🏠 Home 📖 Core Concepts 📐 Formulas ⚡ FBD & Forces 🧩 Problem Types 🔗 Interlinking 📊 PYQ Analysis 🧠 Advanced ⏱ Practice 🎯 Strategy ⚡ Quick Revision

⚡ MODULE 3 — FORCE REPRESENTATION

Free Body Diagrams & Force Analysis

The single most important skill in this chapter. Master FBD and you solve 80% of Newton's Laws problems automatically.

FBD Mastery Force Resolution Multi-body Systems

The single rule that separates toppers from average students: Draw the FBD BEFORE writing any equation. Every single time. No exceptions. Students who do this consistently score 15-20% higher in mechanics.

How to Draw Free Body Diagrams

5-STEP FBD PROTOCOL

Identify the body you're analyzing
Draw the body as a point or box (isolated from surroundings)
Draw ALL forces acting ON that body (not by it)
Label each force with magnitude/symbol and direction
Choose a coordinate system and resolve forces

Types of Forces to Include:

Weight (W = mg): Always downward from center of mass
Normal Force (N): Perpendicular to contact surface, away from surface
Tension (T): Along string, away from body (strings pull)
Friction (f): Along surface, opposing relative motion or tendency
Applied Force (F): In given direction
Spring Force: Along spring, toward natural length position

Including forces that the body EXERTS (reaction forces) in the FBD of a body is the #1 FBD error. FBD shows only forces acting ON the body.

Drawing "centripetal force" as a separate arrow on FBD. Centripetal force is not a separate force — it's the net inward component of the real forces already drawn.

FBD: Block on Horizontal Surface (Rough)

N (Normal)

f (friction)

F (applied)

mg (weight)

Rough Surface (μ)

For this FBD: ΣFy = 0 → N - mg = 0 → N = mg. ΣFx = ma → F - f = ma → F - μN = ma → F - μmg = ma. Solve for a = (F - μmg)/m.

Always write ΣFx = max and ΣFy = may explicitly. JEE markers look for this structure. Missing it = losing method marks even if final answer is correct.

Atwood Machine

⚙️ Pulley (massless, frictionless)

m₁

↓ m₁g

m₂

↓ m₂g

(m₂ > m₁, so m₂ descends)

EQUATIONS (Derived from FBD)

For m₂: m₂g - T = m₂a  ...(1)
For m₁: T - m₁g = m₁a  ...(2)
 a = (m₂-m₁)g/(m₁+m₂) 
T = 2m₁m₂g/(m₁+m₂)

Atwood machine appears in NEET 2-3 times per decade. The shortcut: if m₁ = m₂, a = 0 and T = m₁g = m₂g. At the limit m₂ >> m₁, a → g (like free fall).

Lift / Elevator Problems

FBD of person: N (up) - mg (down) = ma (up).
∴ N = m(g + a). Apparent weight INCREASES.
Person feels heavier. Scale reading increases.

FBD of person: mg (down) - N (up) = ma (down).
∴ N = m(g - a). Apparent weight DECREASES.
If a = g: N = 0 → weightlessness (free fall).

a = 0. N - mg = 0. N = mg. Apparent weight = real weight. No change in scale reading. Constant velocity = no net force needed.

If body hangs from spring/string inside elevator:
Moving up with a: T = m(g+a)
Moving down with a: T = m(g-a)
Free fall: T = 0
The tension replaces normal force concept.

When lift decelerates while moving UP — is apparent weight more or less? Most students say "less" incorrectly. Decelerating while moving UP means acceleration is DOWNWARD → N = m(g-a). Weight appears LESS.

Inclined Plane Analysis

KEY EQUATIONS

Perpendicular: N = mg cosθ
Along incline (sliding down):
ma = mg sinθ - f
= mg sinθ - μN
= mg sinθ - μmg cosθ
a = g(sinθ - μcosθ)

Critical Conditions on Inclined Plane:

mg sinθ ≤ μs × mg cosθ
tanθ ≤ μs
θ ≤ angle of repose
Static friction is sufficient to hold the body.

θ > angle of repose. Friction acts up the incline.
a = g(sinθ - μk cosθ)
Use kinetic friction since body is sliding.

Both weight component (mg sinθ) AND friction (μmg cosθ) oppose the motion (both act down the incline).
Net retardation = g(sinθ + μcosθ). Body decelerates faster when moving up than accelerates when moving down.

F_min = mg sin(θ + λ) / cos(φ - θ - λ)
where λ = angle of friction = arctan(μ)
For horizontal force: F = mg(sinθ + μcosθ)/(cosθ - μsinθ)
This is a standard JEE Advanced problem type.

JEE twists this by giving a block on incline with another block connected by string over pulley at top. You need TWO FBDs — one for each block — then solve simultaneously. The key: constraint relation (both accelerations equal in magnitude).

NEET tests inclined planes in almost every paper. Focus on: (1) condition for sliding, (2) FBD with resolved forces, (3) acceleration formula. These three alone are worth 8 marks in NEET.

Multi-Body & Connected Systems

Two Blocks on Horizontal Surface

F →

m₁

m₂

F applied on m₁, connected via string to m₂

SOLUTION METHOD

Step 1: System acceleration: a = F/(m₁+m₂)

(if no friction)

Step 2: FBD of m₂ alone:

T = m₂ × a = m₂F/(m₁+m₂)

Step 3: Verify with m₁:

F - T = m₁a → F - m₂F/(m₁+m₂) = m₁F/(m₁+m₂) ✓

Writing T = F for connected bodies is wrong. The tension in string is LESS than F. If T = F, block m₁ would have zero acceleration — which contradicts the system moving. Think before you write.

Three Blocks System (JEE Level)

F →

Masses mA, mB, mC on rough surface (μ)

System a = (F - μ(mA+mB+mC)g) / (mA+mB+mC)
T₁ (string AB) = mA × a + μmAg (from A's FBD)
T₂ (string BC) = mC × a + μmCg (from C's FBD)
Verify: B's FBD → T₁ - T₂ - μmBg = mBa ✓

For n-block systems, ALWAYS use the system approach first. Find common acceleration. Then isolate the lightest/simplest block to find tension. Never try to find tensions simultaneously — it creates a system of equations that's hard to solve under pressure.

Constraint Relations:

Inextensible string over pulley: a₁ = a₂ (magnitudes equal)
Wedge + block: a_block on wedge + a_wedge = absolute acceleration of block
Spring constraint: extension = relative displacement

Force Resolution — Master Skill

Any force F at angle θ to x-axis can be resolved into:

Fx = F cosθ  (along x)
Fy = F sinθ  (along y)

Smart Coordinate Choice:

Horizontal surface: x = horizontal, y = vertical
Inclined plane: x = along incline, y = perpendicular to incline (minimizes number of resolutions)
Circular motion: radial direction + tangential direction
Rule: Choose axes so that most forces are along axes (not at angles)

If a force at angle θ to incline surface must be resolved — resolve it with respect to the incline, NOT the horizontal. Using wrong axes makes problem 3x harder.

Weight Components on Incline:

Weight mg (vertically downward) resolved on incline of angle θ:

Along incline (down): mg sinθ
Perpendicular (into surface): mg cosθ

Memory trick: sinθ ↔ component along slope (sin goes with "sliding down"). cosθ ↔ component into surface (cos goes with "compress").

Applied Force at Angle:

ΣFx = F cosθ - f = ma
ΣFy = N + F sinθ - mg = 0
∴ N = mg - F sinθ
f = μN = μ(mg - F sinθ)
Note: N is REDUCED when force is at angle above horizontal!

ΣFx = F cosθ - f = ma
ΣFy = N - mg - F sinθ = 0
∴ N = mg + F sinθ
f = μN = μ(mg + F sinθ)
N is INCREASED when force is angled downward! More friction!

Pulling at angle above horizontal is ALWAYS more efficient than pushing at angle below horizontal for moving a body on rough surface. This is a standard NEET/JEE conceptual question.

FBD Mastered? Now Apply It.

See all 6 problem types with worked solutions using the FBD method.

Next: Problem Types → Jump to Practice →