Free Body Diagrams — Mastery Guide

What a Free Body Diagram Actually Is

A free body diagram (FBD) is a drawing of a single object with every force acting on it shown as an arrow. Nothing else — no surfaces, no other objects, just the body and its forces. Students who get FBDs right rarely lose marks on mechanics. Students who don’t, struggle for years.

The reason is simple. Newton’s second law $F_{\text{net}} = ma$ is a vector equation, and vectors only make sense when we know exactly which forces act on which object. The FBD enforces that discipline. Once we have a clean FBD, the equations almost write themselves.

We’ll build the technique from scratch, work through five graded examples, and then look at the mistakes that cost students marks.

Key Terms & Definitions

Free body: the single object whose motion you’re analysing. Pick one body at a time.

Force: a push or pull on the body. Forces come from contact (normal, friction, tension) or from fields (gravity, electric, magnetic).

Internal forces: forces between parts of the chosen body. These do not appear in the FBD because they cancel internally.

External forces: forces from outside the chosen body. These are what we draw.

Action-reaction pair: by Newton’s third law, every force on body A from body B has an equal-and-opposite partner on B from A. The pair acts on different bodies, so it never appears in a single FBD.

How to Draw an FBD — Five-Step Method

Decide which object you’re analysing. Draw it as a dot or simple shape. Label it.

List every object touching the chosen body. Each contact contributes a normal force, possibly a friction force, and possibly a tension if a string or rod is involved.

Gravity always acts (drawn straight down, magnitude $mg$ ). Electric or magnetic forces if relevant.

Choose convenient axes — usually horizontal-vertical, or along-the-incline if the surface is tilted. Mark them on the diagram.

Write component equations along each axis. Solve.

Solved Examples

Example 1 (Easy, CBSE) — Block on a Horizontal Surface

A $5\,\text{kg}$ block sits on a smooth horizontal floor. A horizontal force of $20\,\text{N}$ pushes it. Find the acceleration. Take $g = 10\,\text{m/s}^2$ .

The FBD has three forces on the block: weight $mg = 50\,\text{N}$ down, normal $N$ up, applied force $20\,\text{N}$ horizontal.

Vertical: $N - mg = 0 \implies N = 50\,\text{N}$ . Horizontal: $20 = ma \implies a = 4\,\text{m/s}^2$ .

Example 2 (Easy, CBSE) — Block on an Incline

A $2\,\text{kg}$ block rests on a frictionless incline of $30°$ . Find its acceleration down the slope.

FBD: weight $mg = 20\,\text{N}$ down, normal $N$ perpendicular to surface. Along the slope: $mg\sin 30° = ma$ , giving $a = g\sin 30° = 5\,\text{m/s}^2$ .

Example 3 (Medium, JEE Main) — Pulley with Two Masses

Masses $m_1 = 3\,\text{kg}$ and $m_2 = 5\,\text{kg}$ hang from a smooth pulley. Find the acceleration and tension.

Two FBDs, one per mass. For $m_1$ : $T - m_1 g = m_1 a$ (upward positive). For $m_2$ : $m_2 g - T = m_2 a$ (downward positive). Adding:

$a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{2 \times 10}{8} = 2.5\,\text{m/s}^2$

$T = m_1(g+a) = 3 \times 12.5 = 37.5\,\text{N}$

For multi-body problems, draw a separate FBD for each body, then write Newton’s second law for each. Tension is the same throughout a massless inextensible string passing over a smooth pulley.

Example 4 (Medium, JEE Main) — Block on Block

A $4\,\text{kg}$ block sits on a $6\,\text{kg}$ block on a frictionless floor. Coefficient of friction between the blocks is $0.4$ . Apply force $F = 50\,\text{N}$ horizontally to the lower block. Find the acceleration of the upper block. Take $g = 10\,\text{m/s}^2$ .

FBD of upper block: friction from lower block is the only horizontal force. Maximum static friction $= 0.4 \times 4 \times 10 = 16\,\text{N}$ .

If they move together, common acceleration $a = 50/10 = 5\,\text{m/s}^2$ . Force needed on the upper block $= 4 \times 5 = 20\,\text{N}$ . But the maximum friction is only $16\,\text{N}$ , so they slip.

Upper block: $a_{\text{upper}} = 16/4 = 4\,\text{m/s}^2$ .

Example 5 (Hard, JEE Advanced) — Wedge with Block

A block of mass $m$ sits on a smooth wedge of mass $M$ and angle $\theta$ , on a smooth floor. The wedge is free to slide. Find the acceleration of the block relative to the ground.

This requires two FBDs and constraint analysis. The block accelerates down the slope relative to the wedge, while the wedge slides backward. Working through the algebra:

$a_{\text{block, vertical}} = \frac{g\sin^2\theta\,(M+m)}{M + m\sin^2\theta}$

The trick is recognising that the block’s motion relative to the ground is not along the incline, because the incline is moving.

Exam-Specific Tips

CBSE marking scheme: A correct FBD with all forces labelled is worth $1$ – $2$ marks even before any equation. Always draw it.

JEE Main: FBD questions usually appear as MCQ, so accuracy matters more than rigour. Practice spotting the correct FBD from four options.

JEE Advanced: Multi-body and constraint problems dominate. Draw one FBD per body, then write the constraint equations carefully.

NEET: Force-on-pulley and block-on-block scenarios are standard. Memorise the standard results.

Common Mistakes to Avoid

Mistake 1: Drawing internal forces (e.g., the force one half of the block exerts on the other half). These cancel and shouldn’t appear.

Mistake 2: Drawing both members of an action-reaction pair on the same FBD. Each pair acts on different bodies.

Mistake 3: Forgetting normal force when the surface isn’t horizontal.

Mistake 4: Drawing friction in the wrong direction. Static friction opposes the tendency of relative motion, not the applied force.

Mistake 5: Using $g$ instead of $mg$ for the weight. Always include the mass.

Practice Questions

Q1. A $10\,\text{kg}$ block is pulled along a horizontal floor with $\mu_k = 0.2$ by a $40\,\text{N}$ force at $30°$ above horizontal. Find the acceleration. Take $g = 10\,\text{m/s}^2$ .

Vertical: $N + 40\sin 30° = 100 \implies N = 80\,\text{N}$ . Friction $= 16\,\text{N}$ . Horizontal: $40\cos 30° - 16 = 10a \implies a \approx 1.86\,\text{m/s}^2$ .

Q2. Two masses $m_1 = 2\,\text{kg}$ and $m_2 = 3\,\text{kg}$ are connected by a string over a smooth pulley. $m_1$ rests on a horizontal floor while $m_2$ hangs. The system is released. Find the acceleration if the floor is smooth.

$a = m_2 g/(m_1 + m_2) = 30/5 = 6\,\text{m/s}^2$ .

Q3. A block of mass $m$ on a $60°$ incline with $\mu_k = 0.3$ slides down. Find its acceleration. Take $g = 10\,\text{m/s}^2$ .

$a = g(\sin 60° - 0.3\cos 60°) = 10(0.866 - 0.15) = 7.16\,\text{m/s}^2$ .

Q4. A car of mass $1000\,\text{kg}$ accelerates at $2\,\text{m/s}^2$ . Find the friction force from the road on the driving wheels.

$F = ma = 2000\,\text{N}$ .

Q5. A boy of mass $40\,\text{kg}$ stands in a lift accelerating upward at $2\,\text{m/s}^2$ . Find his apparent weight. Take $g = 10\,\text{m/s}^2$ .

$N = m(g+a) = 40 \times 12 = 480\,\text{N}$ .

FAQs

Why don’t internal forces appear in the FBD? They are equal and opposite within the body, so they cancel out. Only forces from outside the body affect its motion.

How do I know the direction of friction? Static friction opposes the tendency of relative motion. Kinetic friction opposes the actual relative motion. Always ask: which way would the surfaces slide if there were no friction?

When is the normal force not equal to $mg$ ? When the body accelerates vertically, when the surface is tilted, when an external vertical component is applied, or when the body is on a non-horizontal surface.

Should I draw forces from the centre of mass? For point particles, yes. For rigid bodies in non-rotational problems, all forces can be drawn from the centre. For rotational problems, the line of action matters — draw forces at their actual points of application.

Can I solve a problem without an FBD? Sometimes, for trivial cases. But for any non-trivial problem, the FBD prevents sign errors and missed forces. It’s worth the 30 seconds.