What an FBD is and why it matters
A free body diagram is a sketch of one chosen body — the system — with every external force on it drawn as an arrow. "Free body" does not mean the body is free of forces. It means the body has been mentally cut free from the rest of the assembly so only forces acting on it remain on the page. Surrounding objects, ropes, walls and the Earth are gone; their influence survives only as arrows.
NEET problems hand you a tangle and ask for one number — an acceleration, a tension, a normal reaction. The FBD reduces that tangle to a Newton-law equation. Almost every wrong answer in NEET mechanics traces back to a missing force, an extra force, or a wrong direction on the FBD.
The five-step NCERT procedure
NCERT §4.11 lays out five steps. Memorise them — they apply to every problem in the chapter.
- Draw a schematic of the assembly. Blocks, links, supports, pulleys, springs, surfaces. Geometry only — not yet an FBD.
- Choose a convenient part as the system. One body, or a group of bodies sharing the same acceleration. When in doubt, start with the body whose acceleration you want.
- Draw a separate diagram of the system showing ALL forces ON the system by the rest of the assembly, and by other agencies. Do not include forces the system exerts ON its environment. This is the FBD. "Free body" does not mean zero net force — only isolated from the environment.
- Mark what you know. Weight \(mg\) vertically down. Tension along the string. Normal perpendicular to contact. Leave the rest as unknowns.
- Repeat for another system if needed. Link them via the third law: if A's FBD shows \(\vec F\) due to B, then B's FBD must show \(-\vec F\) due to A.
The six common forces on a NEET FBD
Before you write a single equation, scan the assembly for these six force types. NEET's silent error is forgetting one.
| Force | Direction rule | When it appears |
|---|---|---|
| Gravity \(mg\) | Vertically downward, through centre of mass | Always, for any body of mass \(m\) near Earth |
| Normal reaction \(N\) or \(R\) | Perpendicular to contact surface, pointing away from it | Whenever the system touches another solid surface |
| Friction \(f\) | Parallel to contact surface; opposes relative motion or its tendency | At any rough contact |
| Tension \(T\) | Along the string, pulling the system toward the string's other end | Whenever a string or rope is attached |
| Spring force \(kx\) | Along the spring axis; restores it to natural length | Whenever a deformed spring is attached |
| Applied force \(F\) | As specified in the problem | Externally applied push or pull |
Stop there. NEET inertial-frame FBDs do not contain centrifugal force, "force of motion", "Coriolis force", or imaginary "drag toward the centre of mass". If a force is not in the table above and not a clearly stated pseudo-force in a non-inertial frame, it does not belong on the FBD.
Worked FBD 1 — NCERT Example 4.12 (block + cylinder on yielding floor)
A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block + cylinder together go down with an acceleration of \(0.1~\text{m s}^{-2}\). What is the action of the block on the floor (a) before and (b) after the floor begins to yield? Take \(g=10~\text{m s}^{-2}\).
FBD before the cylinder is placed. System = wooden block alone. Forces on it: weight \(W = 2\times 10 = 20~\text{N}\) downward, normal \(R\) upward from floor. The block is at rest, so \(R - W = 0 \Rightarrow R = 20~\text{N}\). By Newton's third law, the action of the block on the floor is 20 N directed downward.
FBD after the cylinder is placed. System = block + cylinder together (treated as one composite body of mass 27 kg). Forces: total weight \(W' = 27\times 10 = 270~\text{N}\) downward, normal \(R'\) upward. The system accelerates downward at \(0.1~\text{m s}^{-2}\), so Newton's second law (down positive) gives \(270 - R' = 27 \times 0.1\), i.e. \(R' = 270 - 2.7 = 267.3~\text{N}\). The action of the (block + cylinder) on the floor is therefore \(267.3~\text{N}\) downward.
Action–reaction pairs. Part (a): Earth pulls the block down (20 N) ↔ block pulls Earth up (20 N); block pushes floor down (20 N) ↔ floor pushes block up (20 N). Part (b): Earth pulls system down (270 N) ↔ system pulls Earth up (270 N); system pushes floor down (267.3 N) ↔ floor pushes system up (267.3 N). The block–cylinder contact is an internal pair that disappeared once we picked the composite system.
Worked FBD 2 — Atwood machine (NEET 2020 Q.93)
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a frictionless pulley. The acceleration of the system in terms of \(g\) is:
FBD of the 6 kg mass (heavier — will descend). Forces: tension \(T\) upward, weight \(6g\) downward. Choose down as positive: \(6g - T = 6a\).
FBD of the 4 kg mass (lighter — will rise). Forces: tension \(T\) upward (same magnitude as above because the string is massless and the pulley is frictionless), weight \(4g\) downward. Choose up as positive: \(T - 4g = 4a\).
Add the two equations. The tensions cancel: \(2g = 10a \Rightarrow a = g/5\). The tension follows from either equation: \(T = 4g + 4a = 4g + 4g/5 = 24g/5 = 48~\text{N}\) with \(g=10\).
Two clean FBDs, two single-unknown equations, one sum. The Atwood machine is the prototype FBD problem: forget the upward tension on the heavier block and your sign of \(a\) flips; forget the equal-tension assumption and you have two unknowns and one equation.
Worked FBD 3 — Block on an incline (NCERT Example 4.11)
A block of mass \(m\) rests on a horizontal plane which is gradually tilted. At an angle \(\theta_c\) the block just begins to slip. Use the FBD to find the coefficient of static friction.
FBD of the block. Three forces: weight \(mg\) vertically downward; normal \(R\) perpendicular to the inclined surface; static friction \(f_s\) along the surface, up the slope (because the block tends to slide down).
Choose axes along and perpendicular to the incline. This is the discipline of "axes for convenience". Resolving \(mg\) along the incline gives \(mg\sin\theta\) (down the slope) and perpendicular to the incline gives \(mg\cos\theta\) (into the surface).
Perpendicular equilibrium: \(R = mg\cos\theta\). Along the slope, at the limiting angle, the block is on the verge of sliding so \(mg\sin\theta_c = f_s^{\max} = \mu_s R = \mu_s mg\cos\theta_c\). Dividing, \(\boxed{\mu_s = \tan\theta_c}\) — the angle of repose.
Notice three things. Axes along and perpendicular to the incline turned one Newton equation into two single-unknown equations. The direction of \(f_s\) was inferred from physics — friction opposes the tendency of motion, even when nothing slides yet. And \(\mu_s = \tan\theta_c\) folds geometry and dynamics into one equation.
Friction direction can be subtle — see friction (static, kinetic, rolling) for the full \(\mu_s\) vs \(\mu_k\) framework.
Worked FBD 4 — Spring-block cut (NEET 2017 Q.165)
Two blocks A (mass \(3m\)) and B (mass \(m\)) are connected by a string. The whole assembly hangs in equilibrium from a spring. The string between A and B is then cut. Find the accelerations of A and B immediately after the string is cut.
Before the cut the system is in equilibrium. The spring force equals the total weight: \(kx = 3mg + mg = 4mg\), directed upward on A.
The instant the string is cut, the spring's length has not changed yet — it cannot change discontinuously. So at that instant the spring still exerts \(4mg\) upward on block A.
FBD of A immediately after cut. Spring force \(4mg\) upward, weight \(3mg\) downward. Net force \(= mg\) upward. Acceleration \(a_A = \dfrac{mg}{3m} = \dfrac{g}{3}\) upward.
FBD of B immediately after cut. The string is gone, so only weight \(mg\) downward acts. Acceleration \(a_B = g\) downward.
Answer: \(a_A = g/3\) upward, \(a_B = g\) downward.
The trap: students assume the spring "snaps" along with the string. It does not — springs cannot change length in zero time. The FBD keeps this locality clear: draw A in isolation, copy the spring force from a moment before, and the answer falls out.
Choosing the system
NCERT step (ii) is where the cleverness happens. The system can be a single body, two bodies treated as one, or the whole assembly. The choice matters because forces between bodies inside the chosen system are internal and do not enter the FBD.
In Example 4.12, if you choose the block alone you must include the cylinder-on-block contact force — two unknowns. Choose block + cylinder and that contact force becomes internal. Only weight (270 N) and floor reaction \(R'\) remain.
The decision rule: group bodies that share the same acceleration. Two boxes pushed together — group them. A monkey climbing a rope (different accelerations) — do not group.
Sign conventions and axes
- Choose axes for convenience. Incline problems: axes along and perpendicular to the slope. Atwood problems: vertical axes with positive direction chosen so the heavier block has \(+a\). The FBD is unchanged; only components change.
- Default to the ground (inertial) frame. Newton's second law reads \(\sum\vec F = m\vec a\) with no pseudo-forces. Lift problems can be solved in the lab frame or in the lift frame (with pseudo-force \(-ma\) added) — pick one and stay.
- Single-unknown discipline. Align axes with as many forces as possible. One equation per unknown.
Typical FBD errors
Score every FBD against this checklist before writing equations.
- Missing force. Did you scan all six force types? Surface contact → normal. String → tension. Rough contact → friction.
- Extra force. No "centrifugal force" in the lab frame. No "ma arrow" on the FBD — \(ma\) is the answer, not an input.
- Wrong direction. Normal is perpendicular to the surface, not the body. Tension is along the string. Friction opposes relative motion, not the applied force.
- Forces the body exerts on others. Belong on the other body's FBD.
- Equal-and-opposite ≠ action-reaction. Two forces on the same FBD are never a third-law pair.
- Uniform tension assumption. Breaks if the pulley has mass or friction.
FBD discipline in one breath
- Sketch the assembly. Choose a system. Draw all forces on that system. Mark knowns and unknowns. Repeat for other systems if needed.
- Six force types to check: gravity, normal, friction, tension, spring, applied. No more.
- "Free" means cut free from the environment, not zero net force.
- Forces on the same body are never a third-law pair, even when equal and opposite.
- Group bodies with the same acceleration to make contact forces internal and disappear.
- Axes along natural directions (incline, vertical Atwood) give single-unknown equations.