less than half the chain can hang (l'/l < 1/2). If μ = 1, exactly half the chain can hang. If μ > 1 (rough surface), more than half can hang.

What if the chain hangs vertically on both sides of a pulley with friction?

That is a different problem (Atwood machine with rope-on-drum friction). The chain-on-table problem has friction only where the chain touches the table surface (horizontal) and gravity acts vertically to pull the hanging portion off. The two setups use friction differently and must not be confused.

A chain of length l is placed entirely on a table. If you slowly pull it until it slides, it slides when l' = μl/(1+μ). Is this the same as the 'maximum length before sliding' problem?

Yes — both are the same problem. Whether the chain is placed with l' already hanging, or you slowly pull the chain until it begins to slide, the critical condition is identical: l'_critical = μl/(1+μ). At this point, the net force on the chain transitions from zero (friction holding) to positive (chain slides). The 'maximum length before sliding' and 'length at which sliding begins' are the same quantity.

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Maximum Length of Hung Chain

Maximum Length of Hung Chain – Complete Notes, Revision, Important Questions & Downloads

1) Chain Hanging Over Edge — Complete Derivation

Distributed Mass Systems in AP Physics 1

Static Equilibrium with Variable Load in University Physics

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Chain Hanging Over Edge

Subtopics

Chain Hanging Over Edge

NEET > Physics > Laws of Motion Chapters

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Maximum Length of Hung Chain

Overview content

Maximum Length of Hung Chain – Complete Notes, Revision, Important Questions & Downloads

1) Chain Hanging Over Edge — Complete Derivation

Distributed Mass Systems in AP Physics 1

Static Equilibrium with Variable Load in University Physics

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Chain Hanging Over Edge

Subtopics

Chain Hanging Over Edge

NEET > Physics > Laws of Motion Chapters

Comments

Leave a comment

Comments (0)