Graph Between Applied Force and Friction

NEET > Physics > Laws of Motion > Friction > Graph Between Applied Force and Friction

Unit Progress

NEET Physics — Friction

Graph Between Applied Force and Friction – Complete Notes, Revision, Important Questions & Downloads

The graph between applied force (F) and friction (f) captures the entire life cycle of friction in a single visual — consolidated in the subtopic Friction vs Applied Force Graph. NEET tests: (1) identifying segments (OA = static, BC = kinetic), (2) which point is limiting friction, (3) why BC is horizontal. Segment OA (slope 45°): static friction is self-adjusting — as applied force increases from 0 to limiting friction value, static friction matches it exactly (f = F, so the slope is 1). Point A: the peak, where friction equals its maximum value — limiting friction (μ_s R). Segment BC: kinetic friction — once the applied force crosses the limiting value, the object starts to slide and friction drops slightly to kinetic friction (μ_k R), then stays constant regardless of how much the applied force increases. The BC segment is horizontal (parallel to x-axis), because kinetic friction is independent of applied force magnitude. The slope of BC = 0. The key takeaway: limiting friction > kinetic friction, so the peak (A) is always above the plateau (BC). This graph is the single most important visual in the Friction chapter for NEET.

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Friction vs Applied Force GraphFriction Ch.5f_l > f_k

Expected QuestionsQ

0–1

NEET tests the graph via MCQs that show the graph and ask: which segment represents static/limiting/kinetic friction? What is the slope of OA? Why is the BC segment horizontal? What does point A represent? Graph-reading and interpretation questions. Sometimes tested as 'assertion-reason' format about kinetic vs static friction.

Time Required⏱

15 min

5 min to draw and label the 3-segment graph (OA linear static, A peak limiting, BC horizontal kinetic). 5 min to understand what each segment means physically. 5 min to solve NEET-style MCQs on the graph.

Difficulty⚡

Easy

The graph has only 3 regions and 4 key labels (OA, A, AB, BC). The physical interpretation of each segment is straightforward. The most common trap is confusing which peak (limiting friction, point A) and which plateau (kinetic friction, BC). Easy once the graph is drawn from scratch and each segment's meaning is understood.

NRI USA Curriculum GapUS

Low

AP Physics 1 and University Physics both use friction coefficient diagrams. The exact applied-force vs friction graph (showing limiting friction peak followed by kinetic plateau) is a standard illustration in US physics textbooks (Serway, Halliday). US students may call it the 'friction transition graph'. The graph is tested in NEET in a very standardised way — knowing which segment is which is the key skill.

1Subtopics

4+Practice Questions

4Free Downloads

15 minPrep Time

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NEET Weightage — Graph Between Applied Force and Friction

Friction (Chapter 5)

NEET Year	Questions from this Topic	Bar	Marks
2024	0	0 Q	0
2023	1	1 Q	4
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)	1–2		4–8

The applied force vs friction graph has three physically distinct segments: (1) OA — slope = 1, linear increase: static friction is self-adjusting. The friction f exactly equals applied force F (f = F) because the body is stationary (net force = 0). This linear segment holds until static friction reaches its maximum. In this range, the friction force adjusts itself automatically to prevent motion. (2) Point A — the peak of the graph: this is LIMITING FRICTION (maximum static friction), f_l = μ_s R. The applied force at point A is the minimum force needed to set the body in motion. Just before point A: body is stationary, f = F. At point A: body is on the verge of motion. (3) AB — steep drop: as the body starts to slide, friction drops from limiting (μ_s R) to kinetic (μ_k R). Since μ_k < μ_s, kinetic friction is less than limiting friction. This drop is the 'transition' from static to kinetic friction. (4) BC — horizontal plateau, slope = 0: kinetic friction. Once sliding begins, kinetic friction remains constant at μ_k R regardless of applied force magnitude.
Key inequalities from the graph: limiting friction (A) > kinetic friction (BC) because μ_s > μ_k. Static friction (OA segment) ≤ limiting friction (point A). For any applied force F in OA region: f = F exactly. At point A: f = f_l = μ_s R. In BC region: f = f_k = μ_k R (constant, independent of F). The slope of OA is 1 (45° in same-scale axes). The slope of BC is 0 (horizontal). The transition drop AB confirms that kinetic friction is easier to maintain than static: less force is needed to keep a body sliding than to start it sliding.
NEET MCQ pattern for this graph: (1) Part of curve that represents self-adjusting friction = OA. (2) Maximum static friction (limiting friction) = point A. (3) Kinetic friction region = BC. (4) Why is BC horizontal? Because kinetic friction (μ_k R) is constant and does not depend on applied force. (5) Why is A above BC? Because limiting friction > kinetic friction (μ_s > μ_k). (6) What happens at point B? The transition from sliding onset to steady kinetic friction. Some graphs show A and B at the same point (simplified version without the drop).

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0.3

Avg Questions / Year

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Total Marks (6 yrs)

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Direct

Pattern

⚠️

Easy

Difficulty

How to Prepare the Friction Graph for NEET

Practise drawing the graph from scratch in 60 seconds X-axis = Applied force (F). Y-axis = Friction force (f). Step 1: draw OA at 45° (slope = 1, f = F). Step 2: mark point A as the peak (f = μ_s R, limiting friction). Step 3: draw drop from A to B. Step 4: draw BC as horizontal (f = μ_k R, constant). Label: OA = static friction (self-adjusting), A = limiting friction, AB = transition, BC = kinetic friction. Doing this in 60 seconds means you will never confuse the segments in NEET.

Memorise the 4 key properties of each segment OA: slope=1, f=F exactly, body stationary. A: peak, f=μ_sR, body just on verge. AB: drop, transition from static to kinetic. BC: slope=0, f=μ_kR, body sliding. The critical comparative: A is always above BC because limiting (static) friction > kinetic friction. This is both the definition and the graph feature.

Apply graph knowledge to numerical problems Given the graph: read off the limiting friction value (y-coordinate of A) and kinetic friction (y-coordinate of BC). From these, calculate μ_s = f_l/R and μ_k = f_k/R. NEET may give the graph and ask for μ_s or μ_k from the graph coordinates. Example: if limiting friction = 30 N and normal reaction = 50 N: μ_s = 30/50 = 0.6; kinetic friction = 25 N → μ_k = 25/50 = 0.5.

Study Materials — Friction Graph

PDF · Cheat Sheet · MCQ Set · PYQ

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Full Notes

Three-segment graph: OA (static, slope=1), A (limiting friction peak), BC (kinetic, horizontal). Physical meaning of each segment. Slope of OA=1 because f=F (Newton's 1st law for stationary body). BC horizontal because kinetic friction is independent of applied force. Limiting > kinetic (μ_s > μ_k).

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Formula Sheet

OA: f = F (f ≤ μ_s R). Slope of OA = 1. A: f_l = μ_s R (peak). BC: f_k = μ_k R (constant). Slope of BC = 0. Key: μ_s > μ_k → point A above plateau BC.

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MCQ Practice

8 MCQs: graph segment identification, slope values, quantitative reading from graph, limiting vs kinetic friction comparison, assertion-reason on the graph.

8 MCQsGraph-basedSolved

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PYQ

Year-tagged NEET questions on the friction-applied force graph with solutions.

3+ year-tagged Qs2015–2024Solved

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2-Column Table

Column A	Column B
Friction vs Applied Force Graph↗

Rapid Revision — Friction Graph

Concept → Trap → Example

1) Friction vs Applied Force Graph — All Segments Explained

Friction vs Applied Force Graph

The graph of friction force (f) vs applied force (F) has three distinct regions: (1) SEGMENT OA — Static friction region. The line is LINEAR with slope = 1 (45° if axes are equal scale). Static friction is self-adjusting: f = F exactly, because the body is stationary → net force = 0 → friction perfectly balances the applied force. The body stays still for any applied force in this region. Friction 'uses up' as much of its capacity as needed. Runs from F=0 to F=f_l (limiting friction value). (2) POINT A — Limiting friction (maximum static friction). f = f_l = μ_s R. This is the PEAK of the graph. The applied force at point A is the MINIMUM force to start motion. At this instant, friction is at its absolute maximum static value. The body is on the verge of sliding. (3) SEGMENT AB — Transition. As the body just begins to slide, friction drops from limiting (μ_s R) to kinetic (μ_k R). A brief downward slope from peak A to the plateau BC. (4) SEGMENT BC — Kinetic friction region. Horizontal line at height f_k = μ_k R. The slope is ZERO because kinetic friction is constant: it does NOT depend on the magnitude of applied force. The body slides for any applied force in the BC region. CRITICAL FACT: Point A is above BC because μ_s > μ_k (limiting friction > kinetic friction). More force is needed to start sliding than to maintain sliding.

SLOPE ANALYSIS: OA slope = 1 (f increases 1 N for every 1 N increase in F, because f = F for static equilibrium). BC slope = 0 (f is constant = μ_k R regardless of F; kinetic friction has no dependence on applied force magnitude). The slopes encode the physical law: static friction adjusts, kinetic friction does not. Slope of OA = tan(45°) = 1 exactly. Slope of BC = tan(0°) = 0 exactly.
READING VALUES FROM THE GRAPH: y-coordinate of point A = limiting friction = μ_s R. y-coordinate of BC plateau = kinetic friction = μ_k R. From graph: μ_s = (y at A)/R, μ_k = (y at BC)/R. If R (normal reaction) = mg is given: μ_s and μ_k can be calculated directly from graph readings. The x-coordinate of the transition point A = the minimum force needed to start motion = f_l. For applied force less than this x-coordinate: body stays still. For applied force greater: body slides.
PHYSICAL INTERPRETATION OF EACH REGION: OA region: the body is STATIONARY. Even though a force is applied, friction perfectly opposes it. No net force → no acceleration → no motion. The friction force in OA region is not fixed — it equals whatever F is applied. Point A: body is just about to move. One additional unit of force tips the balance. BC region: the body is SLIDING. Friction has dropped to kinetic value and stays constant. Even if F is tripled in BC region, friction does not increase — it stays at μ_k R. This is why μ_k is called constant: it does not increase with applied force.

Example (NEET-style)A block of mass 5 kg (R = 50 N, g = 10 m/s²) is on a rough surface: μ_s = 0.6, μ_k = 0.4. Graph coordinates: OA starts at (0,0), ends at A = (30 N, 30 N) [f_l = μ_s R = 0.6×50 = 30 N]. BC plateau at f_k = μ_k R = 0.4×50 = 20 N (horizontal line from x=30 N onward). Drop from A(30,30) to B(30,20). Verification: if F=20 N (in OA region): body stationary, f=20 N (exactly 20 N, self-adjusting). If F=30 N (at A): body on verge, f=30 N (limiting). If F=40 N (in BC): body sliding, f=20 N (kinetic, constant at 20 N despite F=40 N). Net force on sliding body = 40−20 = 20 N → acceleration = 20/5 = 4 m/s².

US Curriculum Gaps — Friction Graph

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

Applied Force vs Friction Graph in AP Physics 1

AP Physics 1 covers static and kinetic friction but typically does not explicitly require students to draw or analyse the applied-force vs friction graph as a single diagram. The concepts (f < f_max in static regime, f constant in kinetic regime) are covered in Newton's laws units, but the graph as a standalone fixture is more prominent in NCERT and NEET preparation. US physics students who have seen this graph (common in OpenStax University Physics) will find NEET graph questions straightforward. Students who have not seen the unified graph may need to practise recognising all three segments (OA, A, BC) on a single diagram.

AP Physics 1: f ≤ μ_s N (static) and f = μ_k N (kinetic) taught as equations, not always as a graph
NEET: the 3-segment graph (OA linear, A peak, BC horizontal) is a standard NEET figure
US OpenStax University Physics: includes this graph; students who used OpenStax are well-prepared

Self-Adjusting Nature of Static Friction in US Courses

The phrase 'self-adjusting force' for static friction is NCERT-specific. US courses teach the same concept as 'static friction is not a fixed value; it adjusts up to the maximum μ_s N'. The graph's OA segment is the visual representation of this self-adjusting behaviour. Both curricula teach the same physics, but NCERT uses the term 'self-adjusting' explicitly, and the OA segment is directly tied to this term in NEET MCQs. US students should know: OA segment = static friction adjusts; slope = 1; body is stationary throughout OA.

NCERT: explicitly uses 'self-adjusting' for static friction in OA segment
US (AP/OpenStax): same concept: f_s ≤ μ_s N, with f_s adjusting to balance applied force
NEET MCQ likely uses 'Part OA of the graph represents self-adjusting friction' — recognise this phrasing

NEET-Style Practice Questions — Friction Graph

4 Questions

1In the applied force vs friction force graph, the segment BC is horizontal (parallel to the x-axis). This is because:Graph Interpretation

Kinetic friction increases linearly with applied force

Kinetic friction is constant and independent of applied force

The body is stationary in the BC region

Normal reaction decreases in the BC region

'Part OA of the curve represents static friction. At point A the static friction is maximum. This represents limiting friction. Then the portion BC of the curve represents the kinetic friction.' BC is horizontal because kinetic friction f_k = μ_k R is constant — it depends only on μ_k (fixed material property) and normal reaction R (fixed for horizontal surface). It does NOT depend on how large the applied force is. Once the body slides at velocity V, doubling the applied force does not change friction — it only increases the net force and acceleration. The body is SLIDING in BC region (not stationary). Option C is wrong (stationary friction is OA region). Normal reaction doesn't decrease on a horizontal surface. Option B is correct.

2In a friction vs applied force graph, the point A (peak of graph) represents:Point A Identification

The point where kinetic friction begins

The point where friction becomes zero

The maximum value of static friction (limiting friction)

The point where static friction equals kinetic friction

Point A is the peak of the graph where static friction reaches its maximum value — this is the limiting friction (f_l = μ_s R). It is the transition point: for applied forces ≤ f_l, the body is stationary (OA region); at exactly f_l, the body is on the verge of sliding. Beyond f_l, kinetic friction takes over (BC region). The minimum force to START motion = f_l (x-coordinate of point A = y-coordinate = f_l, since slope of OA = 1). Option A is wrong — kinetic friction begins after point A (in BC). Option B is wrong — friction never becomes zero at A. Option D is wrong — after the drop AB, kinetic friction is LESS than limiting friction. Option C is correct.

3A student draws the graph of friction (y-axis) vs applied force (x-axis) for a block on a rough surface. The correctly drawn graph will show:Graph Shape

A straight line through origin with slope 1, then a horizontal line below the peak

A straight line through origin with slope 1, then a horizontal line at the same height

A curved line through origin, then a horizontal line

A horizontal line throughout

The correct graph: (1) OA — straight line from origin with slope = 1 (f = F exactly, body stationary, self-adjusting static friction). (2) Point A — the peak at height f_l = μ_s R. (3) AB — drop from peak to a lower level (since μ_k < μ_s). (4) BC — horizontal line at height f_k = μ_k R. The horizontal line is BELOW point A (not at the same height) because kinetic friction < limiting friction. Option B incorrectly says same height (would imply μ_s = μ_k). Option C is wrong (OA is linear, not curved). Option D is wrong (friction is not constant throughout). Option A is correct (slope 1, then horizontal line BELOW peak).

4For a block on a rough surface, the limiting friction is 24 N and kinetic friction is 18 N. The normal reaction is 40 N. What are μ_s and μ_k?Coefficient Calculation from Graph

μ_s = 0.6, μ_k = 0.45

μ_s = 0.45, μ_k = 0.6

μ_s = 0.6, μ_k = 0.6

μ_s = 0.4, μ_k = 0.3

From the graph: the y-coordinate of peak A = f_l = 24 N = μ_s × R = μ_s × 40. So μ_s = 24/40 = 0.6. The y-coordinate of the BC plateau = f_k = 18 N = μ_k × R = μ_k × 40. So μ_k = 18/40 = 0.45. Check: μ_k < μ_s (0.45 < 0.6) ✓ — kinetic friction is less than limiting friction. This is consistent with the graph showing the BC plateau below peak A. Option B swaps μ_s and μ_k (wrong — μ_s should be larger). Option A is correct.

Practice Problems — Friction Graph

Click "Reveal Answer" after attempting

1A 4 kg block rests on a rough floor. μ_s = 0.5, μ_k = 0.3 (g = 10 m/s²). Identify: (a) limiting friction, (b) kinetic friction, (c) the x-coordinate of point A in the friction vs applied force graph, and (d) what is the friction when applied force = 15 N?

(a) 20 N; (b) 12 N; (c) 20 N; (d) 15 N (static friction matches applied force)

(a) 20 N; (b) 12 N; (c) 20 N; (d) 12 N (kinetic friction applies)

(a) 12 N; (b) 20 N; (c) 12 N; (d) 12 N

(a) 20 N; (b) 12 N; (c) 12 N; (d) 20 N

👁 Reveal Answer

R = mg = 4×10 = 40 N. (a) Limiting friction = μ_s × R = 0.5×40 = 20 N. (b) Kinetic friction = μ_k × R = 0.3×40 = 12 N. (c) x-coordinate of point A = the applied force at which motion starts = f_l = 20 N (since slope of OA = 1, x = y = 20 N at point A). (d) Applied force = 15 N < f_l = 20 N → body is in OA region (stationary). Static friction = applied force = 15 N (self-adjusting). Answer: (a) 20 N; (b) 12 N; (c) 20 N; (d) 15 N.

2From a friction vs applied force graph, the following readings are taken: the graph rises with slope 1 until F = 36 N, then drops to a constant value of 27 N. The block has mass 6 kg (g = 10 m/s²). Find: (a) μ_s, (b) μ_k, (c) acceleration of block when F = 45 N.

(a) 0.6; (b) 0.45; (c) 3 m/s²

(a) 0.6; (b) 0.45; (c) 2.5 m/s²

(a) 0.45; (b) 0.6; (c) 3 m/s²

(a) 0.6; (b) 0.5; (c) 3 m/s²

👁 Reveal Answer

R = mg = 60 N. Limiting friction (peak) = 36 N → μ_s = 36/60 = 0.6. Kinetic friction (plateau) = 27 N → μ_k = 27/60 = 0.45. When F = 45 N (> 36 N limiting): block is sliding. Kinetic friction = 27 N. Net force = F − f_k = 45 − 27 = 18 N. Acceleration = 18/6 = 3 m/s². Answer: (a) μ_s = 0.6; (b) μ_k = 0.45; (c) a = 3 m/s².

3Explain what would happen to the friction vs applied force graph if (a) the surface is polished (μ_s and μ_k both decrease), (b) a heavier block is placed (mass increases, μ unchanged), (c) the block is placed on an incline instead of a horizontal surface.

(a) Both peak A and BC plateau lower; (b) both A and BC move higher (more normal force); (c) graph shape unchanged but scales change (normal force = mg cosθ < mg)

(a) Graph shape unchanged; (b) graph unchanged; (c) graph shape unchanged

(a) Only BC lowers; (b) only OA slope changes; (c) graph moves horizontally

(a) Both A and BC lower; (b) OA slope doubles; (c) BC becomes non-horizontal

👁 Reveal Answer

(a) Polished surface: lower μ_s and μ_k → peak A drops (f_l = μ_s R decreases) and BC plateau drops (f_k = μ_k R decreases). Graph shape same, but both heights decrease. Slope of OA stays 1 (f = F in static region regardless of μ). (b) Heavier block: R = mg increases → f_l = μ_s R increases (A moves up) and f_k = μ_k R increases (BC moves up). OA slope stays 1. Again, graph shape same but both heights increase proportionally. (c) Incline: normal reaction R = mg cosθ < mg → both A and BC lower. Also, gravitational component along incline (mg sinθ) means the body may move even with F=0 if sinθ > μ_s cosθ (i.e., if θ > angle of repose). Answer: Option A is correct.

4A block A (mass 3 kg) is on a rough surface and connected by a string over a pulley to a hanging block B. Gradually increase the mass of B. Describe how friction on block A changes, referencing the friction vs applied force graph segments.

Friction on A = tension in string = weight of B while A is stationary (OA segment). When B reaches limiting mass, A starts sliding; friction on A = kinetic friction (BC segment). Friction on A in BC is less than just before motion.

Friction on A constantly equals μ_k R regardless of B's weight.

Friction on A is always zero until B reaches limiting mass, then jumps to μ_k R.

Friction on A increases beyond kinetic friction as B gets heavier in the sliding phase.

👁 Reveal Answer

As B's mass increases: tension in string = weight of B = m_B × g. For block A: the applied force (tension) increases. Static friction on A adjusts to equal this tension (OA segment: friction = tension = m_B g, self-adjusting). Block A stays still as long as m_B g ≤ f_l (limiting friction on A = μ_s × m_A × g). When m_B = μ_s × m_A: we are at point A (limiting friction). If m_B increases further: block A slides (crosses point A). Friction drops to f_k = μ_k × m_A × g (BC segment, constant). Friction is now LESS than what it was at point A — even though B is heavier. The net force on A = tension − f_k = m_B g − μ_k m_A g (increases as B gets heavier in BC region). Answer: Option A correctly describes OA (self-adjusting, equals weight of B) and BC (constant kinetic friction, less than at onset).

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.

Friction vs Applied Force Graph: Draw the 3-segment graph from scratch: OA (slope=1, f=F, static/self-adjusting, body stationary), A (peak = limiting friction = μ_s R), drop AB, BC (horizontal plateau = kinetic friction = μ_k R, body sliding). Key: slope of OA = 1, slope of BC = 0, A is above BC because μ_s > μ_k. Read μ_s and μ_k from graph using R. Know that in OA region static friction = applied force exactly. BC friction does not increase with applied force.

FAQ — Friction vs Applied Force Graph

Notes · Downloads · Revision · Important Questions

Why is the slope of segment OA exactly 1 in the friction vs applied force graph?

In the OA region, the body is STATIONARY. Newton's 1st law: net force = 0 → friction f = applied force F exactly. For every 1 N of applied force, friction increases by exactly 1 N. This gives a slope of Δf/ΔF = 1 (45° if the axes are on the same scale). This is the mathematical expression of static friction being 'self-adjusting'. If you push with 10 N, static friction is 10 N. Push with 15 N, friction is 15 N. This self-adjustment holds until reaching the limiting friction at point A.

Why does friction DROP from point A to the BC plateau instead of staying constant during the transition?

It drops because static friction is larger than kinetic friction. At point A, the object first starts to move (limiting static friction = μ_s R). Once the object is sliding, the friction mechanism changes — surfaces no longer interlock as deeply, and the friction force falls to kinetic friction (μ_k R). Since μ_k < μ_s, kinetic friction < limiting friction. The drop occurs because the friction TYPE changes from static to kinetic at the onset of motion. Physics: the intermolecular bonding between surfaces breaks at limiting friction; once sliding occurs, only dynamic (kinetic) contact resistance remains, which is lower.

Why is segment BC horizontal and not inclined?

BC is horizontal because kinetic friction f_k = μ_k R depends only on μ_k (fixed for given surface pair) and R (normal reaction, fixed for horizontal surface at given mass). It does NOT depend on the magnitude of the applied force. Whether you push with 50 N or 500 N, kinetic friction stays at μ_k R. Therefore, as applied force (x-axis) increases, kinetic friction (y-axis) stays constant → horizontal line. The only way to change kinetic friction in BC region is to change the normal reaction (e.g., by adding weight) or the surface material (change μ_k).

If I know only the friction vs applied force graph, can I find both μ_s and μ_k?

Yes, if the normal reaction R is also given (or computable from mass and g). From the graph: y-coordinate of peak A = f_l = μ_s R → μ_s = f_l/R. y-coordinate of BC plateau = f_k = μ_k R → μ_k = f_k/R. The graph alone (without R) gives the values of f_l and f_k in Newtons, but to get the dimensionless coefficients μ_s and μ_k, you need to divide by R. The graph also tells you: minimum force to start motion = f_l (x-coordinate of A); friction during sliding = f_k (height of BC); and that μ_s > μ_k (A is above BC).

What does point B in the graph represent (the start of the kinetic friction plateau)?

Point B represents the moment when kinetic friction stabilises at its constant value μ_k R after the transition from limiting static friction. In many NCERT-style graphs, A and B are at the same x-coordinate (the applied force at which motion starts). The segment AB is the brief transition from static (μ_s R) to kinetic (μ_k R). Some graphs simplify this by showing the drop as instantaneous (A = B coincide at x-coordinate). For NEET, the key facts are: (1) A is the peak (limiting friction); (2) BC is horizontal (kinetic friction); (3) A is above BC. The exact shape of the transition drop (AB) is not usually tested.

Does the OA segment start exactly at the origin?

Yes. When applied force F = 0, there is no tendency of the body to slide, so friction = 0. As F increases from 0, static friction increases from 0 to match it. The graph starts at (0, 0) and moves along a 45° line. The origin represents a body at rest with no applied force: zero applied force, zero friction.

How would the graph change if the block is on an inclined surface rather than horizontal?

The shape (OA slope=1, A peak, BC plateau) stays the same, but the absolute values and the OA starting behaviour change. On an incline: the body experiences gravity component along the incline (mg sinθ, pulling it down). Even with zero applied force, the incline gravity acts as a 'built-in applied force'. Normal reaction = mg cosθ (reduced). So f_l = μ_s mg cosθ (lower than horizontal case). The graph has a lower peak A. If θ > angle of repose, the body slides even without applied force — the OA segment might not be explored before the body starts moving at F = 0. The slope of OA remains 1 (Newton's 1st law: if body is in equilibrium, friction + applied force balances gravity component).

What is the physical significance of the applied force value at point A (the x-coordinate)?

The x-coordinate of A = the minimum horizontal applied force needed to START the body sliding on the surface. It equals the limiting friction (f_l = μ_s R), because on the horizontal OA line (slope 1), x-coordinate = y-coordinate at every point. So at peak A: x = y = f_l = μ_s R. This means: applied force needed to start motion = μ_s R. Applied force needed to KEEP motion going (BC region) = f_k = μ_k R (any force exceeding this will accelerate the body; any force equal to f_k maintains constant velocity). Physically: starting friction > sliding friction → you need MORE force to start than to maintain sliding. This is a fundamental, experimentally verified property of solid surfaces.

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Newton's Laws of Motion

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Graph Between Applied Force and Friction

NEET > Physics > Laws of Motion > Friction > Graph Between Applied Force and Friction

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NEET Physics — Friction

Graph Between Applied Force and Friction – Complete Notes, Revision, Important Questions & Downloads

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Friction vs Applied Force GraphFriction Ch.5f_l > f_k

Expected QuestionsQ

0–1

Time Required⏱

15 min

Difficulty⚡

Easy

NRI USA Curriculum GapUS

Low

1Subtopics

4+Practice Questions

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NEET Weightage — Graph Between Applied Force and Friction

Friction (Chapter 5)

NEET Year	Questions from this Topic	Bar	Marks
2024	0	0 Q	0
2023	1	1 Q	4
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)	1–2		4–8

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0.3

Avg Questions / Year

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Total Marks (6 yrs)

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Direct

Pattern

⚠️

Easy

Difficulty

How to Prepare the Friction Graph for NEET

Study Materials — Friction Graph

PDF · Cheat Sheet · MCQ Set · PYQ

📘

Full Notes

1 subtopic1 graphVisual + Conceptual

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Formula Sheet

OA: f = F (f ≤ μ_s R). Slope of OA = 1. A: f_l = μ_s R (peak). BC: f_k = μ_k R (constant). Slope of BC = 0. Key: μ_s > μ_k → point A above plateau BC.

4 identities1 pageQuick reference

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📙

MCQ Practice

8 MCQs: graph segment identification, slope values, quantitative reading from graph, limiting vs kinetic friction comparison, assertion-reason on the graph.

8 MCQsGraph-basedSolved

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📒

PYQ

Year-tagged NEET questions on the friction-applied force graph with solutions.

3+ year-tagged Qs2015–2024Solved

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2-Column Table

Column A	Column B
Friction vs Applied Force Graph↗

Rapid Revision — Friction Graph

Concept → Trap → Example

1) Friction vs Applied Force Graph — All Segments Explained

Friction vs Applied Force Graph

SLOPE ANALYSIS: OA slope = 1 (f increases 1 N for every 1 N increase in F, because f = F for static equilibrium). BC slope = 0 (f is constant = μ_k R regardless of F; kinetic friction has no dependence on applied force magnitude). The slopes encode the physical law: static friction adjusts, kinetic friction does not. Slope of OA = tan(45°) = 1 exactly. Slope of BC = tan(0°) = 0 exactly.
READING VALUES FROM THE GRAPH: y-coordinate of point A = limiting friction = μ_s R. y-coordinate of BC plateau = kinetic friction = μ_k R. From graph: μ_s = (y at A)/R, μ_k = (y at BC)/R. If R (normal reaction) = mg is given: μ_s and μ_k can be calculated directly from graph readings. The x-coordinate of the transition point A = the minimum force needed to start motion = f_l. For applied force less than this x-coordinate: body stays still. For applied force greater: body slides.
PHYSICAL INTERPRETATION OF EACH REGION: OA region: the body is STATIONARY. Even though a force is applied, friction perfectly opposes it. No net force → no acceleration → no motion. The friction force in OA region is not fixed — it equals whatever F is applied. Point A: body is just about to move. One additional unit of force tips the balance. BC region: the body is SLIDING. Friction has dropped to kinetic value and stays constant. Even if F is tripled in BC region, friction does not increase — it stays at μ_k R. This is why μ_k is called constant: it does not increase with applied force.

US Curriculum Gaps — Friction Graph

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

Applied Force vs Friction Graph in AP Physics 1

AP Physics 1: f ≤ μ_s N (static) and f = μ_k N (kinetic) taught as equations, not always as a graph
NEET: the 3-segment graph (OA linear, A peak, BC horizontal) is a standard NEET figure
US OpenStax University Physics: includes this graph; students who used OpenStax are well-prepared

Self-Adjusting Nature of Static Friction in US Courses

NCERT: explicitly uses 'self-adjusting' for static friction in OA segment
US (AP/OpenStax): same concept: f_s ≤ μ_s N, with f_s adjusting to balance applied force
NEET MCQ likely uses 'Part OA of the graph represents self-adjusting friction' — recognise this phrasing

NEET-Style Practice Questions — Friction Graph

4 Questions

1In the applied force vs friction force graph, the segment BC is horizontal (parallel to the x-axis). This is because:Graph Interpretation

Kinetic friction increases linearly with applied force

Kinetic friction is constant and independent of applied force

The body is stationary in the BC region

Normal reaction decreases in the BC region

2In a friction vs applied force graph, the point A (peak of graph) represents:Point A Identification

The point where kinetic friction begins

The point where friction becomes zero

The maximum value of static friction (limiting friction)

The point where static friction equals kinetic friction

3A student draws the graph of friction (y-axis) vs applied force (x-axis) for a block on a rough surface. The correctly drawn graph will show:Graph Shape

A straight line through origin with slope 1, then a horizontal line below the peak

A straight line through origin with slope 1, then a horizontal line at the same height

A curved line through origin, then a horizontal line

A horizontal line throughout

4For a block on a rough surface, the limiting friction is 24 N and kinetic friction is 18 N. The normal reaction is 40 N. What are μ_s and μ_k?Coefficient Calculation from Graph

μ_s = 0.6, μ_k = 0.45

μ_s = 0.45, μ_k = 0.6

μ_s = 0.6, μ_k = 0.6

μ_s = 0.4, μ_k = 0.3

Practice Problems — Friction Graph

Click "Reveal Answer" after attempting

(a) 20 N; (b) 12 N; (c) 20 N; (d) 15 N (static friction matches applied force)

(a) 20 N; (b) 12 N; (c) 20 N; (d) 12 N (kinetic friction applies)

(a) 12 N; (b) 20 N; (c) 12 N; (d) 12 N

(a) 20 N; (b) 12 N; (c) 12 N; (d) 20 N

👁 Reveal Answer

(a) 0.6; (b) 0.45; (c) 3 m/s²

(a) 0.6; (b) 0.45; (c) 2.5 m/s²

(a) 0.45; (b) 0.6; (c) 3 m/s²

(a) 0.6; (b) 0.5; (c) 3 m/s²

👁 Reveal Answer

(a) Both peak A and BC plateau lower; (b) both A and BC move higher (more normal force); (c) graph shape unchanged but scales change (normal force = mg cosθ < mg)

(a) Graph shape unchanged; (b) graph unchanged; (c) graph shape unchanged

(a) Only BC lowers; (b) only OA slope changes; (c) graph moves horizontally

(a) Both A and BC lower; (b) OA slope doubles; (c) BC becomes non-horizontal

👁 Reveal Answer

Friction on A constantly equals μ_k R regardless of B's weight.

Friction on A is always zero until B reaches limiting mass, then jumps to μ_k R.

Friction on A increases beyond kinetic friction as B gets heavier in the sliding phase.

👁 Reveal Answer

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.

FAQ — Friction vs Applied Force Graph

Notes · Downloads · Revision · Important Questions

Why is the slope of segment OA exactly 1 in the friction vs applied force graph?

Why does friction DROP from point A to the BC plateau instead of staying constant during the transition?

Why is segment BC horizontal and not inclined?

If I know only the friction vs applied force graph, can I find both μ_s and μ_k?

What does point B in the graph represent (the start of the kinetic friction plateau)?

Does the OA segment start exactly at the origin?

How would the graph change if the block is on an inclined surface rather than horizontal?

What is the physical significance of the applied force value at point A (the x-coordinate)?

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Graph Between Applied Force and Friction

Overview content

Graph Between Applied Force and Friction – Complete Notes, Revision, Important Questions & Downloads

1) Friction vs Applied Force Graph — All Segments Explained

Applied Force vs Friction Graph in AP Physics 1

Self-Adjusting Nature of Static Friction in US Courses

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Friction vs Applied Force Graph

Types of kinetic friction

Advantages of friction

Two body sticks together due to friction

Brake works on the basis of friction

Subtopics

Friction vs Applied Force Graph

Types of kinetic friction

Advantages of friction

Two body sticks together due to friction

Brake works on the basis of friction

NEET > Physics > Laws of Motion Chapters

Comments

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Graph Between Applied Force and Friction

Overview content

Graph Between Applied Force and Friction – Complete Notes, Revision, Important Questions & Downloads

1) Friction vs Applied Force Graph — All Segments Explained

Applied Force vs Friction Graph in AP Physics 1

Self-Adjusting Nature of Static Friction in US Courses

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Friction vs Applied Force Graph

Types of kinetic friction

Advantages of friction

Two body sticks together due to friction

Brake works on the basis of friction

Subtopics

Friction vs Applied Force Graph

Types of kinetic friction

Advantages of friction

Two body sticks together due to friction

Brake works on the basis of friction

NEET > Physics > Laws of Motion Chapters

Comments

Leave a comment

Comments (0)