100k Followers100k500k Followers500k+1 (510) 706-9331+1 (510) 706-9331
Schedule Your Free Exam Readiness Analysis Session!
Testprepkart Logo
Sign InEnroll NowEnroll
Select an exam to view its content.
  • Blog
  • Download
  • Course
  • Result
  • Video Library
  • Pages
  • Notifications

Loading...

Preparing content

Testprepkart Logo

Enabling students prepare and crack toughest examinations worldwide for over a decade with problem solving aptitude!

Contact Us

Useful Links

  • Connect With Counselor
  • University Admissions
  • Prime Videos
  • Enrollment Form
  • Online Fee Payment
  • Testprepkart Operations
  • Faculty Registration

Our Company

  • Contact Us
  • Work With Us
  • Blogs
  • Facultie
  • Partner

Contact Details

  • Phone: +91 0120 4525484
  • Whatsapp: +1 (510) 706-9331
  • Admission: +91 8800123492
  • E-mail: info@testprepkart.com
  • Head Office: F 377, Sector 63, Noida, Uttar Pradesh, India

Copyright © 2024 CounselKart Educational Services Pvt. Ltd.. All Rights Reserved

Terms of service|Privacy policy|Refund Policy|Login & Register

Resultant Force by Surface on Block

NEET > Physics > Laws of Motion > Friction > Resultant Force by Surface on Block

Unit Progress

0%

Overview content

NEET Physics — Friction

Resultant Force by Surface on Block – Complete Notes, Revision, Important Questions & Downloads

When a block rests on a rough surface, the surface exerts two forces: normal reaction R and friction f. Their resultant S = √(R²+f²) is the Net Contact Force — the single TOC subtopic. NEET tests: (1) S = √(R²+f²), (2) range mg ≤ S ≤ mg√(1+μ²), (3) angle of max S = angle of friction. For a block of mass m on a horizontal surface: R = mg and f varies from 0 to μ_s mg. S_min = mg (f=0); S_max = mg√(1+μ²) (limiting friction). The angle the resultant makes with the normal = angle of friction at maximum. NEET question: 'find range of resultant contact force' → mg to mg√(1+μ²).

⬇ Download Notes PDFView Important Questions →
Net Contact ForceFriction Ch.5mg ≤ S ≤ mg√(1+μ²)
Expected QuestionsQ
0–1
NEET tests this topic as: (1) direct formula S = √(R² + f²), (2) range of contact force (mg to mg√(1+μ²)), (3) connection to angle of friction (angle of S with normal = θ = angle of friction at maximum). Also appears as 'what is the resultant force exerted by the surface on the block?' in various situations.
Time Required⏱
20 min
10 min to understand the vector addition: R (perpendicular) and f (parallel) → resultant S at angle θ. 10 min to derive the range (mg ≤ S ≤ mg√(1+μ²)) and connect to angle of friction.
Difficulty⚡
Medium
The formula derivation is straightforward trigonometry. The challenge is understanding why S is not constant — it depends on the friction force, which varies with applied force. Identifying minimum S (when friction = 0) and maximum S (when friction = limiting friction) requires careful logical analysis.
NRI USA Curriculum GapUS
Medium
AP Physics 1 keeps normal force and friction separate and does not typically compute their resultant S as a named quantity. The concept of 'net contact force' magnitude is occasionally used in problem-solving but is not a standard topic in US courses. NRI students should specifically practise the S = mg√(1+μ²) maximum formula and the range statement.
1Subtopics
4+Practice Questions
4Free Downloads
20 minPrep Time
⬇ Get Free Downloads

NEET Weightage — Resultant Force by Surface on Block

Friction (Chapter 5)
NEET YearQuestions from this TopicBarMarks
20240
 
0 Q
0
20230
 
0 Q
0
20221
 
1 Q
4
20210
 
0 Q
0
20201
 
1 Q
4
20190
 
0 Q
0
6-Year Total (2019–2024)0–2 0–8
BASIC FORMULA: For a block on a horizontal surface with normal reaction R and friction force f: resultant contact force S = √(R² + f²). R and f are perpendicular (R perpendicular to surface, f parallel to surface). The angle S makes with R: tan(angle) = f/R. At limiting friction: f = μ_s R → angle = θ = angle of friction. S = R/cosθ at maximum. NCERT: 'In the above figure resultant force S = √(F² + R²)' where F is the friction force.
RANGE OF S: Friction f varies from 0 (no applied force) to f_max = μ_s R (limiting friction). S = √(R² + f²). Minimum: f = 0 → S_min = √(R² + 0) = R = mg (for horizontal block). Maximum: f = μ_s R = μ_s mg → S_max = √(mg)² + (μ_s mg)²) = mg√(1 + μ_s²). Therefore: mg ≤ S ≤ mg√(1+μ_s²). The lower bound is the normal force alone (smooth surface or no applied force). The upper bound is the full contact force at limiting friction (angle of friction).

CONNECTION TO ANGLE OF FRICTION: The maximum S = mg√(1+μ²) occurs when f = limiting friction. At this point, the angle of S with R = angle of friction θ = tan⁻¹(μ). The resultant S at maximum is exactly the contact force from the surface at limiting friction, which is what was called S in the angle-of-friction derivation. This unifies the two topics: angle of friction is the angle of maximum contact force, and S_max = mg/cosθ = mg√(1+μ²).
📊
0.3
Avg Questions / Year
🎯
8
Total Marks (6 yrs)
📈
Indirect
Pattern
⚠️
Medium
Difficulty

How to Prepare Resultant Contact Force for NEET

1

Understand the vector addition: R perpendicular to f, resultant = S Draw a right-angle triangle: one leg = R (perpendicular to surface, pointing away from surface), other leg = f (parallel to surface, opposing motion). Hypotenuse = S = resultant contact force on block. S = √(R²+f²). The angle S makes with R = tan⁻¹(f/R). At limiting friction: angle = tan⁻¹(μ) = angle of friction.

2

Memorise the range: mg ≤ S ≤ mg√(1+μ²) Min: when applied force = 0, friction = 0, S = R = mg (just the normal force from surface, pointing straight up). Max: at limiting friction, S = mg√(1+μ²). The statement: 'the resultant force exerted by a rough surface on a block varies between mg (smooth contact) and mg√(1+μ²) (rough contact at limiting friction)'. NEET MCQ: 'range of S for a block on rough surface' → mg to mg√(1+μ²).

3

Connect to angle of friction and solve numerical problems Given μ_s = some value and mass m: S_max = mg√(1+μ_s²). Angle of S with normal at this maximum = angle of friction = tan⁻¹(μ_s). For S between min and max: the angle of S with normal is between 0° and θ. If the problem asks for S when a specific friction f is acting: S = √(R²+f²) = √((mg)²+f²).

Study Materials — Resultant Contact Force

PDF · Cheat Sheet · MCQ Set · PYQ
📘
Full Notes
S = √(R²+f²). R and f perpendicular. Range: mg ≤ S ≤ mg√(1+μ²). Max at limiting friction. Min when friction = 0 (S = R = mg). Angle of S with normal = θ (angle of friction) at maximum. S = R/cosθ. Connection to angle of friction.
1 subtopic1 diagramVector + Range
Download Notes
📗
Formula Sheet
S = √(R²+f²). S_min = mg (f=0). S_max = mg√(1+μ²) (f=limiting). Range: mg ≤ S ≤ mg√(1+μ²). Angle at max: θ = tan⁻¹(μ). S = R/cosθ.
4 formulas1 pageQuick reference
Download Sheet
📙
MCQ Practice
8 MCQs: minimum/maximum contact force, range identification, angle of resultant, connection to angle of friction.
8 MCQsRange + AngleSolved
Download MCQs
📒
PYQ
Year-tagged NEET questions on resultant contact force with solutions.
3+ year-tagged Qs2015–2024Solved
Download PYQs

2-Column Table
Column AColumn B
Net Contact Force↗

Rapid Revision — Resultant Contact Force

Concept → Trap → Example

1) Net Contact Force — Range and Angle

Net Contact Force

A block of mass m rests on a rough horizontal surface (μ_s = coefficient of static friction). The surface exerts two perpendicular contact forces on the block: (1) NORMAL REACTION R: perpendicular to surface, pointing away from surface. For horizontal surface: R = mg (vertical equilibrium). (2) FRICTION f: parallel to surface, in direction opposing tendency of motion. Self-adjusting: 0 ≤ f ≤ μ_s R = μ_s mg. RESULTANT CONTACT FORCE S: Since R ⊥ f: S = √(R² + f²) = √((mg)² + f²). The direction of S: angle θ_actual = tan⁻¹(f/R) from normal. RANGE OF S: Case 1 — No applied force: f = 0. S = √(mg)² + 0 = mg (minimum). S is vertical, equal to weight, at 0° from normal. Case 2 — Applied force, but block stationary (partial friction): 0 < f < μ_s mg. S = √(mg)² + f² → between mg and mg√(1+μ_s²). Case 3 — At limiting friction: f = μ_s mg (maximum). S = √(mg)² + (μ_s mg)² = mg√(1 + μ_s²) (maximum). Angle of S from normal = θ = tan⁻¹(μ_s) = ANGLE OF FRICTION. NCERT states: 'Hence the range of S can be given by: mg ≤ S ≤ mg√(μ_s² + 1).' The lower bound: smooth surface contact (f=0; S=R=mg). Upper bound: maximum roughness contact at limiting friction.

  • VECTOR GEOMETRY: R and f are always perpendicular (R perpendicular to surface; f parallel to surface). Their resultant S = √(R²+f²) always. The angle of S with R: tan(angle) = f/R. As f increases from 0 to μ_s R: the angle increases from 0° to tan⁻¹(μ_s) = angle of friction. S increases from R to R√(1+μ_s²). The resultant S rotates from along-the-normal toward the friction direction as friction increases. The MAXIMUM angle S can make with the normal = angle of friction θ.
  • FORMULAS AND ALTERNATIVE EXPRESSIONS: S_max = mg√(1+μ_s²) = mg/cosθ = R/cosθ. Using sinθ = μ_s/√(1+μ_s²): S_max = mg/cosθ. The angle at maximum: θ = tan⁻¹(μ_s). For a specific applied force F_applied on the block: if block stationary, friction f = F_applied (static). S = √((mg)²+(F_applied)²). If at limiting friction: f = μ_s mg → S = mg√(1+μ_s²). NEET FACT: on a smooth surface (μ_s = 0): S = mg always (only normal force; no friction). On rough surface: S varies between mg and mg√(1+μ_s²) depending on applied force.
  • EQUILIBRIUM CHECK: The resultant S of contact forces (R and f) must balance all external forces for the block to remain stationary. If a horizontal force F is applied: equilibrium requires S to balance the resultant of mg (downward) + F (horizontal). For equilibrium: horizontal component of S = f = F, vertical component of S = R = mg. S = √(F²+(mg)²) = √F² + m²g². This is only possible as long as S ≤ S_max = mg√(1+μ_s²). If S required > S_max: block slides (equilibrium impossible). This is the complete equilibrium criterion using contact force resultant.
Example (NEET-style)A 3 kg block (μ_s = 0.6, g = 10 m/s²) is on a rough floor. Find: (a) S when no force applied, (b) S when horizontal force F = 12 N acts (block stationary), (c) S at limiting friction. (a) f=0: S = mg = 30 N. (b) Block stationary → f = F = 12 N. R = mg = 30 N. S = √(30²+12²) = √(900+144) = √1044 = 6√29 ≈ 32.3 N. Angle with normal: tan⁻¹(12/30) = tan⁻¹(0.4) ≈ 21.8°. (c) f_l = μ_s mg = 0.6×30 = 18 N. S_max = √(30²+18²) = √(900+324) = √1224 ≈ 34.99 N ≈ 35 N. Check: = mg√(1+μ²) = 30√(1+0.36) = 30√1.36 ≈ 30×1.166 ≈ 34.97 N ✓. Range: 30 N ≤ S ≤ 35 N. At (c), angle with normal = tan⁻¹(18/30) = tan⁻¹(0.6) ≈ 30.96° = angle of friction ✓.

US Curriculum Gaps — Resultant Contact Force

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

Net Contact Force as Named Quantity in AP Physics 1

AP Physics 1 treats normal force and friction as separate components of the contact interaction. The concept of their vector resultant S as a named quantity ('resultant force exerted by surface on block') is not standard in US AP Physics 1. However, the vector addition is straightforward: S = √(N²+f²). US students comfortable with vector addition will easily compute S. The key NEET-specific knowledge: the range mg ≤ S ≤ mg√(1+μ²) and the equality S_max direction = angle of friction. These are NCERT-specific named results that NRI students should explicitly memorise.

  • AP Physics 1: normal force and friction are kept separate; their vector sum not typically computed
  • NEET: S = √(R²+f²) is expected; range mg to mg√(1+μ²) is a standard exam result
  • Connecting S to angle of friction (S_max direction = θ) is an NCERT-unique unifying concept

Range of Contact Force in University Physics

The range statement mg ≤ S ≤ mg√(1+μ²) — that the surface contact force varies between mg (frictionless limit) and mg√(1+μ²) (maximum friction) — is presented in NCERT as a explicit result from the Angle of Friction discussion. US university physics (Halliday & Resnick) does not present this range as a standalone named result, though all physics follows from the same principles. For NEET, the range formula and the connection to limiting friction are required knowledge.

  • NCERT: mg ≤ S ≤ mg√(1+μ²) is a stated range in the angle of friction / contact force section
  • Halliday & Resnick: same physics derivable from vector addition; not presented as a named range
  • NEET may ask: 'range of contact force on block' → know immediately this range formula

NEET-Style Practice Questions — Resultant Contact Force

4 Questions
1A block of mass 4 kg (μ_s = 0.5, g = 10 m/s²) is on a rough floor with no applied force. The resultant force exerted by the floor on the block is:S when f=0
20 N
40 N
20√5 N
0 N
No applied horizontal force → no tendency to slide → friction f = 0. Only the normal reaction R acts: R = mg = 4×10 = 40 N (vertical, away from floor). Resultant S = √(R²+f²) = √(40²+0) = 40 N. The resultant contact force is just the normal reaction = 40 N. Option B is correct. Option A (20 N) is wrong. Option C (20√5) would be S_max. Option D (0 N) is wrong — the floor supports the block.
2The maximum resultant contact force of a rough surface on a 5 kg block (μ_s = 0.6, g = 10 m/s²) is:S Maximum
50 N
30 N
10√34 N
50√(1.36) N
S_max = mg√(1+μ_s²) = 5×10×√(1+0.36) = 50√1.36 = 50×√(34/25) = 50×√34/5 = 10√34 N ≈ 10×5.831 ≈ 58.3 N. This occurs at limiting friction. Check: R = 50 N, f_l = μ_s R = 0.6×50 = 30 N. S_max = √(50²+30²) = √(2500+900) = √3400 = 10√34 ✓. Option C (10√34) is correct. Option A (50=mg) is the minimum S. Option B (30=f_l) is just the friction. Option D is the same value written differently: 50√1.36 = 50×√(34/25) = 10√34. Option D is equivalent to C but Option C is the simpler exact form.
3For a block on a rough horizontal surface, the range of the resultant contact force S from the surface is:Range of S
0 ≤ S ≤ mg√(1+μ²)
mg ≤ S ≤ mg√(1+μ²)
μmg ≤ S ≤ mg
S is always equal to mg
Minimum S: when no horizontal force is applied, friction f = 0, S = R = mg (normal reaction only). Maximum S: at limiting friction, f = μmg, S = √(mg²+(μmg)²) = mg√(1+μ²). Therefore: mg ≤ S ≤ mg√(1+μ²). Option A is wrong (lower bound should be mg, not 0 — the surface always exerts at least the normal force mg to support the block's weight). Option C is wrong (μmg is just friction, not the minimum S). Option D is wrong (S is constant at mg only for a smooth surface). Option B is correct.
4A block of mass 2 kg is on a rough horizontal table. A horizontal force of 8 N acts on it and the block remains stationary (g = 10 m/s²). Find the resultant contact force on the block from the table.S for Stationary Block with Applied Force
20 N
8 N
√464 N ≈ 21.5 N
28 N
Block is stationary. Normal reaction R = mg = 2×10 = 20 N (vertical). Friction force f = applied force = 8 N (horizontal, static friction adjusts to exactly balance applied force). Resultant S = √(R²+f²) = √(20²+8²) = √(400+64) = √464 = 4√29 ≈ 21.5 N. Option A (20 N) is just R. Option B (8 N) is just friction. Option D (28 N) is wrong (scalar sum, not vector sum). Option C (√464 ≈ 21.5 N) is correct.

Practice Problems — Resultant Contact Force

Click "Reveal Answer" after attempting
1A 6 kg block (μ_s = 0.4, μ_k = 0.3, g = 10 m/s²) is on a rough floor. Find the resultant contact force from the floor when: (a) no applied force, (b) applied horizontal force F = 15 N (block stays still), (c) applied force = 30 N (block slides).
(a) 60 N; (b) √(60²+15²) = √3825 ≈ 61.8 N; (c) √(60²+18²) = √3924 ≈ 62.6 N
(a) 60 N; (b) 15 N; (c) 18 N
(a) 60 N; (b) √3825 N; (c) 30 N
(a) 60 N; (b) 75 N; (c) 48 N
👁 Reveal Answer
R = mg = 60 N. (a) f = 0: S = 60 N. (b) F = 15 N, block stays (check: limiting f = μ_s×60 = 24 N > 15 N ✓). Static friction f = 15 N. S = √(60²+15²) = √(3600+225) = √3825 = 15√17 ≈ 61.8 N. (c) F = 30 N > 24 N → block slides. Kinetic friction f_k = μ_k×60 = 18 N. S = √(60²+18²) = √(3600+324) = √3924 ≈ 62.6 N. Note: in (c) S is still > S in (a) because kinetic friction adds to R. Answer: Option A.
2For a block of mass 3 kg (μ_s = 1/√3, g = 10 m/s²): (a) find the angle of friction, (b) find S_max, (c) verify S_max = mg/cosθ.
(a) 30°; (b) S_max = 30√(4/3) = 20√3 ≈ 34.6 N; (c) 30/cos30° = 30/(√3/2) = 20√3 ✓
(a) 60°; (b) S_max = 30√3; (c) 30/cos60° = 60 N
(a) 30°; (b) S_max = 30√3; (c) 30/cos30° = 20√3
(a) 30°; (b) 40 N; (c) 30/cos30° ≠ 40 N
👁 Reveal Answer
(a) θ = tan⁻¹(μ_s) = tan⁻¹(1/√3) = 30°. (b) S_max = mg√(1+μ_s²) = 30×√(1+1/3) = 30×√(4/3) = 30×2/√3 = 60/√3 = 20√3 ≈ 34.6 N. (c) mg/cosθ = 30/cos30° = 30/(√3/2) = 60/√3 = 20√3 ≈ 34.6 N ✓. Both methods give the same value. Answer: Option A.
3Show that the minimum resultant contact force on a block is mg (when no external horizontal force acts). Then find μ_s if S_max = 2×S_min for a 5 kg block.
S_min = mg = 50 N; from S_max = 2S_min: mg√(1+μ²) = 2mg → √(1+μ²) = 2 → μ_s = √3
S_min = 0; μ_s = √3
S_min = mg = 50 N; μ_s = 1
S_min = mg = 50 N; μ_s = 2
👁 Reveal Answer
Minimum S: no applied force → f = 0. S_min = √(R²+0²) = R = mg. For 5 kg: S_min = 50 N ✓. Given S_max = 2×S_min: mg√(1+μ_s²) = 2mg → √(1+μ_s²) = 2 → 1+μ_s² = 4 → μ_s² = 3 → μ_s = √3. Angle of friction: tan⁻¹(√3) = 60°. Answer: Option A.
4A block of mass 5 kg is pushed against a rough vertical wall with a horizontal force P (g = 10 m/s²). The coefficient of static friction between block and wall is 0.5. What minimum P is needed to keep the block from sliding down? What is the resultant contact force on the block from the wall?
P = 100 N; S = √(100²+50²) = 50√5 N
P = 100 N; S = √(P²+(mg)²) = √(100²+50²) = 50√5 N
P = 100 N; S = mg = 50 N
P = 50 N; S = 50√5 N
👁 Reveal Answer
Block against vertical wall. Normal force from wall = P (horizontal, perpendicular to wall). Friction from wall = μ_s×P (vertical, upward — opposing tendency to slide down). For block not to slide: friction ≥ weight: μ_s P ≥ mg → P ≥ mg/μ_s = 50/0.5 = 100 N minimum. At minimum P = 100 N: friction = 0.5×100 = 50 N (upward) = mg ✓ (just balanced). Resultant contact force from wall on block: S = √(P² + (friction)²) = √(100²+50²) = √(10000+2500) = √12500 = 50√5 N ≈ 111.8 N. This resultant S points diagonally outward and upward from the wall. Answer: Option B.

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.

Tip: Mark a chapter complete only after revising formulas, solving PYQs, and reviewing your error log for that chapter.

FAQ — Resultant Contact Force

Notes · Downloads · Revision · Important Questions
What is the resultant force exerted by a surface on a block?
The surface exerts two forces on a block: (1) Normal reaction R (perpendicular to surface), (2) Friction f (parallel to surface). Since these are perpendicular, their resultant S = √(R²+f²). For a block on a horizontal surface: R = mg (for vertical equilibrium) and f varies from 0 (no applied force) to μ_s mg (limiting friction). The resultant S varies from mg (minimum, when f=0) to mg√(1+μ_s²) (maximum, at limiting friction).
When is the resultant contact force minimum and when is it maximum?
Minimum: when there is no horizontal applied force (and block is stationary). Friction = 0. S = √(R²+0) = R = mg. The surface exerts only the normal reaction. Maximum: when the block is on the verge of sliding (friction at limiting value). f = μ_s R = μ_s mg. S_max = √(mg)²+(μ_s mg)²) = mg√(1+μ_s²). Range: mg ≤ S ≤ mg√(1+μ²).
Why is the minimum contact force equal to mg and not zero?
Even on a smooth surface (or with no applied force), the surface must support the block's weight. The normal reaction R = mg is always present as long as the block rests on the surface (it is the surface's response to the block's weight). The friction component can be zero (when no horizontal force is applied), but the normal reaction cannot be zero for a block resting on the surface. So the minimum resultant S = √(mg² + 0) = mg, not zero.
What angle does the maximum resultant contact force make with the normal?
At maximum S (limiting friction): f = μ_s R. The angle of S with R: tan(angle) = f/R = μ_s R/R = μ_s. Angle = tan⁻¹(μ_s) = angle of friction θ. This is why the angle of friction is the angle of the maximum resultant contact force with the normal. It is the maximum angle the contact force can make — at any smaller friction, the angle is less than θ.
How does the resultant contact force change as applied force increases?
As applied horizontal force F increases from 0: friction self-adjusts from 0 to F (static friction). Normal reaction R = mg stays constant. Resultant S = √(mg²+F²) increases from mg toward mg√(1+μ_s²). The angle of S from normal increases from 0° toward θ. At F = μ_s mg (limiting friction): S = S_max = mg√(1+μ_s²), angle = θ. If F > μ_s mg: block slides. Kinetic friction f_k = μ_k mg (< μ_s mg) takes over. S drops slightly: S = √(mg²+(μ_k mg)²) = mg√(1+μ_k²) (slightly less than S_max). The resultant S during sliding is constant at mg√(1+μ_k²).
For a block pushed against a vertical wall, how does the contact force analysis change?
When a horizontal force P pushes a block against a vertical wall: normal reaction from wall R = P (the wall's normal force is horizontal, balancing the applied push). Friction from wall = vertical (upward, opposing the block's tendency to slide down due to gravity). For the block not to slide down: friction ≥ mg → μ_s P ≥ mg → P ≥ mg/μ_s. Resultant contact force S = √(P²+(friction)²). At minimum P (= mg/μ_s): friction = μ_s × (mg/μ_s) = mg. S = √((mg/μ_s)² + (mg)²) = mg√(1/μ_s² + 1) = mg√((1+μ_s²)/μ_s²) = mg√(1+μ_s²)/μ_s. This is a vertical wall scenario where the formulas adapt to the changed geometry.
Is the resultant contact force the same as the reaction force mentioned in Newton's 3rd law?
Yes. The resultant contact force S (combination of normal reaction R and friction f) is the total force the surface exerts on the block. By Newton's 3rd law, the block exerts an equal and opposite force S on the surface. The 'net reaction force from the surface' = S = √(R²+f²). This is why computing S is useful: it tells you the complete interaction force between the block and surface, not just the perpendicular component (R) or the parallel component (f) individually.
Does the range mg ≤ S ≤ mg√(1+μ²) apply to all surfaces?
This range applies specifically to a HORIZONTAL surface with a block of mass m, where vertical equilibrium gives R = mg. For other configurations: (a) Inclined plane: R = mg cosφ (normal force reduced). S_max = R√(1+μ²) = mg cosφ × √(1+μ²). (b) Block in an elevator (accelerating up): R = m(g+a). S_max = m(g+a)√(1+μ²). The principle S = √(R²+f²) and S_max = R√(1+μ²) always hold, but the value of R changes with the geometry and accelerations. Always compute R first from the specific force balance, then apply S_max = R√(1+μ²).
For NRI / OCI / U.S.-Based Families

NEET NRI Counseling & Admission eBook Download

A practical guide covering sponsor rules, document checklist, verification traps, NRI quota reality, and step-by-step counselling flow. Designed to prevent last-minute rejections and wrong choice filling.

Sponsor + Proof ClarityDocuments ChecklistState-wise Traps
↓ Download eBook (PDF)→ See What's Inside
Tip: Keep this eBook open during verification + choice filling week for quick cross-checking.
NEET Prep (India + NRI-USA)

Schedule Trial Session For NEET Prep

Get a short diagnostic + study roadmap: syllabus gaps (NCERT vs U.S. curriculum), weak chapters, and the exact weekly plan needed to improve accuracy under time.

Gap MappingWeekly PlanAccuracy Fix
→ Book Trial Session→ WhatsApp Us
Best for: Students in Grade 10–12 (U.S. / India) who want a clear NEET timeline and daily practice structure.

Net Contact Force

By polishing

By lubrication

By proper selection of material

By streamlining the shape of the body

By using ball bearing

Subtopics

Net Contact Force

By polishing

By lubrication

By proper selection of material

By streamlining the shape of the body

By using ball bearing

Previous
Resultant Force by Surface on Block > By using ball bearing
Next
Net Contact Force

Loading tests...

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.

ChapterStatusProgress

Newton's Laws of Motion

Weightage: 02.2K
0%

Friction

Weightage: 02.2K
0%

Comments

Leave a comment

0/2000Comments are moderated

You can comment without logging in. We'll ask for your name and email before submitting.

Comments (0)

No comments yet. Be the first to comment!