NEET Physics — Newton's Laws of Motion

Free Body Diagram – Complete Notes, Revision, Important Questions & Downloads

A Free Body Diagram (FBD) is the foundational problem-solving tool in Newton's Laws of Motion. The object of interest is isolated from its surroundings, and every force acting on it — gravity, normal reaction, tension, friction, applied force — is represented as an arrow at its point of application. NEET tests FBDs through multi-body problems, inclined plane force analysis, and Atwood machine setups where correct force identification on each body determines whether the equation of motion is set up correctly. The core procedure: isolate the body, identify all contact and field forces, draw arrows, choose axes, write F_net = ma (or ΣF = 0 for equilibrium).

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Problem-Solving MethodNewton's Laws Ch.4ΣF = ma on isolated body

Expected QuestionsQ

1–2

FBD is the foundation of most Newton's Laws problems in NEET. It rarely appears as a standalone question but is the key step in every multi-body, inclined-plane, or connected-block problem. Incorrect FBD setup is responsible for most wrong answers in the Laws of Motion chapter.

Time Required⏱

50 min

20 min to understand force identification rules (which forces act on a body vs. which forces the body exerts on others). 20 min to practice FBDs for: block on a surface, block on an incline, two-block system, Atwood machine, hanging mass. 10 min for NEET-style FBD-based numerical problems.

Difficulty⚡

Easy–Medium

Conceptually simple — draw forces on an isolated body. The difficulty lies in systematic force identification: forgetting the normal force, including internal forces, or drawing forces on the wrong body. For multi-body problems and pulley systems, consistent sign conventions across FBDs are essential.

NRI USA Curriculum GapUS

Low

AP Physics 1 explicitly teaches free body diagrams as a central skill and they appear on the AP exam as standalone questions ('draw and label a FBD for...'). The coverage is comparable. NEET-specific additions: FBDs for non-inertial frames (pseudo force inclusion), wedge-and-block systems with constraint equations, and Atwood machines with multiple pulleys.

0Subtopics

4+Practice Questions

4Free Downloads

50 minPrep Time

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NEET Weightage — Free Body Diagram

Newton's Laws of Motion (Chapter 4)

NEET Year	Questions from this Topic	Bar	Marks
2024	1	1 Q	4
2023	1	1 Q	4
2022	0	0 Q	0
2021	1	1 Q	4
2020	0	0 Q	0
2019	1	1 Q	4
6-Year Total (2019–2024)	2–4		8–16

Definition: In a free body diagram, the object of interest is isolated from its surroundings and the interactions between the object and the surroundings are represented in terms of forces. Choose the axes and write equation of motion.
FBD procedure: (1) Identify the body of interest. (2) Isolate it — mentally remove all contacts and replace with force arrows. (3) Draw all external forces: gravity (mg downward), normal reaction (perpendicular to surface), tension along string, friction parallel to surface. (4) Choose a coordinate system. (5) Write ΣF = ma along each axis.
Critical rule: Include only forces ON the body. Newton's third law pairs (the forces the body exerts on others) must NOT appear in the FBD of the body. Including reaction forces on the same body is the most common FBD error.

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0.7

Avg Questions / Year

🎯

Total Marks (6 yrs)

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Indirect

Pattern

⚠️

Easy–Medium

Difficulty

How to Prepare Free Body Diagram for NEET

Master the force identification hierarchy For any body in NEET Physics, systematically check for: (1) Gravitational force — always present, mg downward through the center. (2) Normal force — present at every solid surface contact, perpendicular to the surface and away from it. (3) Tension — along the string, toward the pulley or connecting point. (4) Friction — at surface contacts where relative motion (or tendency) exists, parallel to surface and opposite to motion (kinetic) or tendency (static). (5) Applied external force — direction and magnitude given in the problem. (6) Buoyancy — only in fluid problems. Never skip step (1) and step (2). Normal force direction is NOT always vertical — it is always perpendicular to the contact surface.

Practice FBDs for the five canonical NEET configurations Five configurations appear in virtually every Newton's Laws NEET question: A) Block on a horizontal surface with applied force at an angle. B) Block on a smooth or rough inclined plane. C) Two blocks in contact on a surface. D) Two blocks connected by a string (horizontal). E) Atwood machine (masses over a pulley). For each, draw the FBD of EVERY body separately. Write equations: ΣF_x = ma_x, ΣF_y = ma_y. Solve for the unknown. Verify: acceleration should be in the direction of net unbalanced force. If all forces balance, a = 0 (equilibrium).

Sign convention discipline across multiple bodies For multi-body problems: choose a consistent positive direction for the entire system (usually the direction of acceleration). For a two-block system pulled by force F, if both blocks accelerate rightward at 'a', write F − T = m₁a for block 1 and T = m₂a for block 2. Tension T appears as a forward force on the back block and a backward force on the front block — consistent with Newton's third law. Never mix sign conventions within the same problem. For Atwood machines: one block's upward direction = positive for that body, and the other block's downward direction = positive for that body, so both equations have the same 'a'.

Study Materials — Free Body Diagram

PDF · Cheat Sheet · MCQ Set · PYQ

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Full Notes

FBD definition and procedure. Force identification hierarchy. Worked FBDs for: block on surface, inclined plane, two blocks in contact, blocks connected by string, Atwood machine. Non-inertial frame pseudo force addition to FBD.

Single topic3 pagesConceptual + Procedure

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Formula Sheet

FBD checklist: gravity (mg↓), normal (⊥ surface), tension (toward anchor), friction (opposing motion), applied force. Key equations: ΣF_x = ma_x; ΣF_y = ma_y; ΣF = 0 (equilibrium).

Force checklist1 pageQuick reference

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MCQ Practice

15 questions: FBD identification for various configurations, force direction MCQs, multi-body FBD force calculation, assertion-reason on Newton's third law pairs, FBD in non-inertial frames.

15 MCQsAll configurationsSolved

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PYQ

Year-tagged NEET questions requiring free body diagram analysis in Newton's Laws application problems.

6+ year-tagged Qs2015–2024Solved

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2-Column Table

Column A	Column B

Rapid Revision — Free Body Diagram

Concept → Trap → Example

1) Definition and Core Procedure

Core

FBD: The object of interest is isolated from its surroundings and the interactions between the object and the surroundings are represented in terms of forces. Choose the axes and write equation of motion. Five-step procedure: (1) Choose the body. (2) Isolate it. (3) Draw ALL external forces on it (not the forces it exerts on others). (4) Set up coordinate axes (usually along and perpendicular to acceleration direction). (5) Write F_net = ma for each axis.

Only include forces ACTING ON the body. The force the body exerts on another object (Newton's third law pair) must appear in that other object's FBD, not in this one. Including both action and reaction on the same body is the single most common FBD error in NEET problems.
Normal force is ALWAYS perpendicular to the contact surface. On a horizontal surface, it's vertical. On an inclined plane at angle θ, it acts at angle θ from the vertical (i.e., perpendicular to the slope). On a wedge, the normal from the wedge on the block is perpendicular to the wedge surface.
NEET application: A block on a horizontal surface with an applied force F at angle θ upward. FBD of block: (1) mg downward. (2) Normal N upward (perpendicular to surface = vertical here). (3) Applied force F at angle θ (components: F cosθ horizontal, F sinθ upward). (4) Friction f horizontal (opposing motion). Equation along vertical: N + F sinθ − mg = 0 → N = mg − F sinθ. Along horizontal: F cosθ − f = ma.

Example (NEET-style)A 10 kg block rests on a frictionless horizontal surface. A 50 N force is applied at 30° above horizontal. Draw FBD and find acceleration. FBD: mg = 100 N downward, N upward, F = 50 N at 30°. Vertical: N + 50 sin30° = 100 → N + 25 = 100 → N = 75 N. Horizontal: F cos30° = ma → 50 × (√3/2) = 10a → 43.3 = 10a → a = 4.33 m/s². The normal force is reduced (75 N) because the upward component of the applied force partly supports the weight. This is why a pull at an angle is more efficient than a horizontal push — it reduces N and hence friction.

2) FBD for Multi-Body and Pulley Systems

High Priority

For multi-body problems: draw a separate FBD for EACH body. Connect the bodies through Newton's laws: the tension in a string is equal and opposite on the two bodies it connects. Acceleration of bodies connected by an inextensible string over a frictionless pulley is the same in magnitude (constraint equation). For Atwood machine: draw FBD for m₁ (lighter, going up): T − m₁g = m₁a. FBD for m₂ (heavier, going down): m₂g − T = m₂a. Solve simultaneously.

For two blocks A and B on a surface connected by a string, pulled by force F on B: FBD of B: F (forward) − T (backward, string pulls B) = m_B × a. FBD of A: T (forward, string pulls A) = m_A × a. Solving: a = F/(m_A + m_B); T = m_A × F/(m_A + m_B). The string tension is always LESS than the applied force when masses are positive.
For three blocks in contact (no string): the mutual contact force between blocks 1 and 2 equals m_contact × a = (m₂ + m₃) × F/(m₁ + m₂ + m₃). The contact force between the front block and the middle block equals m₃ × F/(m₁ + m₂ + m₃). These contact forces must appear in BOTH FBDs (as an equal and opposite pair).
NEET trap for Atwood machine: when the pulley is inside a lift accelerating at 'a', replace g with g_eff = g ± a in all Atwood machine formulas. This is because the FBD of each mass in the lift frame includes a pseudo force if analyzed in the non-inertial frame, or the weight term changes in the inertial frame analysis.

Example (NEET-style)Atwood machine: m₁ = 3 kg on left, m₂ = 5 kg on right, over a massless pulley. Find acceleration and tension (g = 10 m/s²). FBD m₁: T (up) − m₁g (down) = m₁a (up) → T − 30 = 3a. FBD m₂: m₂g (down) − T (up) = m₂a (down) → 50 − T = 5a. Adding: 50 − 30 = 8a → 20 = 8a → a = 2.5 m/s². T = 30 + 3×2.5 = 30 + 7.5 = 37.5 N. Verify: m₂ side check: 50 − 37.5 = 5×2.5 = 12.5 ✓. The key: both equations use the SAME 'a' because the string is inextensible. The heavier mass accelerates down and the lighter mass accelerates up, both at 2.5 m/s².

3) FBD on Inclined Plane and Non-Inertial Frames

Application

On an inclined plane at angle θ: align ONE axis along the incline (positive direction = down the incline or up, depending on motion) and the other perpendicular to it. Forces: mg along (mg sinθ down the slope) and perpendicular (mg cosθ into surface). Normal N = mg cosθ. Acceleration along slope = g sinθ (frictionless). For non-inertial frame (lift or accelerating surface): add a pseudo force = ma_frame opposite to the frame's acceleration direction.

Inclined plane FBD axis choice: always resolve gravity into components along and perpendicular to the incline. Along incline: mg sinθ (down the slope). Perpendicular to incline: mg cosθ (into the slope). Normal force N = mg cosθ (perpendicular equilibrium). Net force along incline = mg sinθ (frictionless) → a = g sinθ.
Block on an accelerating inclined plane (incline has horizontal acceleration b): in the lab frame, draw FBD with N (perpendicular to incline), mg (downward). The block's actual acceleration has components both along and perpendicular to the incline. Along incline: m × a_along = mg sinθ − mb cosθ → a_along = g sinθ − b cosθ. Condition for block at rest on incline: a_along = 0 → b = g tanθ.
Non-inertial frame (accelerating lift): draw FBD of block inside lift. If the lift accelerates upward at 'b', add pseudo force mb downward on the block (opposite to lift's acceleration). This transforms the non-inertial problem into an effective-g problem where g_eff = g + b. For an inclined plane inside such a lift: effective component along incline = (g + b) sinθ.

Example (NEET-style)A 5 kg block slides on a smooth incline of 30°. Draw FBD and find acceleration. FBD: weight mg = 50 N vertically downward. Normal N perpendicular to incline surface (outward). Resolve gravity: along incline (downward) = mg sin30° = 50 × 0.5 = 25 N. Perpendicular to incline (into surface) = mg cos30° = 50 × (√3/2) = 43.3 N. Perpendicular equation: N = mg cos30° = 43.3 N. Along incline: ma = mg sin30° → a = g sin30° = 10 × 0.5 = 5 m/s². If the incline is now accelerated horizontally at b = 5 m/s² (tilted away): a_along = g sinθ − b cosθ = 5 − 5 × (√3/2) = 5 − 4.33 = 0.67 m/s² (still slides, but slower than free incline).

US Curriculum Gaps — Free Body Diagram

Topics in this section are tested in NEET but organised differently in standard US physics courses.

FBD in Non-Inertial Frames with Pseudo Force (AP Physics C Gap)

AP Physics 1 covers FBDs in inertial frames. NEET regularly requires FBDs drawn in non-inertial frames (accelerating lifts, accelerating inclined planes, rotating frames) where a pseudo force (= mass × frame acceleration, opposite to frame's acceleration) must be included. AP Physics C: Mechanics covers non-inertial frames briefly, but AP Physics 1 does not. In NEET, questions like 'a block is on an incline inside a lift accelerating upward at a' require the student to either shift to the lab frame or add a pseudo force in the lift frame — a technique not systematically taught in AP Physics 1.

NEET: block inside lift accelerating upward → pseudo force mb downward → effective g_eff = g + b
NEET: block on accelerating incline → a_along = g sinθ − b cosθ (lab frame)
AP Physics 1: FBDs in inertial frames only; non-inertial frames introduced only in AP Physics C

Systematic Force Identification for Multi-Body Contact Problems (AP Gap)

NEET problems routinely involve three or more bodies in contact (blocks stacked, blocks in a line with applied forces) requiring a separate FBD for each body and tracking the contact forces. AP Physics 1 tests this concept but NEET problems have higher complexity: three blocks in a row (A-B-C), force on A, asked for contact force between B and C. Students must write three FBDs and three equations of motion simultaneously. The systematic tabular approach to FBDs for three bodies in contact is emphasized in NEET preparation but not at the same depth in AP Physics 1.

NEET: three blocks A-B-C on surface, force F on A, find contact force between B and C
NEET: contact force = (mass of C) × a = m_C × F/(m_A + m_B + m_C)
AP Physics 1: two-body systems emphasized; three-body FBDs less common

NEET-Style Practice Questions — Free Body Diagram

4 Questions

1A 10 kg block is being pulled along a frictionless horizontal surface by a force F = 40 N applied at 30° above horizontal. The normal reaction of the surface on the block is (g = 10 m/s²):FBD Application

100 N

80 N

60 N

120 N

Draw FBD of block: (1) mg = 100 N downward. (2) N upward (normal from surface). (3) F = 40 N at 30° — components: F sinθ = 40 × sin30° = 40 × 0.5 = 20 N upward, F cosθ = 40 × cos30° horizontal. Vertical equilibrium (no vertical acceleration): N + F sinθ − mg = 0 → N + 20 − 100 = 0 → N = 80 N. The applied force's upward component reduces the normal reaction from 100 N to 80 N. Trap: choosing N = mg = 100 N (forgetting the upward component of the applied force). N = mg only when the applied force is purely horizontal. Answer: 80 N.

2In an Atwood machine, masses 4 kg and 6 kg are connected by a light, inextensible string over a frictionless pulley. The acceleration of the system is (g = 10 m/s²):FBD — Atwood Machine

1 m/s²

2 m/s²

4 m/s²

g/5

FBD of m₁ = 4 kg (lighter, goes up): T − m₁g = m₁a → T − 40 = 4a ...(1). FBD of m₂ = 6 kg (heavier, goes down): m₂g − T = m₂a → 60 − T = 6a ...(2). Add (1) + (2): 60 − 40 = 10a → 20 = 10a → a = 2 m/s². Using the formula directly: a = (m₂ − m₁)g / (m₁ + m₂) = (6−4)×10 / (6+4) = 20/10 = 2 m/s². Answer: 2 m/s². Tension: T = m₁(g+a) = 4×(10+2) = 48 N. Verify: m₂: m₂(g−a) = 6×(10−2) = 48 N ✓.

3Three blocks A (2 kg), B (3 kg), and C (5 kg) are placed in contact on a smooth horizontal surface. A horizontal force F = 20 N is applied on A. What is the contact force between B and C?Multi-body FBD

10 N

14 N

6 N

20 N

System acceleration: a = F/(m_A + m_B + m_C) = 20/(2+3+5) = 20/10 = 2 m/s². FBD of C alone: contact force from B on C = m_C × a = 5 × 2 = 10 N. Alternatively, using the formula for blocks in contact: contact force between B and C = (m_C × F)/(m_A + m_B + m_C) = (5 × 20)/10 = 10 N. The contact force is the only force on block C (frictionless surface), so it equals m_C × a. Trap: using total force F = 20 N (that only applies to A). Answer: 10 N.

4Assertion: In a free body diagram of a block on a surface, the weight mg and the normal force N are Newton's third law action-reaction pairs. Reason: N and mg are equal and opposite, so they must form an action-reaction pair.Assertion-Reason

Both A and R are true and R is the correct explanation of A

Both A and R are true but R does NOT correctly explain A

A is false and R is true

Both A and R are false

Assertion: FALSE. N and mg are NOT Newton's third law action-reaction pairs. Newton's third law pairs act on DIFFERENT bodies. The reaction pair of 'Earth pulls block downward (mg)' is 'block pulls Earth upward (mg)'. The reaction pair of 'surface pushes block upward (N)' is 'block pushes surface downward (N)'. So N and mg appear in the same FBD (forces ON the block) but are from different interactions — one from gravity and one from the contact. They happen to be equal in magnitude only when acceleration is zero. Reason: FALSE. N = mg only in the static case; if the surface is accelerating, N ≠ mg. Two forces being equal and opposite DOES NOT make them an action-reaction pair — they must be from the same interaction (same type of force on two different bodies). Answer: D.

Practice Problems — Free Body Diagram

Click "Reveal Answer" after attempting

1A block of mass 8 kg is on a rough inclined plane of angle 30°. The coefficient of kinetic friction is μ_k = 0.3. (a) Draw the FBD of the block. (b) Find the normal force. (c) Find the net force and acceleration along the incline (downward). (g = 10 m/s²)

N = 69.3 N; a = 2.4 m/s² down the incline

N = 80 N; a = 3 m/s² down the incline

N = 69.3 N; a = 5 m/s² down the incline

N = 40 N; a = 2 m/s² down the incline

👁 Reveal Answer

FBD of block: (1) mg = 80 N vertically downward. (2) Normal N perpendicular to incline surface (outward). (3) Friction f up the incline (opposing downward tendency). Components perpendicular to incline: N − mg cos30° = 0 → N = 80 × (√3/2) = 80 × 0.866 = 69.3 N. Friction force: f = μ_k × N = 0.3 × 69.3 = 20.8 N (up the incline). Net force along incline (downward positive): F_net = mg sin30° − f = 80 × 0.5 − 20.8 = 40 − 20.8 = 19.2 N. Acceleration: a = F_net / m = 19.2 / 8 = 2.4 m/s² down the incline. The FBD is the essential first step: resolving mg into perpendicular (→ N) and parallel (→ net driving force) components. Without the FBD, students forget the friction term or incorrectly use mg instead of mg cosθ for the normal force.

2Two blocks, m₁ = 5 kg and m₂ = 3 kg, are connected by a light string and placed on a smooth horizontal surface. A force F = 16 N pulls m₂ along the surface. (a) Find the acceleration of the system. (b) Find the tension T in the string connecting the blocks. (c) If m₁ and m₂ are swapped (F now pulls m₁), how does T change?

(a) a = 2 m/s²; (b) T = 10 N; (c) T = 6 N

(a) a = 2 m/s²; (b) T = 6 N; (c) T = 10 N

(a) a = 4 m/s²; (b) T = 20 N; (c) T = 12 N

(a) a = 2 m/s²; (b) T = 8 N; (c) T = 8 N

👁 Reveal Answer

(a) System acceleration: a = F/(m₁+m₂) = 16/8 = 2 m/s². (b) FBD of m₁ (force F not on this block): T = m₁ × a = 5 × 2 = 10 N. OR equivalently, FBD of m₂: F − T = m₂ × a → 16 − T = 3 × 2 = 6 → T = 10 N. ✓ (c) Swap: F pulls m₁ = 5 kg (now in front). FBD of m₂ (now at back): T = m₂ × a = 3 × 2 = 6 N. The tension in the string depends on which mass is being dragged by the string vs. which is being pushed by the applied force. Tension = (mass of block NOT directly pulled) × a. Pulling from the m₁ side: T = m₂ × a = 6 N. Pulling from the m₂ side: T = m₁ × a = 10 N. Tension is higher when the heavier block is 'behind' and must be pulled by the string.

3A light string passes over a frictionless pulley fixed to the ceiling. On one side hangs mass m₁ = 4 kg and on the other m₂ = 6 kg. Additionally, a constant force F = 10 N is pulling m₁ downward. Find the acceleration and the tension in the string (g = 10 m/s²).

a = 3 m/s²; T = 52 N

a = 2 m/s²; T = 48 N

a = 3 m/s²; T = 46 N

a = 4 m/s²; T = 40 N

👁 Reveal Answer

FBD of m₁: m₁g + F (both downward) − T (upward) = m₁ × a (downward). So: 40 + 10 − T = 4a → 50 − T = 4a ...(1). FBD of m₂: m₂g − T (downward net) = m₂ × a (downward acceleration of m₂ side). Wait — if extra force pulls m₁ down, m₁ goes down (heavier effective side). m₁ goes down, m₂ goes up: T − m₂g = m₂a → T − 60 = 6a ...(2). Wait, let's reconsider direction: if m₁ (4 kg + extra 10 N = effectively 5 kg-equivalent) goes down, m₂ (6 kg × 10 = 60 N) goes up? Net downward pull on m₁ side = 40 + 10 = 50 N. Net downward pull on m₂ side = 60 N. m₂ side is heavier → m₂ goes DOWN, m₁ goes UP. FBD m₂ (going down): m₂g − T = m₂a → 60 − T = 6a ...(1). FBD m₁ (going up): T − m₁g − F = m₁a → T − 40 − 10 = 4a → T − 50 = 4a ...(2). From (1): T = 60 − 6a. Sub in (2): 60 − 6a − 50 = 4a → 10 = 10a → a = 1 m/s². T = 60 − 6 = 54 N. Verify (2): 54 − 50 = 4 × 1 = 4 ✓. Answer: a = 1 m/s², T = 54 N. (Note: closest to option format — the key process is correctly identifying which side accelerates downward.)

4A block of mass 3 kg is resting on a 2 kg wedge that sits on a frictionless floor. The wedge angle is 45°. The wedge is pushed to the right with force F so that the block does NOT slide on the wedge (they move together). What force F is required? (g = 10 m/s², tan45° = 1)

F = 15 N

F = 30 N

F = 50 N

F = 20 N

👁 Reveal Answer

FBD of block (mass 3 kg) — it moves with acceleration a to the right. Forces on block: mg = 30 N down, N from wedge (perpendicular to wedge surface, at 45° to vertical → components: N sin45° horizontal right, N cos45° upward). For block to accelerate right without sliding: Vertical equilibrium: N cos45° = mg → N × (1/√2) = 30 → N = 30√2 N. Horizontal (provides acceleration): N sin45° = 3a → 30√2 × (1/√2) = 3a → 30 = 3a → a = 10 m/s². FBD of entire system (block + wedge, total mass = 5 kg): F = (m_block + m_wedge) × a = 5 × 10 = 50 N. Answer: F = 50 N. This problem requires FBDs of the block alone (to find the constraint that gives 'a') and then of the total system (to find F). Drawing the FBD of the block correctly — with the normal from the wedge at 45° — is the critical step.

Physics — Newton's Laws of Motion Revision Checklist

Check off chapters as you revise

Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

Free Body Diagram procedure mastery: From a problem description, draw the FBD of EACH body: isolate the body, add gravity (mg down), add normal force (perpendicular to surface), add tension (along string toward anchor), add friction (parallel, opposing motion), add any applied force. Write ΣF_x = ma_x and ΣF_y = ma_y. Verify: does the net force direction match the acceleration direction? Check for included reaction forces that should NOT be in the FBD.

Newton's 3rd law pair identification: For any force in a FBD, identify its Newton's third law pair. 'Earth pulls block with mg' → pair is 'block pulls Earth with mg upward'. 'Surface pushes block with N' → pair is 'block pushes surface with N'. The pair NEVER appears in the same FBD as the original force. N and mg in the same FBD are NOT a Newton's third law pair — they are from different interactions.

Multi-body FBD practice (Atwood + blocks in contact): Practice writing separate FBDs for: (1) Two-block Atwood machine — T and weight on each block, same 'a' constraint. (2) Two blocks on surface connected by string and pulled — tension appears as backward force on pulled block and forward force on following block. (3) Three blocks in contact — contact forces equal and opposite between adjacent blocks. Solve for 'a' from the system equation, then substitute back to find internal forces.

FAQ — Free Body Diagram

Notes · Downloads · Revision · Important Questions

What is a Free Body Diagram?

A Free Body Diagram (FBD) is a representation in which the object of interest is isolated from its surroundings, and all forces acting on it from the surroundings are drawn as arrows at their points of application. The FBD allows us to apply Newton's second law (ΣF = ma) systematically by accounting for every force on the body without confusion from forces on other bodies.

What forces should I include in a FBD?

Include ALL external forces ACTING ON the body: (1) Gravitational force (mg, vertically downward through the center of mass). (2) Normal force (perpendicular to and away from every solid contact surface). (3) Tension (along the string, toward the pulley/anchor). (4) Friction (parallel to the surface, opposing relative motion or tendency). (5) Applied external forces (direction and magnitude as given). Do NOT include internal forces between parts of the body, and do NOT include forces that the body exerts on other objects.

Why should Newton's third law pair forces not both appear in the same FBD?

Newton's third law pairs act on DIFFERENT bodies. The FBD of a single body should include only the forces acting ON that body. The force the body exerts on another object is a force on that other object — it belongs in that object's FBD, not the first body's. If you included both the action and reaction on the same body, the net force would always appear to be zero, which would incorrectly imply all bodies are always in equilibrium. The 'reaction' to a force on body A always acts on body B (the source of the force).

How is normal force direction determined?

Normal force is always perpendicular to the contact surface and directed away from the surface into the body. On a horizontal surface, it's vertical (upward). On an incline at angle θ, it's perpendicular to the incline surface (at angle θ from the vertical). On a vertical wall, it's horizontal. On a curved surface, it points toward the center of curvature. NEET trap: students often draw N straight up even on an incline — this is wrong. N must be perpendicular to the surface of contact, not perpendicular to the ground.

How do I handle FBDs for connected-body problems?

Draw a SEPARATE FBD for each body. For bodies connected by an inextensible string: (1) Tension T appears in both FBDs — as a forward force on the trailing body and a backward force on the leading body (or upward on the lighter mass and downward on the heavier in a pulley). (2) Acceleration magnitude 'a' is the same for all connected bodies (constraint of inextensible string). (3) Write ΣF = ma for each body. (4) Solve the simultaneous equations. The internal force (tension or contact force) can only be found by analyzing individual body FBDs, not the system as a whole.

How is FBD used for bodies in equilibrium?

If a body is in equilibrium (a = 0), draw the FBD and set ΣF = 0. This means all force components must balance: ΣF_x = 0 and ΣF_y = 0 (and ΣF_z = 0 in 3D). For a hanging mass at rest: T = mg. For a block on an incline at rest (with friction): friction up the slope = mg sinθ; normal force = mg cosθ. The FBD makes clear which forces must balance and what the unknown force magnitudes must be.

What is the correct approach to FBD on an inclined plane?

Step 1: Draw the block on the incline and the incline surface. Step 2: Draw mg vertically downward from the block's center. Step 3: Draw N perpendicular to the incline surface (not vertical). Step 4: Draw friction parallel to the incline (up the slope if block slides/tends to slide down). Step 5: Rotate the coordinate axes so x is along the incline and y is perpendicular to it. Step 6: Resolve mg into components: mg sin θ (along incline, downward direction) and mg cos θ (perpendicular to incline, into surface). Step 7: Write ΣF_y = 0 → N = mg cos θ. ΣF_x = ma → mg sin θ − friction = ma.

When is a pseudo force added to a FBD?

A pseudo force (fictitious force) is added to the FBD when you choose to analyze the problem in a non-inertial (accelerating) reference frame. Its magnitude = mass × frame acceleration; its direction is OPPOSITE to the frame's acceleration. Example: analyzing a block inside a lift accelerating upward at 'a' from the lift's frame — add pseudo force ma downward on the block. This converts the non-inertial frame problem into an equilibrium or constant-acceleration problem in the frame. Result: effective g = g + a (lift going up) or g − a (lift going down). In NEET, working in the inertial (lab) frame is usually simpler and avoids pseudo force errors.

What are the most common FBD errors in NEET?

Top 5 FBD errors: (1) Drawing normal force vertically up even on an inclined plane (should be perpendicular to inclined surface). (2) Forgetting the normal force on a horizontal surface when an applied force has an upward or downward component (N changes!). (3) Including Newton's third law reaction forces in the same FBD as the original force. (4) Drawing friction in the wrong direction (friction always opposes relative motion or tendency — if block tends to slide right, friction acts LEFT). (5) In Atwood machine, using different 'a' for the two masses — they must have the same magnitude 'a' (inextensible string constraint).

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Free Body Diagram

Free Body Diagram – Complete Notes, Revision, Important Questions & Downloads

1) Definition and Core Procedure

2) FBD for Multi-Body and Pulley Systems

3) FBD on Inclined Plane and Non-Inertial Frames

FBD in Non-Inertial Frames with Pseudo Force (AP Physics C Gap)

Systematic Force Identification for Multi-Body Contact Problems (AP Gap)

Physics — Newton's Laws of Motion Revision Checklist

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Instantaneous velocity of the rocket

Rocket propulsion

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Instantaneous velocity of the rocket

Rocket propulsion

NEET > Physics > Laws of Motion Chapters

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Free Body Diagram

Overview content

Free Body Diagram – Complete Notes, Revision, Important Questions & Downloads

1) Definition and Core Procedure

2) FBD for Multi-Body and Pulley Systems

3) FBD on Inclined Plane and Non-Inertial Frames

FBD in Non-Inertial Frames with Pseudo Force (AP Physics C Gap)

Systematic Force Identification for Multi-Body Contact Problems (AP Gap)

Physics — Newton's Laws of Motion Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Instantaneous velocity of the rocket

Rocket propulsion

Subtopics

Instantaneous velocity of the rocket

Rocket propulsion

NEET > Physics > Laws of Motion Chapters

Comments

Leave a comment

Comments (0)