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Minimum Mass Hung From String

NEET > Physics > Laws of Motion > Friction > Minimum Mass Hung From String

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NEET Physics — Friction

Minimum Mass Hung From String – Complete Notes, Revision, Important Questions & Downloads

Minimum Mass Hung From String covers the TOC subtopic Horizontal and Inclined Cases. In the horizontal case, mass m₁ lies on a rough horizontal surface connected via a string over a frictionless pulley to a hanging mass m₂. The minimum m₂ to just start m₁ moving is m₂ = μm₁. In the inclined case where m₁ is on a rough incline, the minimum hanging mass becomes m₂ = m₁[sinθ + μcosθ]. NEET tests both configurations as direct numericals — given m₁, μ, and surface geometry, find m₂_min; or given m₂ and m₁, find μ.

⬇ Download Notes PDFView Important Questions →
Horizontal and Inclined CasesFriction Ch.5Atwood-Friction System
Expected QuestionsQ
0–1
NEET tests this typically as: (1) find minimum hanging mass to just start motion on horizontal surface given m₁ and μ, (2) find coeeficient of friction from hanging mass condition, (3) inclined surface variant finding m₂ = m₁[sinθ + μcosθ]. The reverse — find μ from m₁, m₂, and θ — is equally common: μ = [m₂/m₁cosθ − tanθ]. Both configurations are tested.
Time Required⏱
25 min
10 min to derive both formulas from first principles using limiting equilibrium (T = F_l). 15 min to practise 4–5 numerical problems: both horizontal and inclined configurations, finding m₂, finding μ, and the reverse problem of finding the angle at which a block just starts to slide. This is a direct-formula topic once the limiting equilibrium setup is understood.
Difficulty⚡
Easy
The derivation uses limiting equilibrium: at the threshold of motion, the tension in the string (= m₂g) equals the limiting static friction (= μm₁g for horizontal; = m₁g[sinθ + μcosθ] for inclined). The main trap is the inclined surface: normal force is m₁gcosθ, not m₁g, so limiting friction = μm₁gcosθ. Also, on the incline, gravity component m₁gsinθ assists the hanging mass in pulling m₁ — this REDUCES the required m₂ if m₁ is on upward slope; the formula m₂ = m₁[sinθ + μcosθ] applies when pulling m₁ UP the incline against both gravity and friction.
NRI USA Curriculum GapUS
Low
AP Physics 1 covers Atwood machine variants with friction. The condition T = f_s_max at limiting equilibrium is standard. The compact NCERT result m₂ = μm₁ (horizontal) is effectively the same as T = μN derived in US physics. The inclined case m₂ = m₁[sinθ + μcosθ] is also derivable from FBD. NRI students may not have the formula memorised but can derive it in under 2 minutes.
1Subtopics
4+Practice Questions
4Free Downloads
25 minPrep Time
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NEET Weightage — Minimum Mass Hung From String

Friction (Chapter 5)
NEET YearQuestions from this TopicBarMarks
20240
 
0 Q
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20231
 
1 Q
4
20220
 
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0
20211
 
1 Q
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20200
 
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20190
 
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6-Year Total (2019–2024)1–2 4–8
HORIZONTAL CASE: m₁ on rough horizontal surface, m₂ hanging via string and pulley. At limiting condition: Tension T = m₂g (for m₂). Limiting friction on m₁: F_l = μm₁g (normal = m₁g on horizontal). At threshold: T = F_l → m₂g = μm₁g → m₂ = μm₁. This is the minimum m₂ to just initiate motion of m₁. Note: coefficient of friction μ = m₂/m₁ (can be found experimentally this way).
INCLINED CASE: m₁ on rough inclined plane (angle θ), m₂ hanging. At limiting condition for m₁ to just start moving UP the incline. T must overcome gravity component (m₁gsinθ down the incline) AND limiting friction (μm₁gcosθ down the incline). From m₂: T = m₂g. From m₁ on incline: T = m₁gsinθ + μm₁gcosθ = m₁g[sinθ + μcosθ]. So minimum m₂ = m₁[sinθ + μcosθ].

REVERSE FORMULA (finding μ): From horizontal case: μ = m₂/m₁. From inclined case: m₂ = m₁[sinθ + μcosθ] → μ = [m₂/m₁ − sinθ]/cosθ = [m₂/(m₁cosθ)] − tanθ. Both are direct NEET calculation formulas. Check: when θ = 0, inclined formula reduces to m₂ = μm₁ (horizontal case). At θ = angle of repose (μ = tanθ), sinθ + μcosθ = 2sinθ → m₂ = 2m₁sinθ for just-motion on rough incline at repose angle.
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0.3
Avg Questions / Year
🎯
8
Total Marks (6 yrs)
📈
Direct
Pattern
⚠️
Easy
Difficulty

How to Solve Minimum Mass Hung Problems for NEET

1

Step 1 — Identify surface geometry and direction of impending motion Is m₁ on horizontal or inclined surface? If inclined, is m₁ being pulled UP or DOWN the incline by m₂? For standard NEET problems, m₂ tries to pull m₁ UP the incline. This means both gravity component (m₁gsinθ, down slope) AND friction (μm₁gcosθ, opposing upward motion so also down slope) both oppose the tension.

2

Step 2 — Apply limiting equilibrium (T = F_l at threshold) At minimum m₂, the system is exactly at the threshold of motion (static limiting condition). Set T = m₂g (from m₂'s equation, since m₂ just barely not accelerating). Set T = F_l (net resistant force on m₁). For horizontal: F_l = μm₁g. For incline up: F_l = m₁gsinθ + μm₁gcosθ. Equate: m₂g = F_l and solve for m₂.

3

Step 3 — Check angle-of-repose condition as a sanity check For θ = 0 (horizontal), inclined formula gives m₂ = m₁[0 + μ×1] = μm₁ ✓. As θ → 90°, sinθ → 1 and cosθ → 0, giving m₂ → m₁ (gravity component alone must be overcome, friction negligible on vertical surface). At θ = angle of repose (tanθ = μ): m₂ = m₁[sinθ + tanθ×cosθ] = m₁×2sinθ.

Study Materials — Minimum Mass Hung From String

PDF · Cheat Sheet · MCQ Set · PYQ
📘
Full Notes
Horizontal case: m₂_min = μm₁; reverse: μ = m₂/m₁. Inclined case (m₁ up incline): m₂_min = m₁[sinθ + μcosθ]; reverse: μ = m₂/(m₁cosθ) − tanθ. Derivation: limiting equilibrium T = F_l. Normal force = m₁gcosθ on incline (not m₁g). Friction = μN = μm₁gcosθ on incline.
1 subtopic2 configurationsFormulas + Derivations
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📗
Formula Sheet
Horizontal: m₂ = μm₁; μ = m₂/m₁. Inclined (m₁ up slope): m₂ = m₁[sinθ + μcosθ]; μ = m₂/(m₁cosθ) − tanθ. Check: θ=0 reduces inclined to horizontal. At angle of repose: m₂ = 2m₁sinθ.
4 formulas1 pageQuick reference
Download Sheet
📙
MCQ Practice
10 MCQs: find m₂ given m₁ and μ (horizontal); find μ given m₁ and m₂; inclined version; finding angle; system acceleration when m₂ exceeds minimum.
10 MCQs2 configurationsSolved
Download MCQs
📒
PYQ
Previous year NEET questions on limiting mass conditions with string-and-pulley friction systems, including inclined variants.
3+ year-tagged Qs2015–2024Solved
Download PYQs

Subtopics — Minimum Mass Hung From String

2-Column Table
Column AColumn B
Horizontal and Inclined Cases↗

Rapid Revision — Minimum Mass Hung From String

Concept → Trap → Example

1) Horizontal and Inclined Cases — Minimum Hanging Mass Formula

Horizontal and Inclined Cases

SETUP: mass m₁ on a surface, connected via massless string over a frictionless pulley to mass m₂ hanging vertically. HORIZONTAL CASE: Normal force on m₁ = m₁g. Limiting friction F_l = μ_s × m₁g = μm₁g. String tension T = m₂g (from m₂'s equilibrium when just at threshold). At limiting condition: T = F_l → m₂g = μm₁g → m₂ = μm₁. This is the minimum mass to just start motion. Coefficient can be found: μ = m₂/m₁. INCLINED CASE (m₁ on incline angle θ, m₂ pulls m₁ UP the slope): Normal force N = m₁gcosθ. Friction force (opposing upward motion, directed DOWN slope): F_l = μN = μm₁gcosθ. Gravity component along incline (opposing upward motion, directed DOWN slope): m₁gsinθ. For m₁ on incline: T = m₁gsinθ + F_l = m₁gsinθ + μm₁gcosθ = m₁g[sinθ + μcosθ]. From m₂: T = m₂g. At threshold: m₂g = m₁g[sinθ + μcosθ] → m₂ = m₁[sinθ + μcosθ]. Reverse formula: μ = [m₂/m₁ − sinθ]/cosθ = m₂/(m₁cosθ) − tanθ. CHECK: θ = 0 → m₂ = m₁[0 + μ×1] = μm₁ ✓ (reduces to horizontal case).

  • INCLINED CASE — m₂ pulling m₁ DOWN slope (m₁ on incline, m₂ hangs, m₁ about to slide DOWN): Here gravity component aids downward motion. Friction now acts UP the slope (opposing downward motion). For m₁ about to slide down: m₁gsinθ − T − μm₁gcosθ = 0 → T = m₁g[sinθ − μcosθ]. Since T = m₂g: m₂ = m₁[sinθ − μcosθ]. NOTE: For m₁ to actually slide down, we need sinθ > μcosθ (tanθ > μ), i.e., angle exceeds angle of repose. If tanθ < μ, the body won't slide down under its own weight even without m₂.
  • NORMAL FORCE ON INCLINE: The most common error is using N = m₁g instead of N = m₁gcosθ. On incline, N = mgcosθ because the weight component perpendicular to incline surface is mgcosθ (the surface cancels only the perpendicular component of gravity). Using N = mg gives friction = μmg (wrong) and leads to m₂ = m₁[sinθ + μ], which is incorrect.
  • STRING AND PULLEY ASSUMPTIONS: The derivation assumes (a) string is massless and inextensible, (b) pulley is frictionless. If pulley has friction, the tension on the two sides of the string differs. If rope has mass, the analysis must include rope's weight. NEET problems typically state 'frictionless pulley' and 'massless string' — confirm these before applying m₂ = m₁[sinθ + μcosθ] directly.
Example (NEET-style)A 5 kg block (m₁) rests on a horizontal surface with μ = 0.3. What minimum mass m₂ must hang to just start m₁ moving? m₂ = μm₁ = 0.3 × 5 = 1.5 kg. If the same block is on a 30° incline (sin30°=0.5, cos30°=0.866): m₂ = m₁[sinθ + μcosθ] = 5×[0.5 + 0.3×0.866] = 5×[0.5+0.2598] = 5×0.7598 = 3.8 kg. Reverse check: μ = m₂/(m₁cosθ) − tanθ = 3.8/(5×0.866) − 0.577 = 0.877 − 0.577 = 0.3 ✓.

US Curriculum Gaps — Minimum Mass Hung From String

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

Limiting Friction Threshold in AP Physics 1

AP Physics 1 covers the Atwood machine and inclined plane with friction as separate topics. The limiting equilibrium condition T = μN at the threshold of motion is covered, but the combined result m₂ = m₁[sinθ + μcosθ] is typically derived per-problem rather than memorised. For NEET speed, commit both the horizontal formula (m₂ = μm₁) and inclined formula (m₂ = m₁[sinθ + μcosθ]) to memory.

  • AP Physics 1: derives limiting condition from FBD each time; no named 'minimum mass' result
  • NEET: direct formula recall of m₂ = μm₁ (horizontal) and m₂ = m₁[sinθ + μcosθ] (inclined) expected
  • Reverse formula μ = m₂/(m₁cosθ) − tanθ is equally testable in NEET — AP students may not have this memorised

Pulley-and-Incline Friction Systems in University Physics

Halliday & Resnick covers pulley-string-incline systems in Chapter 6 (Force and Motion II). However, the specific scenario of finding the minimum hanging mass to just initiate motion is not presented as a named formula. NRI students should focus on the limiting equilibrium setup: at the instant of 'just about to move,' static friction reaches its maximum value μsN, and the system is in equilibrium (net force = 0 on each mass).

  • US textbooks: solve pulley-incline problems using component equations, not named minimum-mass formulas
  • NEET: requires direct formula application; m₂ = m₁[sinθ + μcosθ] is a commonly tested result
  • Coefficient derivation μ = m₂/(m₁cosθ) − tanθ is a common NEET MCQ sub-type not emphasized in US courses

NEET-Style Practice Questions — Minimum Mass Hung From String

4 Questions
1A block of mass 8 kg lies on a rough horizontal surface. It is connected by a string passing over a frictionless pulley to a hanging mass. The coefficient of static friction is 0.4. What is the minimum mass of the hanging body to just move the block? (g = 10 m/s²)Horizontal Case
3.2 kg
8 kg
0.4 kg
4 kg
Using the horizontal case formula: m₂_min = μm₁ = 0.4 × 8 = 3.2 kg. At this mass, the string tension T = m₂g = 3.2×10 = 32 N exactly equals the limiting friction F_l = μm₁g = 0.4×8×10 = 32 N. Option A (3.2 kg) is correct. Option B (8 kg) confuses m₂ with m₁. Option C (0.4 kg) uses μ as mass (wrong). Option D (4 kg) would correspond to μ = 0.5 (wrong μ). The coefficient of friction is found as μ = m₂/m₁ = 3.2/8 = 0.4 ✓.
2A 10 kg block is on a 30° rough inclined plane (μ = 0.2). A hanging mass m₂ is connected via string and pulley to pull m₁ up the incline. Find the minimum m₂ to start motion. (sin30°=0.5, cos30°=0.866, g=10 m/s²)Inclined Case
6.73 kg
5 kg
1.73 kg
8.66 kg
Using the inclined formula: m₂_min = m₁[sinθ + μcosθ] = 10×[0.5 + 0.2×0.866] = 10×[0.5 + 0.1732] = 10×0.6732 = 6.732 ≈ 6.73 kg. Option A is correct. Option B (5 kg) = m₁sinθ only (ignores friction). Option C (1.73 kg) = μm₁cosθ only (ignores gravity component). Option D (8.66 kg) = m₁cosθ (wrong formula). Verify: T = m₂g = 6.73×10 = 67.3 N. Force needed to pull m₁ up: m₁gsinθ + μm₁gcosθ = 50 + 17.32 = 67.32 N ≈ 67.3 N ✓.
3In a horizontal surface experiment, m₁ = 6 kg and the minimum hanging mass to start motion is found to be 1.8 kg. What is the coefficient of static friction?Finding μ — Horizontal
0.3
0.18
3.33
0.6
From horizontal case: μ = m₂/m₁ = 1.8/6 = 0.3. Option A (0.3) is correct. This is a common method for experimentally measuring coefficient of static friction — suspend mass m₂ from a pulley connected to m₁ on horizontal surface, increase m₂ until m₁ just starts to slide, then μ = m₂/m₁. Option B (0.18) = m₂/(m₁ × 10) (wrong normalization). Option C (3.33) = m₁/m₂ (inverted). Option D (0.6) is double the correct value.
4A 4 kg block on a 45° rough incline (μ = 0.3) is connected to hanging mass m₂ by a string. The coefficient of kinetic friction between the hanging mass and something else is irrelevant. Find m₂ for the block to just start moving up. (sin45° = cos45° = 0.707, g = 10 m/s²)Inclined Case — Finding μ
4.07 kg
2.83 kg
1.24 kg
5.66 kg
m₂_min = m₁[sinθ + μcosθ] = 4×[0.707 + 0.3×0.707] = 4×0.707×[1 + 0.3] = 4×0.707×1.3 = 4×0.9191 = 3.676 ≈ 3.68 kg. Recalculating: sinθ + μcosθ = 0.707 + 0.3×0.707 = 0.707×1.3 = 0.9191. m₂ = 4×0.9191 = 3.676 kg. Hmm, but Option A says 4.07 kg; let me recheck: 4×(0.707 + 0.3×0.707) = 4×(0.707 + 0.2121) = 4×0.9191 = 3.676 kg. Close to option A (4.07 kg) may use slightly different values. With sin45=cos45=1/√2=0.7071: m₂=4×0.7071×1.3=3.677 kg. The correct nearest value is approximately 3.68 kg. This is the key formula: m₂ = m₁[sinθ + μcosθ] where for 45°, sinθ = cosθ so m₂ = m₁×cosθ×(tanθ + μ) = m₁cosθ(1+μ) = 4×0.707×1.3=3.68 kg.

Practice Problems — Minimum Mass Hung From String

Click "Reveal Answer" after attempting
1A 12 kg block on a rough horizontal plane is connected by a string over a frictionless pulley to a 3 kg hanging mass. The system is in equilibrium. What is the minimum coefficient of static friction? If μ_k = 0.2, find the acceleration when the system is released. (g = 10 m/s²)
μ_s_min = 0.25; a = 0.5 m/s² (for μ_k = 0.2)
μ_s_min = 0.25; system won't move since μ_s ≥ 0.25 equals threshold
μ_s_min = 0.4; system won't move
μ_s_min = 0.25; a = 1 m/s²
👁 Reveal Answer
Minimum μ_s for equilibrium: m₂ ≤ μ_s m₁ → 3 ≤ μ_s × 12 → μ_s ≥ 3/12 = 0.25. So minimum μ_s = 0.25. If μ_s = exactly 0.25, system is at threshold of motion. For acceleration with μ_k = 0.2 (< μ_s = 0.25): m₂g − μ_k m₁g = (m₁ + m₂)a → 3×10 − 0.2×12×10 = 15×a → 30 − 24 = 15a → a = 6/15 = 0.4 m/s². Answer: μ_s_min = 0.25; once moving with μ_k = 0.2, a = 0.4 m/s².
2A block of mass 6 kg is on a 37° incline (sin37°=0.6, cos37°=0.8, μ=0.3). A hanging mass m₂ is connected to pull the block UP the slope. Find m₂_min. Then find m₂_min to just prevent the block from sliding DOWN the slope.
m₂_up = 5.04 kg; m₂_prevent_down = 2.16 kg
m₂_up = 3.6 kg (gravity only); m₂_prevent_down = 2.16 kg
m₂_up = 5.04 kg; m₂_prevent_down = 1.44 kg
m₂_up = 7.2 kg; m₂_prevent_down = 2.4 kg
👁 Reveal Answer
FOR PULLING UP: m₂_min = m₁[sinθ + μcosθ] = 6×[0.6 + 0.3×0.8] = 6×[0.6+0.24] = 6×0.84 = 5.04 kg. FOR PREVENTING SLIDE DOWN (m₂ hangs to hold m₁ on incline): Gravity pulls m₁ down (m₁gsinθ = 36 N). Friction now acts UP (opposing tendency to slide down): F_l = μm₁gcosθ = 0.3×6×10×0.8 = 14.4 N. For m₁ not to slide down: T + F_l ≥ m₁gsinθ → m₂g ≥ m₁gsinθ − F_l → m₂ = m₁[sinθ − μcosθ] = 6×[0.6−0.24] = 6×0.36 = 2.16 kg. Answer: Option A.
3In an inclined-surface experiment, m₁ = 5 kg, θ = 30°, and the measured minimum hanging mass is m₂ = 4 kg. Find the coefficient of friction. (sin30°=0.5, cos30°=0.866)
μ = 0.578
μ = 0.3
μ = 0.462
μ = 0.8
👁 Reveal Answer
From m₂ = m₁[sinθ + μcosθ]: 4 = 5×[0.5 + μ×0.866] → 4/5 = 0.5 + 0.866μ → 0.8 = 0.5 + 0.866μ → 0.3 = 0.866μ → μ = 0.3/0.866 ≈ 0.346. Wait: μ = (m₂/m₁ − sinθ)/cosθ = (4/5 − 0.5)/0.866 = (0.8 − 0.5)/0.866 = 0.3/0.866 ≈ 0.346. Rounding to 3 sig figs: μ ≈ 0.346. The closest provided option is 0.3 but the exact answer is 0.346. Answer: μ = (m₂/m₁cosθ) − tanθ = (4/5×0.866) − (0.5/0.866) = 0.924 − 0.577 = 0.347 ≈ 0.346. Use formula μ = m₂/(m₁cosθ) − tanθ to solve directly.
4A 3 kg mass m₂ hangs by a string over a pulley connected to a 5 kg mass m₁ on an inclined plane (μ = 0.4). The system just moves. Find the angle of the inclined plane. (g = 10 m/s²)
θ = 18.43° (tanθ = 1/3)
θ = 30°
θ = 37°
θ = 53°
👁 Reveal Answer
At threshold of motion: m₂ = m₁[sinθ + μcosθ] → 3 = 5[sinθ + 0.4cosθ] → sinθ + 0.4cosθ = 0.6. Solving: write as R×sin(θ + φ) = 0.6 where R = √(1² + 0.4²) = √1.16 = 1.077, tanφ = 0.4 → φ = 21.8°. sin(θ+21.8°) = 0.6/1.077 = 0.557 → θ+21.8° = 33.85° → θ = 12.05°. Alternatively, try θ = 18.43°: sin18.43° = 0.3162, cos18.43° = 0.9487. sinθ + 0.4cosθ = 0.3162+0.3795 = 0.696 ≠ 0.6. The exact solution requires iterative or numerical approach. Best method: m₂/m₁ = sinθ + μcosθ → 0.6 = sinθ + 0.4cosθ. At θ=10°: 0.1736+0.3939=0.567; at θ=12°: 0.2079+0.3928=0.601≈0.6. So θ ≈ 12°. The exact formula is 0.6 = sinθ + 0.4cosθ, solved numerically.

Physics — Friction Revision Checklist

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FAQ — Minimum Mass Hung From String

Notes · Downloads · Revision · Important Questions
What is the minimum hanging mass formula on a horizontal surface?
m₂_min = μm₁. At the threshold of motion, tension T = m₂g equals limiting static friction F_l = μm₁g (since normal force = m₁g on horizontal). Setting T = F_l: m₂g = μm₁g → m₂ = μm₁. This directly gives the coefficient of friction as μ = m₂/m₁, which is how the experiment can be used to measure μ.
What is the minimum hanging mass formula on an inclined surface?
m₂_min = m₁[sinθ + μcosθ], where θ is the angle of inclination. On the incline, normal force N = m₁gcosθ and friction F_l = μm₁gcosθ (opposing upward motion, directed DOWN slope). Gravity component opposing upward motion = m₁gsinθ (DOWN slope). Total resistance = m₁g[sinθ + μcosθ]. At threshold: T = m₂g equals this total resistance.
How does the formula change if m₁ is being pulled DOWN the incline?
If m₁ is about to slide DOWN: gravity (m₁gsinθ) aids motion, friction (μm₁gcosθ) now acts UP the slope opposing downward motion. For m₂ to just prevent sliding down: T + μm₁gcosθ = m₁gsinθ → m₂ = m₁[sinθ − μcosθ]. This requires sinθ > μcosθ (θ > angle of repose) for m₁ to actually have a tendency to slide down. If θ < angle of repose, s the block stays stationary even without m₂.
Why is normal force mgcosθ on an incline, not mg?
On an inclined surface, the block's weight mg acts vertically. Only the component perpendicular to the incline surface is balanced by the normal force: N = mgcosθ (not mg). The other component mgsinθ acts along the surface (downhill). Since kinetic friction = μN, on incline it equals μmgcosθ, not μmg. Ignoring the cosθ overestimates friction and gives wrong answers.
What does 'frictionless pulley' mean for this problem?
A frictionless pulley means the tension in the string is the same on both sides of the pulley. Without this assumption, T₁ ≠ T₂ due to friction at the pulley axle. In NEET problems, unless explicitly stated otherwise, pulley is assumed frictionless and string massless, so T = m₂g on both sides.
How do you find θ given all other values?
From m₂ = m₁[sinθ + μcosθ]: divide by m₁ to get m₂/m₁ = sinθ + μcosθ. Write as A×sin(θ + φ) where A = √(1+μ²) and tanφ = μ. Solve sin(θ + φ) = (m₂/m₁)/√(1+μ²) → θ + φ = arcsin(...) → θ = arcsin(...) − arctan(μ). For simpler cases in NEET, try standard angles (30°, 37°, 45°, 53°) and check which one satisfies the equation.
What happens if m₂ exceeds the minimum value?
If m₂ > m₂_min (= m₁[sinθ + μcosθ] for inclined), the system accelerates. The net force = m₂g − m₁g[sinθ + μcosθ] (replacing static by kinetic friction μ_k). Acceleration a = [m₂g − T_resist]/(m₁+m₂) = [m₂ − m₁(sinθ + μ_k cosθ)]g/(m₁+m₂). Note: once moving, use μ_k (kinetic), not μ_s (static), which is typically slightly smaller.
What is the condition for the incline formula to be valid?
The formula m₂ = m₁[sinθ + μcosθ] applies when m₂ is trying to pull m₁ UP the incline. It is valid for any θ from 0° to 90°. At θ=0°, it reduces to m₂=μm₁ ✓. At θ=90° (vertical wall), m₂=m₁[1+0]=m₁ (must support m₁'s full weight). If θ=angle of repose (tanθ=μ), then m₂=m₁×2sinθ (gravity component equals friction; m₂ must overcome twice the gravity component along the incline).
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Horizontal and Inclined Cases

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Horizontal and Inclined Cases

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Horizontal and Inclined Cases

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NEET > Physics > Laws of Motion Chapters

Review your status and progress for each chapter in this unit. Use the slider to set progress or click "Mark as Done" to complete.

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Newton's Laws of Motion

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Friction

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