Motion of Two Bodies One on Other – Complete Notes, Revision, Important Questions & Downloads
When body A (mass m) rests on body B (mass M), applying force F on A leads to the Four Cases Analysis — the single TOC subtopic. NEET tests: (1) Case II condition F ≤ μ₁mg(M+m)/M for combined motion, (2) separate accelerations a_A=(F−μ₁mg)/m and a_B=μ₁mg/M when sliding, (3) Case IV floor friction = μ₂(M+m)g. CASE I — Both smooth: a_A=F/m, a_B=0. CASE II — Rough A-B, smooth floor: combined if F≤limit, else separate accelerations. CASE III — Smooth A-B, rough floor: a_A=F/m, a_B=0. CASE IV — Both rough: floor friction=μ₂(M+m)g; B moves only if μ₁mg>μ₂(M+m)g. One of the most commonly tested friction scenarios in NEET.
NEET Weightage — Motion of Two Bodies One on Other
Friction (Chapter 5)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 1 | 4 | |
| 2021 | 0 | 0 | |
| 2020 | 1 | 4 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 3–5 | 12–20 |
CASE II — ROUGH A-B, SMOOTH FLOOR (μ₁ between A and B; floor is smooth): Friction between A and B: limiting friction f_l = μ₁mg. Sub-case IIa: F ≤ μ₁mg → A and B move together. Common acceleration: a = F/(M+m). Friction on B from A (forward, accelerating B): f = Ma = MF/(M+m). This f must ≤ μ₁mg. Condition: F ≤ μ₁g(M+m). Sub-case IIb: F > μ₁mg (or more precisely F > μ₁g(M+m)) → A slides on B. Friction on A from B = μ₁mg (backward, kinetic). Friction on B from A = μ₁mg (forward, Newton's 3rd law). a_A = (F − μ₁mg)/m. a_B = μ₁mg/M [since floor is smooth, only force on B = kinetic friction from A]. NCERT: 'The body A will not slide on body B till F < F_l' where F_l is limiting friction.
CASE IV — BOTH SURFACES ROUGH (μ₁ between A and B; μ₂ between B and floor): When A slides on B: Kinetic friction from B on A = μ₁mg (backward). Kinetic friction from A on B = μ₁mg (forward, 3rd law). Normal force of B on floor = (M+m)g (B carries A's weight). Floor friction on B = μ₂(M+m)g (backward). Net force on B = μ₁mg − μ₂(M+m)g. For B to move: μ₁mg > μ₂(M+m)g. If B moves: a_B = [μ₁mg − μ₂(M+m)g]/M. If μ₁mg ≤ μ₂(M+m)g: B stays stationary despite A sliding on it (floor friction sufficient to hold B). For A: a_A = (F − μ₁mg)/m (regardless of whether B moves). Moving together condition: F − μ₂(M+m)g = (M+m)×a AND internal friction is sufficient.
How to Solve Two-Body Stacked Problems for NEET
Step 1 — Identify which case the problem falls into Read the problem carefully: (a) Is the floor smooth or rough? (b) Is the A-B interface smooth or rough? (c) Where is force F applied (on A or B)? These three decisions determine which of the four cases applies. Smooth floor and rough A-B → Case II. Both rough → Case IV. Then: check if the bodies move together or separately (compare F to limiting friction).
Step 2 — Draw SEPARATE FBDs for A and B Do NOT draw a single FBD for the combined system when finding internal forces. FBD of A: forces are F (if applied to A), weight mg (down), normal from B (up = mg), friction from B (horizontal; backward if A moves right). FBD of B: weight Mg (down), normal from A (down = mg), normal from floor (up = (M+m)g = Mg+mg), floor friction (horizontal; backward), friction from A on B (horizontal; forward = Newton's 3rd law pair of friction on A from B). Apply Newton's 2nd law to each body separately: F_net on A = ma_A; F_net on B = Ma_B.
Step 3 — Use common acceleration first, then verify; if sliding, use separate accelerations NEET SHORTCUT: First assume both A and B move with common acceleration a = F_applied_net/(M+m) (or (F−floor_friction)/(M+m)). Then compute the friction needed between A and B to maintain this. If needed friction ≤ limiting friction μ₁mg: confirmed, they move together. If needed friction > limiting friction: A slides on B — rewrite equations with kinetic friction and find a_A and a_B separately.
Study Materials — Two-Body Stacked Friction
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Rapid Revision — Two-Body One on Other (Four Cases)
Concept → Trap → Example1) Four Cases Analysis — Body A on Body B
Four Cases AnalysisSETUP: Body A (mass m) rests on body B (mass M). B is on the floor. Horizontal force F is applied to A. μ₁ = friction coefficient between A and B. μ₂ = friction coefficient between B and floor. NORMAL FORCES: A on B: force = mg (downward, A's weight). B on A: N_AB = mg (upward reaction). B on floor: force = (M+m)g (B carries its own weight M and A's weight m). CASE I — Both surfaces smooth (μ₁=0, μ₂=0): No friction anywhere. For A: only F acts horizontally. a_A = F/m. For B: no horizontal force from A (no friction at A-B interface), no friction from floor. a_B = 0. A slides over B; B stays stationary. CASE II — A-B rough (μ₁), floor smooth (μ₂=0): Limiting friction between A and B: f_l = μ₁ × N_AB = μ₁mg. Sub-case IIa: F ≤ μ₁mg → static friction can hold; A-B interface does not slip. A and B move together with common acceleration: a = F/(M+m). Friction on B from A (forward, driving B): f_AB = Ma = MF/(M+m). Verify: f_AB ≤ μ₁mg → MF/(M+m) ≤ μ₁mg → F ≤ μ₁g(M+m)/1 × (M+m)/M... actually condition is more accurately: combined system condition. More directly: for no sliding, required friction on B from A = Ma. For A: F − f = ma → f = F − ma = F(M/(M+m)). Check: f ≤ μ₁mg → FM/(M+m) ≤ μ₁mg → F ≤ μ₁g(M+m). So the exact condition for NO SLIDING in Case II: F ≤ μ₁g(M+m). Sub-case IIb: F > μ₁g(M+m) → A slides on B. Kinetic friction on A from B: f = μ₁mg (backward). Kinetic friction on B from A: f = μ₁mg (forward). For A: F − μ₁mg = ma_A → a_A = (F − μ₁mg)/m. For B: μ₁mg = Ma_B → a_B = μ₁mg/M. (Floor smooth: no floor friction on B.) CASE III — A-B smooth, floor rough (μ₁=0, μ₂≠0): A slides freely over B (no friction at A-B). For A: a_A = F/m. For B: no horizontal force from A (smooth A-B). Floor friction on B = μ₂(M+m)g (backward). But B has no forward force to overcome this floor friction → B stays at rest. a_B = 0. Same result as Case I, but now floor friction is available (it just has nothing to oppose). CASE IV — Both surfaces rough (μ₁≠0, μ₂≠0): First check if system moves at all. For combined system to start moving: F > μ₂(M+m)g (must overcome floor friction). Assume A slides on B (F is large). From A: a_A = (F − μ₁mg)/m. From B: forces = μ₁mg (forward, reaction from A) − μ₂(M+m)g (backward, floor friction). Net force on B = μ₁mg − μ₂(M+m)g. B moves only if μ₁mg > μ₂(M+m)g. If yes: a_B = [μ₁mg − μ₂(M+m)g]/M. If μ₁mg ≤ μ₂(M+m)g: floor friction holds B; a_B = 0 even though A slides on it. NCERT states: 'When a body A of mass m is resting on a body B of mass M then two conditions are possible: (i) Combined system moves (ii) A slides over B.'
- CONDITION SUMMARY TABLE: Case I (μ₁=0, μ₂=0): a_A=F/m; a_B=0; relative motion f_AB=0. Case II (μ₁≠0, μ₂=0), F ≤ μ₁g(M+m): a_A=a_B=F/(M+m); friction on B from A = MF/(M+m). Case II (μ₁≠0, μ₂=0), F > μ₁g(M+m): a_A=(F−μ₁mg)/m; a_B=μ₁mg/M; A slides forward on B. Case III (μ₁=0, μ₂≠0): a_A=F/m; a_B=0; B held by floor. Case IV (μ₁≠0, μ₂≠0), A slides: a_A=(F−μ₁mg)/m; a_B=[μ₁mg−μ₂(M+m)g]/M if positive, else 0. Key condition: B moves in Case IV only if μ₁m > μ₂(M+m).
- MAXIMUM F FOR NO SLIDING (Case II): Non-sliding condition: F ≤ μ₁g(M+m). Note: this limit is μ₁g(M+m), NOT just μ₁mg. The limiting friction between A and B is μ₁mg (only depends on A's weight and the A-B interface friction), but whether they slide depends on the required friction to give B common acceleration as compared to μ₁mg. From Newton's law on B: f to accelerate B = M×[F/(M+m)]. For no slide: M×F/(M+m) ≤ μ₁mg → F ≤ μ₁mg(M+m)/M. This is slightly different from μ₁g(M+m). Careful derivation: from A: F−f = ma; from B: f=Ma (floor smooth). Combined: a=F/(M+m) and f=MF/(M+m). Condition: f ≤ μ₁mg → F ≤ μ₁mg(M+m)/M. This is the correct NEET-level condition. Note: some textbooks simplify by stating F ≤ μ₁mg (only limiting friction on A) — this is also common in NEET problems. Use the form given in the problem context.
- FRICTION ON B FROM A (DRIVING FORCE FOR B): In Case IIa (no-sliding): friction on B from A = MF/(M+m) [forward]. This is the only force accelerating B (floor smooth). In Case IIb (sliding): friction on B from A = μ₁mg [forward] — now kinetic, constant. In Case IV (sliding, B moves): net force on B = μ₁mg − μ₂(M+m)g. If this net > 0: B accelerates. If it is ≤ 0: a_B = 0 (static friction from floor keeps B still). Important: when A is sliding on B, the normal force on B from A is STILL mg (not reduced) — A's weight acts fully on B regardless of relative sliding. Therefore floor friction = μ₂(M+m)g throughout.
US Curriculum Gaps — Two-Body Stacked Problems
Topics in this section are in NEET but may be framed differently in US physics courses.Stacked Block Friction in AP Physics 1
AP Physics 1 does cover stacked block problems (top block, bottom block) with friction. However, the structured four-case analysis with tabulated conditions is an NCERT-specific pedagogical format. AP students solve these case-by-case using FBD, but may not have practised the generalised condition F_max = μ₁mg(M+m)/M for no-sliding between stacked blocks. NRI students should explicitly practise this formula and the Case IV analysis — where both A-B and B-floor friction coexist and B may or may not move. The key NEET-specific complexity: floor carries weight (M+m)g, making floor friction larger when A is also present.
- AP Physics 1: solves stacked blocks with FBD; no tabulated four-case format
- NEET: the four-case analysis is directly tested; conditions and formulas must be memorised
- Critical detail: floor friction = μ₂(M+m)g (includes A's weight) even when A slides independently
Condition for Combined vs Separate Motion in University Physics
The condition F ≤ μ₁mg(M+m)/M for no-sliding between stacked blocks is derived in NCERT and forms a direct NEET answer. In Halliday & Resnick, this same condition emerges from FBD analysis but is not stated as a standalone named result. NRI students trained in US textbooks may know the physics but slow down on NEET MCQs because they re-derive rather than recall. The no-sliding condition and the four-case pattern should be memorised as ready-to-apply formulas for NEET speed.
- US textbooks: derive condition from FBD each time; no named formula for 'max F for no-sliding'
- NEET: expects immediate formula recall: F_max = μ₁g × m(M+m)/M for no sliding on smooth floor
- Case IV (both rough): B moves only if μ₁m > μ₂(M+m) — this is a testable NEET condition
NEET-Style Practice Questions — Two-Body Stacked
4 QuestionsPractice Problems — Two-Body One on Other
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Physics — Friction Revision Checklist
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Motion of Two Bodies One on Other
Notes · Downloads · Revision · Important QuestionsWhat are the four cases for motion of two stacked bodies?
What is the condition for both bodies to move together (without sliding)?
Why is the floor friction equal to μ₂(M+m)g and not just μ₂Mg?
When does body B move in Case IV (both surfaces rough)?
How do I find the friction force between A and B when they move together?
What changes if the force F is applied to B (lower block) instead of A (upper block)?
Can body A be stationary while B moves?
What is the relative acceleration of A with respect to B in Case IIb?
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