How to Draw Free Body Diagrams — A Step-by-Step Method

A free body diagram (FBD) is the single most useful tool in mechanics. Every solved problem in Newton’s laws, statics, dynamics, or friction starts with one. If your FBD is wrong, the physics is wrong — there’s no recovering from a missing force or a misdirected arrow.

This guide builds the skill from scratch, starting with why FBDs work and ending with multi-body systems that appear in JEE Main and CBSE Class 11.

What Is a Free Body Diagram?

An FBD is a simplified sketch of a single object showing all the forces acting on that object. We draw the object in isolation — “free” from its surroundings — and represent every interaction with the environment as a force vector (arrow).

The key word is on. We draw forces acting on our chosen body, not forces that body exerts on others. Newton’s third law creates pairs of forces, but each force in the pair acts on a different body. In an FBD, we choose one body and show only the forces on that body.

Why FBDs Work

Newton’s second law says $\vec{F}_{net} = m\vec{a}$ for the body. But $\vec{F}_{net}$ is the vector sum of all real forces acting on the body. You cannot apply $F = ma$ without knowing all the forces.

An FBD forces you to inventory every force before writing any equation. It’s a checklist that prevents you from forgetting gravity, normal force, tension, friction, or any applied force.

Key Terms

Weight (W): The gravitational force on the object, $W = mg$ . Acts vertically downward, from the object’s centre of mass. Always present (on Earth).

Normal force (N): The contact force perpendicular to a surface. Acts away from the surface. Adjusts its value to prevent penetration — it’s a reaction, not a fixed force.

Tension (T): The pulling force in a string, rope, or rod. Acts along the string, towards the string (always pull, never push for a flexible string).

Friction (f): The contact force parallel to a surface. Opposes relative motion (kinetic friction) or impending motion (static friction). Direction must be determined from context.

Applied force (F): Any external push or pull. Acts in the direction specified.

Methods: How to Draw an FBD

The 5-Step Method

Step 1 — Choose the body. Decide which object you’re analysing. Draw it as a simple box, dot, or rough outline. Label it.

Step 2 — Identify all contacts. List every surface, string, or agent in contact with the body. Each contact can exert a force.

Step 3 — Add non-contact forces. Gravity is always present. Add $mg$ pointing straight down from the centre.

Step 4 — Add contact forces. For each contact: a surface exerts a normal force (perpendicular) and possibly friction (parallel). A string exerts tension (along the string, towards the string’s far end).

Step 5 — Check completeness. Have you accounted for all contacts? Does each force have a direction and a label? Are you drawing forces ON this body only (not forces FROM this body onto others)?

Solved Examples

Example 1 — Block on a flat surface (CBSE Class 9 level)

A 5 kg block rests on a horizontal table. Draw the FBD.

Contacts: Table (below the block). Forces:

Weight $W = mg = 5 \times 10 = 50$ N, downward
Normal force $N$ , upward (table pushes up)

Since the block is at rest, $N = W = 50$ N.

        ↑ N
   [block]
        ↓ W = 50 N

No friction (no horizontal forces, so no horizontal friction needed).

Example 2 — Block on an inclined plane (CBSE Class 11 / JEE Main)

A block of mass $m$ sits on a smooth inclined plane at angle $\theta$ . Draw the FBD and find the normal force.

Contacts: The inclined surface. Forces:

Weight $mg$ vertically downward
Normal force $N$ perpendicular to the incline surface (at angle $\theta$ from vertical)

Resolve weight into components:

Perpendicular to incline: $mg\cos\theta$
Along incline (down the slope): $mg\sin\theta$

Since there’s no acceleration perpendicular to the surface:

N = mg\cos\theta

The block accelerates down the slope: $a = g\sin\theta$ (smooth surface, so no friction).

         N (perpendicular to surface, outward)
         ↑⁻
    [block]
   /θ  ↙ mg (vertical, down)
  /____/

Example 3 — Two blocks on a surface (JEE Main level)

Two blocks A (2 kg) and B (3 kg) are on a frictionless horizontal surface. A horizontal force $F = 10$ N is applied to block A, which pushes block B. Find acceleration and contact force.

System approach first: Treat A + B as one system.

F = (m_A + m_B) \times a \implies 10 = 5a \implies a = 2 \text{ m/s}^2

FBD of block B alone: The only horizontal force on B is the contact force $N_{AB}$ from A pushing B.

N_{AB} = m_B \times a = 3 \times 2 = 6 \text{ N}

FBD of block A alone: Force $F = 10$ N forward, contact force $N_{BA} = 6$ N backward (B pushes back on A by Newton’s 3rd law).

F_{net\,on\,A} = 10 - 6 = 4 \text{ N}

a_A = 4/2 = 2 \text{ m/s}^2 \checkmark

Example 4 — Block on a surface with friction (CBSE Class 11)

A 10 kg block is pushed by a horizontal force $F = 60$ N on a surface with $\mu = 0.4$ , $g = 10$ m/s². Find acceleration.

FBD of block:

Weight: $W = 100$ N downward
Normal: $N = 100$ N upward (horizontal surface, no vertical acceleration)
Applied force: $F = 60$ N forward
Kinetic friction: $f_k = \mu N = 0.4 \times 100 = 40$ N backward

F_{net} = 60 - 40 = 20 \text{ N}

a = F_{net}/m = 20/10 = 2 \text{ m/s}^2

Example 5 — Atwood’s machine (JEE Main)

Two masses $m_1 = 4$ kg and $m_2 = 2$ kg are connected by a string over a frictionless pulley. Find the acceleration.

FBD of $m_1$ (heavier, moves down):

Weight $m_1 g = 40$ N downward
Tension $T$ upward
Net force downward: $m_1 g - T$

FBD of $m_2$ (lighter, moves up):

Weight $m_2 g = 20$ N downward
Tension $T$ upward
Net force upward: $T - m_2 g$

Since they’re connected: same $|a|$ and same $T$ .

m_1 g - T = m_1 a \quad \text{...(1)}

T - m_2 g = m_2 a \quad \text{...(2)}

Adding: $(m_1 - m_2)g = (m_1 + m_2)a$

a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(4-2) \times 10}{4+2} = \frac{20}{6} \approx 3.3 \text{ m/s}^2

Exam-Specific Tips

CBSE (Class 9–11): FBDs appear in 3–5 mark questions. The examiner awards marks for the diagram separately from the equation. Draw the diagram clearly, label all forces with symbols ( $N$ , $T$ , $f$ , $W$ ), and show direction with arrows. Even if your equation has a calculation error, a correct FBD earns 1–2 marks.

JEE Main: FBDs are usually implicit — you need to draw them yourself before writing equations. Questions involving blocks on inclines, Atwood machines, or systems with friction almost always require multi-body FBDs. The key skill is correctly identifying which forces act on which body, especially the contact (normal) force between two touching bodies.

Common Mistakes to Avoid

Mistake 1 — Drawing Newton’s 3rd law pairs on the same FBD. If block A pushes block B with force $F$ , then B pushes A with $-F$ . In A’s FBD, show $-F$ on A. In B’s FBD, show $+F$ on B. Never show both on the same diagram — that’s a system diagram, not an FBD.

Mistake 2 — Forgetting normal force on an incline. Students sometimes draw only $mg$ and friction on an incline and forget the normal force. The incline does push on the block — perpendicular to the surface. Without $N$ , you cannot calculate friction ( $f = \mu N$ ).

Mistake 3 — Wrong direction for friction. Friction opposes relative motion (or impending motion). If a block is about to slide down an incline, friction acts up the incline. If a block is being pushed to the right, kinetic friction acts to the left. Think about which way the block is moving (or trying to move) before drawing friction.

Mistake 4 — Mixing up the body. In a multi-body problem, students sometimes put the tension in a string on the wrong body’s FBD, or add the applied force to the wrong block. For each FBD, ask: “Is this force acting ON this specific body?”

Mistake 5 — Treating the normal force as always equal to $mg$ . $N = mg$ only for a horizontal surface with no vertical acceleration and no other vertical forces. On an incline, $N = mg\cos\theta$ . If someone pushes down on a block, $N = mg + F_{push}$ . Normal force is a reactive force — it equals whatever it needs to be to prevent penetration.

Practice Questions

Q1. A 3 kg block hangs from a spring scale attached to the ceiling. Draw the FBD of the block and find the spring scale reading.

Forces on block: Weight $W = 30$ N downward; Tension $T$ from scale upward. Equilibrium: $T = W = 30$ N. Scale reads 30 N (= 3 kg).

Q2. A 4 kg block is on a 30° incline with $\mu = 0.3$ . Find the acceleration if the block slides down. ( $g = 10$ m/s²)

$N = mg\cos 30° = 40 \times (\sqrt{3}/2) \approx 34.6$ N. Friction $f = \mu N = 0.3 \times 34.6 = 10.4$ N (up the incline). Component of gravity down the incline: $mg\sin 30° = 20$ N. Net force down = $20 - 10.4 = 9.6$ N. Acceleration = $9.6/4 = 2.4$ m/s².

Q3. Blocks of 5 kg and 3 kg are connected by a light string and placed on a frictionless horizontal table. A 16 N force pulls the 5 kg block. Find tension in the string.

System acceleration: $a = 16/8 = 2$ m/s². FBD of 3 kg block: only tension $T$ acts horizontally. $T = 3 \times 2 = 6$ N.

Q4. A 2 kg block sits on a 5 kg block which sits on a frictionless table. A force $F = 14$ N is applied to the 5 kg block. Friction between the two blocks is $\mu = 0.2$ . Draw FBDs for both and find the acceleration of each block.

Max static friction on 2 kg block from 5 kg: $f = \mu m_{top} g = 0.2 \times 2 \times 10 = 4$ N. If they move together: $a = 14/7 = 2$ m/s². Check: force needed on 2 kg block = $2 \times 2 = 4$ N = max static friction. They just barely move together. Acceleration of both = 2 m/s². (If $F$ were larger, the top block would slide.)

FAQs

Do I always need to draw an FBD? Yes, for any problem involving force or acceleration. Even experienced physicists draw FBDs — the diagram prevents errors that no amount of experience can substitute for.

What if I don’t know the direction of friction? Assume a direction, solve, and check the sign of your answer. A negative friction force in your equations means friction acts opposite to your assumption — just flip the arrow.

Should I use a dot or a box for the object? Either works. A dot is fine for simple problems. Use a realistic shape only if the object’s orientation matters (like a ladder leaning against a wall).

Can I have two normal forces on one body? Yes. A block sitting on a floor and against a wall has a normal force from each surface, in two different directions.

Is tension the same throughout a string? Only for a massless, inextensible string over a frictionless pulley. In real strings, tension varies. In most CBSE and JEE problems, strings are assumed massless and pulleys frictionless — so tension is uniform.

What’s the difference between an FBD and a force diagram? They’re the same thing. “Free body diagram” emphasises that the body is drawn “freed” from its surroundings, with environment effects replaced by force arrows.