Newton's First Law

A bus starts with a jolt and every standing passenger lurches backward. The driver did nothing to them — no hand pushed them, no rope pulled them — yet they staggered as if shoved. What actually happened is that the floor of the bus accelerated forward under their feet while their upper bodies, obeying no command but physics, tried to keep doing exactly what they were already doing: standing still. That stubbornness of matter, its refusal to change its own state of motion unless something outside it forces the issue, is inertia, and Newton's First Law is the precise, testable statement of that stubbornness. It sounds almost too simple to be a "law" — a body at rest stays at rest, a body in motion keeps moving in a straight line at constant speed, unless a net external force intervenes. Aristotle, for two thousand years the reigning authority on motion, believed the opposite: that force is needed to sustain motion, that a cart stops because it "wants" to stop once you quit pushing. Galileo's thought experiments with balls rolling down ever-smoother inclines, and Newton's later codification of them, overturned that entirely. Friction, not some inner tiredness of matter, is what stops the cart. Left alone, motion — like rest — simply continues. This article builds the First Law from the ground up: what inertia measures, how the law defines force itself, why it only holds in special reference frames, and its most exam-tested application — deciding when a set of forces on a body is balanced, including equilibrium of a particle under several concurrent forces.

Key takeaways

Newton's First Law (law of inertia): a body continues in its state of rest or of uniform motion in a straight line unless compelled by a net external force to change that state.
Inertia is the natural tendency of a body to resist any change in its velocity; mass is the quantitative measure of inertia — more mass, more reluctance to change speed or direction.
The law secretly defines force: force is whatever, when unbalanced, changes a body's state of rest or of uniform motion.
The First Law holds only in an inertial frame — one that is itself not accelerating; in an accelerating frame, objects appear to move without any real force acting.
Equilibrium of a particle is the direct working consequence of the First Law: if a body has zero acceleration, the vector sum of all forces on it must be zero, whether the forces number two, three, or many.
For exactly three concurrent, coplanar forces holding a particle in equilibrium, the balance condition reduces to Lami's theorem, a ratio of each force to the sine of the angle between the other two.
Key idea: a body's velocity stays exactly constant, including zero, unless a nonzero net external force acts on it, and this single fact both defines what a force is and gives the working condition — net force equals zero — for equilibrium of a particle under any number of concurrent forces.

The core insight

A body keeps its state of rest or uniform straight-line motion unless a net external force acts, and this is simultaneously the definition of force, the definition of an inertial frame, and the working condition (net force zero) for equilibrium of a particle.

Where it fits in the syllabus

Newton's First Law opens CBSE Class 11 Unit III, Laws of Motion, immediately after the intuitive idea of force is introduced. It is fully in the current 2025-26 prescription — this unit has not been trimmed. You are expected to state the law precisely, connect it to inertia and mass, and use it as the working test for equilibrium of a particle (net force zero), including the special case of three concurrent forces solved by Lami's theorem or a triangle of forces. This section deliberately stays on inertia, the law itself, and equilibrium; momentum, the Second Law's $F = ma$ , and Newton's Third Law are built out in the sibling nodes that follow directly from here, so a numerical value of acceleration under an unbalanced force is only touched here to contrast with the balanced case.

The physical picture

Newton's First Law is usually written as a conditional: if the net external force is zero, the velocity does not change.

\text{If } \vec{F}_{\text{net}} = \sum \vec{F}_i = 0, \quad \text{then } \vec{v} = \text{constant (in magnitude and direction).}

where: $\vec{F}_{\text{net}}$ is the vector sum of every external force acting on the body, and $\vec{v}$ is its velocity. "Constant velocity" is a single idea that covers two everyday-looking cases: $\vec{v} = 0$ (the body stays at rest) and $\vec{v} \ne 0$ but unchanging in both speed and direction (the body cruises in a straight line forever). Rest is not a separate, special state — it is simply uniform motion with speed zero. Reading the arrow the other way is just as important for problem-solving: whenever you see a body moving at constant velocity, or sitting still, you are licensed to conclude its net force is zero, even before you know what the individual forces are. Two deep consequences follow from this one line. First, the law is really Newton's definition of force: a force is precisely that external influence which, if left unbalanced, changes a body's speed, its direction, or both. No net force, no change in velocity — full stop. Second, the resistance every body shows to having its velocity changed is called inertia, and its quantitative measure is mass. A loaded truck needs a much larger push to speed up, or to stop, than an empty scooter of the same shape, precisely because it has more inertia, i.e. more mass. "Heavier objects have more inertia" is not folklore; it is the operational meaning of mass in this law. The law also comes with a silent condition: it holds only in an inertial frame, one that is not itself accelerating. A coin balanced on your palm inside a bus cruising at a steady $50\ \text{km/h}$ behaves exactly as the First Law predicts, staying put with no force needed beyond gravity and your palm's support. The instant the driver brakes hard, the coin appears to slide forward "on its own." Nothing mysterious happened to the coin — it is still trying to keep moving at the bus's old speed — but the braking bus itself is now a non-inertial frame, and the First Law in its simple form no longer describes what an observer inside sees. For Class 11 the fix is not to invent a pseudo-force but simply to treat the ground, and any frame moving at constant velocity relative to it, as a good inertial frame for ordinary problems.

Figure

A bus accelerating forward with a standing passenger who lags behind because of inertia while a coin at rest on a table beside a stationary bus stays exactly where it is

Equilibrium of a particle under several forces

If a particle is at rest, or moving with constant velocity, the First Law tells us its net force must vanish, whether two forces act on it or twenty. Working this backward is the single most exam-tested application of the First Law: given a body known to be in equilibrium, find an unknown force from the requirement that everything must balance. In two dimensions the vector condition $\vec{F}_{\text{net}} = 0$ splits into two scalar equations, one for each perpendicular axis:

\sum F_x = 0, \qquad \sum F_y = 0.

where: $\sum F_x$ and $\sum F_y$ are the algebraic sums of the $x$ - and $y$ -components of every force acting on the particle, with a sign convention fixed once and used consistently. A free-body diagram — the particle drawn alone, with every external force on it shown as an arrow with correct direction, and nothing else — is the standard tool for setting these sums up correctly; skipping it is the single biggest source of sign errors in equilibrium problems. A particularly common exam configuration is exactly three concurrent forces holding a particle in equilibrium — for instance a lamp hanging from two strings meeting a ceiling at different angles. For that special case there is a shortcut, Lami's theorem, derived below.

Derivation AEquilibrium under three concurrent forces — Lami's theorem

The goal here is to turn the bare statement "net force is zero" into a usable numerical tool for the common exam picture of a particle held in place by exactly three forces acting along different directions but all meeting at one point — a hanging lamp pulled by two strings, or a ring held by three cords. We want a relation between the three force magnitudes and the angles between them, using nothing beyond the First Law's equilibrium condition and the sine rule for a triangle, so that an unknown tension or weight can be read off directly without resolving into components by hand.

Step 1. Start from the First Law's equilibrium condition. Since the particle is at rest (zero acceleration), the net external force is zero: $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$ , where $\vec{F}_1$ , $\vec{F}_2$ , $\vec{F}_3$ are the three concurrent forces, all acting through the same point.

Step 2. Turn the vector sum into a closed triangle. Three vectors that sum to zero can always be redrawn head-to-tail as a closed triangle: place $\vec{F}_1$ , then attach the tail of $\vec{F}_2$ to the head of $\vec{F}_1$ , then $\vec{F}_3$ from the head of $\vec{F}_2$ back to the tail of $\vec{F}_1$ . Because the sum is zero, this path closes exactly, forming an ordinary triangle whose three sides have lengths $F_1$ , $F_2$ , $F_3$ .

Step 3. Identify the triangle's angles with the angles between the original force directions. When forces are redrawn head-to-tail, the interior angle of the triangle opposite a given side turns out to be $180^\circ$ minus the angle between the other two original forces as they act at the point. Working through the geometry, the interior angle facing side $F_1$ equals $180^\circ - \alpha$ , where $\alpha$ is the angle between $\vec{F}_2$ and $\vec{F}_3$ at the point of concurrency (and similarly for the other two sides with angles $\beta$ , between $\vec{F}_3$ and $\vec{F}_1$ , and $\gamma$ , between $\vec{F}_1$ and $\vec{F}_2$ ).

Step 4. Apply the sine rule to the closed triangle. For any triangle, each side divided by the sine of its opposite angle gives the same constant: $\dfrac{F_1}{\sin(180^\circ-\alpha)} = \dfrac{F_2}{\sin(180^\circ-\beta)} = \dfrac{F_3}{\sin(180^\circ-\gamma)}$ . Since $\sin(180^\circ - \theta) = \sin\theta$ for any angle $\theta$ , this simplifies at once to

\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}.

Step 5. Read off Lami's theorem. Each force is proportional to the sine of the angle between the other two forces — not the angle it makes with them itself. This is exactly Lami's theorem: for three concurrent, coplanar forces in equilibrium, $F_1/\sin\alpha = F_2/\sin\beta = F_3/\sin\gamma$ , with $\alpha$ , $\beta$ , $\gamma$ the angles opposite each force and summing to $360^\circ$ around the point. Given any two forces and the three angles, the third force is found by one ratio — no separate component resolution needed.

Worked examples

Board level - a car cruising at constant speed

A car travels along a straight, level road at a steady $60\ \text{km/h}$ . The total resistance opposing it — road friction plus air drag — is $800\ \text{N}$ . What forward driving force must the engine supply through the tyres, and is this an example of the First Law or the Second Law?

Step 1. Recognise the equilibrium condition. The speed is constant ( $60\ \text{km/h}$ , unchanging), so the acceleration is zero. By the First Law, a body with zero acceleration has zero net force acting on it — this is a First Law problem even though an engine is clearly doing work, because what matters for the force balance is only that velocity is not changing.

Step 2. Set up the one-dimensional force balance along the direction of motion. Two forces act along this line: the forward drive force $F_{\text{drive}}$ from the tyres, and the $800\ \text{N}$ resistance opposing motion. Equilibrium requires $F_{\text{drive}} - 800 = 0$ .

Step 3. Solve. $F_{\text{drive}} = 800\ \text{N}$ . The engine must supply exactly 800 N of forward driving force — no more, no less — for the car to hold a constant 60 km/h. Notice the actual speed value never entered the calculation; only the fact that it is constant mattered. The engine's job at cruising speed is not to cause motion but purely to cancel the resistive forces, exactly as the First Law demands for a body with unchanging velocity.

Competitive level - a lamp held by two strings at different angles

A lamp of weight $W = 50\ \text{N}$ hangs in equilibrium from two strings attached to a ceiling. String $A$ makes an angle of $30^\circ$ with the ceiling and string $B$ makes an angle of $60^\circ$ with the ceiling, on opposite sides of the lamp, so the two strings are perpendicular to each other. Find the tensions $T_A$ and $T_B$ .

Step 1. Draw the free-body diagram and identify the three concurrent forces at the knot: the weight $W = 50\ \text{N}$ acting vertically down, tension $T_A$ along string $A$ , and tension $T_B$ along string $B$ . Since $A$ makes $30^\circ$ with the horizontal ceiling and $B$ makes $60^\circ$ on the other side, and $30^\circ + 60^\circ = 90^\circ$ , the two strings are indeed perpendicular to each other — a useful check before applying Lami's theorem.

Step 2. Find the angle each force makes with the other two, as Lami's theorem requires. The angle between $T_A$ and $T_B$ (the two strings) is $90^\circ$ . The angle between $T_A$ and $W$ is $90^\circ + 60^\circ = 150^\circ$ (since $W$ points straight down and $A$ is inclined $30^\circ$ above the horizontal on its side). Similarly, the angle between $T_B$ and $W$ is $90^\circ + 30^\circ = 120^\circ$ .

Step 3. Apply Lami's theorem: $\dfrac{T_A}{\sin(\text{angle opposite }T_A)} = \dfrac{T_B}{\sin(\text{angle opposite }T_B)} = \dfrac{W}{\sin(\text{angle opposite }W)}$ . The angle opposite $T_A$ is the one between $T_B$ and $W$ , namely $120^\circ$ ; the angle opposite $T_B$ is the one between $T_A$ and $W$ , namely $150^\circ$ ; the angle opposite $W$ is the one between $T_A$ and $T_B$ , namely $90^\circ$ . So

\frac{T_A}{\sin 120^\circ} = \frac{T_B}{\sin 150^\circ} = \frac{50}{\sin 90^\circ}.

Step 4. Solve numerically. Since $\sin 90^\circ = 1$ , the common ratio is simply $50$ . Then $T_A = 50 \times \sin 120^\circ = 50 \times 0.866 = 43.3\ \text{N}$ , and $T_B = 50 \times \sin 150^\circ = 50 \times 0.5 = 25.0\ \text{N}$ . The string closer to the horizontal (string $A$ , at $30^\circ$ ) carries the larger tension, $T_A \approx 43.3\ \text{N}$ , while the steeper string $B$ carries $T_B = 25.0\ \text{N}$ . Timing note: aim for under three minutes. The whole problem collapses to one line once the three angles opposite each force are identified correctly — that identification, not the arithmetic, is where marks are lost under time pressure. As a quick sanity check, the more nearly vertical string should always carry the smaller share of the weight; here $B$ (steeper, closer to vertical) correctly carries less tension than $A$ .

Common traps

Treating "net force zero" as "no forces at all." Any number of forces can act on a body in equilibrium; the First Law only demands that their vector sum vanishes. A book resting on a table has both gravity and a normal reaction acting on it, two nonzero forces that happen to cancel.
Believing a moving body must have a forward force pushing it. A body cruising at constant velocity has zero net force by definition. Forward motion is a leftover of a force applied earlier, not evidence of a force acting right now.
Calling inertia a force. Inertia is a property of a body, its resistance to a change of velocity, not a push or a pull. Writing an "inertia force" on a free-body diagram in an inertial frame is incorrect and costs marks.
Applying Lami's theorem to the wrong angle. The angle used with a given force is the angle between the other two forces, not the angle that force itself makes with an axis. Mixing these up is the most common slip.
Forgetting the inertial-frame condition. Inside an accelerating lift, car, or bus, objects can appear to move without any real force. The First Law as stated here needs an inertial frame.

You should be able to…

State Newton's First Law precisely, including that "uniform motion" means unchanging speed and direction, and that rest is the special case of zero velocity.
Explain how the First Law defines force as whatever changes a body's state of rest or of uniform motion.
Connect inertia to mass, and explain why a larger mass resists a change in velocity more strongly.
Say what an inertial frame is, and identify situations (a braking bus, a turning car) where analysis inside the frame needs care.
Set up $\sum F_x = 0$ and $\sum F_y = 0$ from a correctly drawn free-body diagram for a particle in equilibrium.
Apply Lami's theorem to find an unknown force among three concurrent, coplanar forces holding a particle in equilibrium.

Ask the tutor: stuck on setting up a free-body diagram or applying Lami's theorem — open the AI tutor from this page and walk through it force by force.

If Newton's First Law just restates that things keep doing what they were doing, why is it called a law at all?

Because it was genuinely revolutionary when Newton, building on Galileo, stated it. For nearly two thousand years the accepted view, going back to Aristotle, was that force is needed to keep something moving, and motion naturally dies out on its own. The First Law flips this: motion, like rest, persists forever unless something external interferes, and friction, not any inner tendency to stop, is what brings everyday moving objects to rest. It is also the law that tells us what a force even is, by saying it is whatever changes a state of rest or of uniform motion — overturning old intuition and defining a core concept at once.

Why does the First Law only work in an inertial frame? What goes wrong otherwise?

Because "change in velocity" always has to be measured relative to some frame of reference. If you stand in a frame that is itself accelerating — a braking bus, a turning car, a lift starting to move — objects at rest relative to the ground will appear, from inside that frame, to speed up, slow down, or curve, even though no real force is acting on them. The law would then look false. It is not false; the frame chosen was simply not inertial. For Class 11, the fix is not to invent pseudo-forces but to always analyse motion from the ground, or any frame moving at constant velocity relative to it.

In the lamp-and-two-strings problem, how do I quickly find which string carries more tension without doing the full calculation?

A fast qualitative check: the string more nearly vertical usually carries the larger share of the weight, since it opposes gravity more directly — but use this only as a sanity check, not a shortcut, because the exact split depends on both angles together. In the worked example, string $B$ was steeper yet carried the smaller tension, since its tension is set by the angle opposite it (between $T_A$ and $W$ ), a reminder that the theorem's angle assignment must be checked carefully every time.

Does the First Law only apply to bodies at rest, or also to bodies already moving?

It applies to both. The law is a statement about velocity staying constant, and constant velocity includes the special value zero (rest) just as much as a steady $20\ \text{m/s}$ in a straight line. A car cruising at fixed speed and a book sitting motionless on a shelf are, physically, in exactly the same situation for the First Law: zero net force, zero acceleration, unchanging velocity. Only the particular constant value of that velocity differs.