Motion of Massive String – Complete Notes, Revision, Important Questions & Downloads
Motion of Massive String addresses the non-uniform tension distribution in a string that has mass. Unlike a massless string (where tension is the same throughout), a massive string has tension that varies continuously from point to point. NEET tests: (1) the tension at a point that is at distance x from the free end of a massive string of mass M and length L, pulled by force F — T_x = F(L−x)/L; (2) the acceleration of the system (same as massless string: a = F/(M+m) where m is any attached block); (3) conceptual understanding that tension is maximum at the pulling end and zero at the free end. This topic clarifies why real ropes experience different internal stresses at different points — important for structural physics applications.
NEET Weightage — Motion of Massive String
Newton's Laws of Motion (Chapter 4)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 0 | 0 | |
| 2023 | 0 | 0 | |
| 2022 | 0 | 0 | |
| 2021 | 1 | 4 | |
| 2020 | 0 | 0 | |
| 2019 | 0 | 0 | |
| 6-Year Total (2019–2024) | 0–1 | 0–4 |
Boundary conditions: at x = 0 (free end): T = F(L−0)/L = F. Wait — if x=0 is the FREE end (no force applied there), tension should be 0 there. Let me re-clarify: if x is measured from the FREE end and F is applied at the OTHER end: at x = 0 (free end), T = F×(L−0)/L = F — this is WRONG. The convention in the textbook: x is measured from the END where F is applied. At x = 0 (pulled end): T = F. At x = L (free end): T = F(L−L)/L = 0. The tension is MAXIMUM at the pulled end and MINIMUM (zero) at the free end.
Block attached: if a block of mass m is attached to the free end and the whole system (block + string) is pulled by F: a = F/(M+m). Tension at distance x from the PULLED end: the part beyond x has mass m + M(L−x)/L. T_x = [m + M(L−x)/L] × a = [m + M(L−x)/L] × F/(M+m).
How to Prepare Motion of Massive String for NEET
Derive the tension formula using the 'beyond the cut' principle To find tension at a cross-section P of a massive string: cut the string at P. Consider all masses BEYOND P (on the side away from the applied force F). These masses are accelerated by only the tension T_P. Newton's second law: T_P = (total mass beyond P) × a. For a uniform string (mass M, length L) pulled horizontally by F on one end (taking x from the pulled end): mass beyond x = (L−x)/L × M (fraction of string from x to free end). Acceleration a = F/M. T_x = [(L−x)/L × M] × (F/M) = F(L−x)/L.
Identify x correctly — always measure from the pulled end The formula T_x = F(L−x)/L uses x measured from the END where F is applied (the pulled end). At x = 0: T = F (maximum, equals applied force — makes sense, string just below the pull point carries full F). At x = L: T = 0 (free end, nothing to pull there). Midpoint (x = L/2): T = F/2. For a block attached at the free end with force F on the block: x measured from F, tension in string at position x = F(L−x)/L (same regardless of how mass is distributed at the ends, for smooth surface, massless block analogy).
Know the vertical massive rope case for quick comparison Vertical rope of mass M, length L, hanging from a fixed point. A block of mass m hangs at the bottom. System at rest (acceleration = 0). Tension T at distance x from TOP: T_x = (m + M(L−x)/L) × g. At the top (x = 0): T = (m + M)g (maximum — supports everything). At the bottom (x = L): T = mg (supports only the block). Gradient: dT/dx = −Mg/L (tension decreases going down by Mg/L per unit length). No acceleration — the 'beyond the cut' principle still applies.
Study Materials — Motion of Massive String
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Rapid Revision — Motion of Massive String
Concept → Trap → Example1) Horizontal Massive String — Tension at a Point
CoreA uniform string of mass M and length L lies on a smooth horizontal surface. Force F applied at one end (the 'pulled end'). Acceleration of string: a = F/M. To find tension T at a point P at distance x from the pulled end: cut the string at P. Segment from P to free end has length (L−x) and mass m_P = M(L−x)/L. This segment is accelerated entirely by tension T_P (the only external force on it, since surface is smooth). T_P = m_P × a = [M(L−x)/L] × (F/M) = F(L−x)/L. Note: T is a linear function of x, decreasing from F at x=0 to 0 at x=L.
- At the midpoint (x = L/2): T_mid = F(L − L/2)/L = F/2. The string tension at the midpoint is exactly half the applied force. This is the most commonly asked NEET point for massive string tension. For a block of mass m attached at the free end: T_mid = [m + M/2]×F/(M+m). At m = M: T_mid = [M + M/2]×F/2M = (3M/2)×F/2M = 3F/4.
- The 'beyond the cut' principle is universal: the tension at any cross-section equals the net force needed to accelerate all mass BEYOND that cross-section. 'Beyond' = on the side away from the applied force. For horizontal string: T = (mass beyond × a). For vertical string: T = (mass beyond × g) when static, or T = (mass beyond × g_eff) when accelerating. This principle generalises to ALL multi-body dynamics problems with strings.
- Tension distribution shape: since T_x = F(L−x)/L, tension varies LINEARLY from F (at pulled end) to 0 (at free end). The tension profile is a straight line. This is different from a real rope with sag (catenary) where the profile curves. For horizontal, no-sag case (string on surface or taut horizontal), the linear profile holds exactly.
2) Block Attached at Free End — Modified Tension Formula
High PriorityBlock of mass m attached to the free end of a massive string (mass M, length L). Force F applied at the pulled end. System acceleration: a = F/(M+m). Tension at distance x from the pulled end: segment beyond x has mass = m (the block) + M(L−x)/L (string beyond x). T_x = [m + M(L−x)/L] × F/(M+m). At x = L (where string meets block): T_L = m × F/(M+m) = (force needed to accelerate block alone). This is the tension at the string's free end — it equals the tension in a massless string connecting the same block and pulled by same system force.
- At x = 0 (pulled end): T₀ = [m + M] × F/(M+m) = F. Tension at the pulled end equals the applied force (entire system has to be pulled). At x = L: T_L = m × F/(M+m) (only block beyond this cut). This is the value you would compute for a massless string system. The tension at the junction (string-end / block-beginning) is the same as in the massless string approximation — the massive string affects internal tension but NOT the tension at the junction with the block.
- Limiting cases: as M → 0 (massless string): T_x = m × F/(M+m) → m × F/m = F for all x? No — with M → 0: T_x = F(L−x)/L × ... wait. With block: T_x = [m + 0×(L−x)/L] × F/(0+m) = m × F/m = F. So all points have T = F (massless string transmits tension uniformly — standard result ✓). As m → 0 (no block): T_x = [0 + M(L−x)/L] × F/M = F(L−x)/L (back to the pure string formula ✓).
- The T_x formula with block gives a linear profile from F (at pulled end) to mF/(M+m) (at free end/block junction). The slope of T vs. x is: dT/dx = −M×F/[L(M+m)] = −Ma/L. This slope magnitude = (mass per unit length) × (acceleration) — the rate at which tension decreases per unit length equals linear mass density × acceleration. Physically: each small element dm of string 'absorbs' dm×a of force, reducing the tension by that amount.
3) Vertical Hanging Rope and Practical Applications
ApplicationVertical uniform rope (mass M, length L) hanging from a fixed point. Block m₀ hanging from the bottom. System at rest (acceleration = 0). Tension T at distance x from the top: the segment from x to the bottom has mass = M(L−x)/L + m₀. Weight of this segment = [M(L−x)/L + m₀]g. Since a = 0, T_x = weight of everything below = [M(L−x)/L + m₀]g. At the top (x=0): T = (M+m₀)g (supports everything). At the bottom of the rope (x=L, above the block): T = m₀g (supports only the block). In the rope itself: tension increases linearly going from bottom to top.
- If the system accelerates upward at a (rope is pulled up with block at bottom): T_x = [M(L−x)/L + m₀](g+a). Effective gravity g_eff = g+a. Tension at the top = (M+m₀)(g+a). If the system is in free fall (a downward = g): effective g = g−g = 0. All tensions vanish. The rope becomes 'weightless' — any element of rope would float in place if cut. This is the astronaut in orbit analogy.
- Stress in a rope: tension at a cross-section divided by the cross-sectional area gives stress. For a uniform rope: tension varies linearly, so stress varies linearly. The top of a hanging rope experiences maximum stress. For long ropes (cables holding bridges, elevator cables, climbing ropes): the mass of the rope itself contributes significantly to the tension at the top — the rope mass cannot be ignored. NEET makes this point conceptually.
- String or rope breaking: if the rope has a breaking tension T_max, it will break at the point where T_x = T_max. For a vertical rope being pulled up: T is maximum at the top. The rope breaks at the top if the applied force is too large. For a horizontal rope being pulled horizontally (sliding system): T is maximum at the pulled end. NEET may ask: 'At what point does the rope break if T_max is given?'
US Curriculum Gaps — Motion of Massive String
Topics in this section are tested in NEET but organised differently in standard US physics courses.Varying Tension in Massive String (AP Physics 1 Gap)
AP Physics 1 exclusively uses massless string (uniform tension throughout). The concept of a string with mass, and the resulting variation in tension along the string's length, is NOT covered in AP Physics 1. NEET introduces this as a natural extension: apply Newton's second law to a differential element (or a sub-segment) of the string to derive T_x = F(L−x)/L. AP Physics C (Mechanics) introduces continuous mass systems via calculus, but AP Physics 1 students have no exposure to this concept. NEET tests both the formula and the linear tension profile.
- NEET: tension at position x from pulled end = F(L−x)/L for horizontal massive string
- NEET: tension profile is linear, maximum at pulled end, zero at free end
- AP Physics 1: all strings assumed massless; uniform tension; this scenario not standard
Tension Profile in Vertical Hanging Rope (AP Physics 1 Gap)
NEET also tests the tension distribution in a vertical rope hanging with a block at the bottom (static case). The formula T_x = [m + M(L−x)/L]g requires understanding that tension at any point supports the weight of everything below that point. AP Physics 1 covers tension in massless ropes and the weight of a hanging block but does not explicitly test the tension distribution within a massive rope at an arbitrary point. This is taught in AP Physics C and introductory university mechanics.
- NEET: tension at the midpoint of a vertical hanging rope carrying a block at the bottom
- NEET: tension increases linearly from bottom to top in a vertical hanging rope
- AP Physics 1: only tension at the ends of a rope (block weight and ceiling reaction) is typically tested
NEET-Style Practice Questions — Motion of Massive String
4 QuestionsPractice Problems — Motion of Massive String
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Physics — Newton's Laws of Motion Revision Checklist
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Motion of Massive String
Notes · Downloads · Revision · Important QuestionsWhy is tension in a massive string NOT uniform, unlike a massless string?
What does 'x measured from the pulled end' mean practically?
How does the massive string formula change if friction is present?
What is the practical significance of tension varying in a massive string?
In the vertical rope with a block at the bottom, the tension is minimum at the bottom. Why does the rope not break at the bottom?
What is the tension at the junction between the string and the attached block?
Can the formula T_x = F(L−x)/L be derived using calculus?
Is the 'motion of massive string' topic in NEET syllabus, or only in JEE?
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