Newton's Second Law

Every machine that moves — every car, rocket, robot arm, and cricket ball — obeys one three-symbol sentence: $F = ma$ . Newton's second law is the engine room of classical mechanics, and this article is about driving it. Give me the forces on an object and its mass, and I will tell you its acceleration — and from acceleration, the whole future of its motion. That is not a figure of speech: $\vec F = m\vec a$ is a second-order differential equation for the position $\vec r(t)$ , and once the forces are listed and the initial position and velocity are fixed, the entire trajectory is determined. When engineers land a probe on Mars or tune a high-speed train's braking curve, the computation under the hood is this page's recipe run millions of times. Curiously, the famous formula appears nowhere in Newton's Principia. What Newton actually wrote is stranger and stronger: the change of motion — mass times velocity, what we now call momentum — is proportional to the motive force impressed. Force is whatever changes momentum, at a rate equal to the force itself: $\vec F = \mathrm{d}\vec p/\mathrm{d}t$ . For a body of fixed mass the two statements are identical, but the momentum form survives where the schoolbook version fumbles — rockets that lose mass as they burn, crashes where the force spikes unmeasurably while the momentum ledger stays exact, and ultimately relativity itself. What you will practise here is the physicist's four-step ritual — isolate the object, draw every force, sum them by components, divide by the mass — applied to the scenarios exams love: elevators that make you feel heavy, connected masses over pulleys, and blocks sliding down rough inclines. Master the ritual and no force problem is ever truly new.

Key takeaways

Newton's second law: the net force on an object equals mass times acceleration, $\vec F_{\text{net}} = m\vec a$ — a vector equation, true component by component, valid in inertial frames.
Newton's own, deeper statement is the momentum form $\vec F_{\text{net}} = \mathrm{d}\vec p/\mathrm{d}t$ : force is the rate of change of momentum. For constant mass it collapses to $F = ma$ ; for rockets and crashes it is the version that keeps telling the truth.
$F_{\text{net}}$ is the vector sum of all forces acting on the object; feeding a single force into the law while others still stand is a bookkeeping error.
The acceleration points along the net force — not necessarily along the velocity (a ball at the top of its flight has $v = 0$ and $a = g$ downward).
Mass is the exchange rate: same force, double the mass, half the acceleration. The newton: $1\ \mathrm{N} = 1\ \mathrm{kg\,m/s^2}$ .
Integrated over time the law becomes the impulse–momentum theorem $\vec J = \vec F\,\Delta t = \Delta\vec p$ — the physics of airbags, crumple zones, and soft hands.
Key idea: net force is not what makes things move — it is what makes velocity change. $\vec F_{\text{net}} = m\vec a$ is a second-order differential equation for $\vec r(t)$ that, together with initial position and velocity, determines all future motion.

The core insight

Net force is not what makes things move — it is what makes velocity change: $\vec F_{\text{net}} = m\vec a$ (Newton's deeper $\vec F = \mathrm{d}\vec p/\mathrm{d}t$ , specialised to constant mass) converts a list of forces into an acceleration and, together with initial position and velocity, determines all future motion.

Where this lives today

Every physics engine — in games, film effects, and robotics simulators — is a second-law integrator: at each tick it sums the forces on every body, divides by mass, and steps velocity and position forward. When a game character's cape swirls convincingly, thousands of $F = ma$ evaluations per frame are doing the tailoring. The same loop, hardened and certified, flies real hardware: a quadcopter's flight controller estimates the forces on the airframe hundreds of times a second and solves for the motor thrusts that produce the acceleration the pilot commanded — the elevator problem of Derivation B, generalised to three dimensions and running on a chip the size of a stamp. The momentum form runs the safety and propulsion industries. Crash engineering is applied equation (6) with a budget of billions: crumple zone, belt pre-tensioner, then the airbag inflating within ~30 ms — every stage engineered to stretch $\Delta t$ and shave the force peak, validated on instrumented dummies. And on the launch pad the law runs in reverse: a Falcon 9 climbs because its engines pump roughly 2,500 kg of exhaust per second backward at about 3 km/s, and the $\Delta p/\Delta t$ of that thrown mass is the ~7.5 MN of thrust. Three centuries on, Newton's proportion between force and change of motion remains the most heavily executed law in engineering.

The physical picture

Push a stalled car and an empty shopping trolley with the same effort. The trolley leaps; the car barely creeps. Same cause, wildly different effect — the difference is mass, matter's built-in resistance to having its velocity changed. Newton's second law makes the comparison quantitative: acceleration is force per unit of that reluctance, $a = F_{\text{net}}/m$ . It is a steering relation as much as a speeding one: the acceleration vector always points where the net force points, growing the velocity if aligned with it, eating the velocity if opposed, bending it if aslant. The phrase that earns its keep is net force. An object at rest on a table has two forces on it — gravity down, table up — summing to zero, so it stays put. A car cruising at a steady 100 km/h has thrust and drag in exact cancellation: zero net force, zero acceleration, velocity frozen. The second law never asks how many forces act, or how large each is alone; it asks for one thing — the vector sum — and repays you with the acceleration. That is why every serious solution starts with a free-body diagram: one object, every force drawn as an arrow, nothing else in the picture.

Figure

Hero — the free-body ritual: four forces in, one acceleration out

The car-versus-trolley comparison is worth graphing, because the two proportionalities in $a = F_{\text{net}}/m$ have opposite shapes — and both are the experiment's actual output:

Figure

Same force, different mass — and same mass, different force

There is a second, deeper way to picture the same law. Momentum $p = mv$ is an object's budget of motion — the honest answer to "how hard is this thing to stop?" — and force is a pump that fills or drains that budget at a definite rate. Catch a cricket ball with rigid arms and you drain its momentum in a hundredth of a second — the pump must run ferociously, and it hurts. Pull your hands back as you catch and the same drain is spread over ten times longer: a tenth of the force. This bookkeeping view — audit the momentum totals, stay humble about the messy middle — is what makes the second law usable in the chaos of kicks, crashes, and rocket burns, where the instantaneous force is unknowable but the totals are exact.

The math

Start with the law in its working form for constant mass. Everything the free-body diagram collects — pulls, pushes, weight, friction — funnels into a single vector sum, and the law converts that sum into the object's response, with mass as the conversion rate:

\vec F_{\text{net}} = m\,\vec a \qquad (1)

where: $\vec F_{\text{net}}$ — the vector sum of every force acting on the object (N); $m$ — its mass (kg); $\vec a$ — its acceleration ( $\mathrm{m/s^2}$ ), always parallel to $\vec F_{\text{net}}$ . Read it as a machine with one input and one output: forces in, acceleration out. The proportions run both ways — double the net force and the response doubles; keep the force and double the mass and the response halves — and nothing else about the object (its speed, its history, its direction of travel) enters the equation at all. That last omission is the trap the warning box below springs. Because (1) is a vector equation, it is really two (or three) independent scalar equations riding under one roof — the $x$ -forces must account for the $x$ -acceleration all by themselves, and likewise for $y$ :

F_{\text{net},x} = m\,a_x \qquad F_{\text{net},y} = m\,a_y \qquad (2)

where: the subscripts pick out components along your chosen axes. This is a superpower: choose axes wisely (along and perpendicular to the motion, or along a ramp) and a messy 2-D problem splits into two clean 1-D ones. The block in the hero figure is (2) in miniature — the vertical equation quietly reports "cancelled," and the whole problem collapses into the horizontal line, $a = 28/m$ . What is a newton, exactly? With mass fixed in kilograms and acceleration in $\mathrm{m/s^2}$ , equation (1) leaves no freedom — the unit of force is defined by the law itself:

1\ \mathrm{N} = 1\ \mathrm{kg\cdot m/s^2} \qquad (3)

where: one newton is the force that gives one kilogram an acceleration of one metre per second squared. An apple weighs about one newton — a coincidence Newton would have appreciated. The most common entry on any free-body diagram earns its own line. Release any object and gravity alone accelerates it at $\vec g$ ; run (1) backwards and the force gravity must have been supplying is mass times that acceleration:

\vec F_{\text{grav}} = m\,\vec g \qquad (4)

where: $\vec g$ — the free-fall acceleration, magnitude $\approx 9.8\ \mathrm{m/s^2}$ near Earth's surface, pointing down. Weight is a force (newtons); mass is not (kilograms). A 70 kg person weighs about 686 N — and it is this 686 N, not the 70 kg, that belongs on a free-body diagram. Now the law as Newton actually stated it. His question was never "what keeps things moving?" — it was "what, precisely, does a force do?" His answer: a force fills or drains momentum $\vec p = m\vec v$ at a rate equal to the force itself. Push twice as hard and the budget changes twice as fast:

\vec F_{\text{net}} = \frac{\mathrm{d}\vec p}{\mathrm{d}t} \qquad (5)

where: $\vec p = m\vec v$ — linear momentum, in kg·m/s; the derivative is the rate at which it changes. A newton is one kg·m/s of momentum delivered per second. Equation (5) earns its keep read in reverse: crash forces are unmeasurable in the chaos of the moment, but momentum-before, momentum-after, and the stopping time are all easy to know — divide, and the average force falls out. Derivation A below shows how (1) hides inside (5), and what survives when the mass itself changes. Multiplying (5) by the time available flips the law from a rate into a total — the full dose of push delivered, called the impulse:

\vec J = \vec F\,\Delta t = \Delta \vec p \qquad (6)

where: $\vec J$ — impulse, in newton-seconds ( $1\ \mathrm{N\,s} = 1\ \mathrm{kg\,m/s}$ — impulse and momentum are the same currency); $\vec F$ — the average net force over the interval $\Delta t$ . A crash fixes $\Delta p$ — your mass and speed decide it before the impact even starts. The one thing design can change is $\Delta t$ : stretch the stop over more time and the average force falls in exact proportion. Crumple zones buy you tenths of a second; airbags buy the tenths your skull needs.

Figure

Two ways to drain the same momentum — same area, very different heights

Finally, the mathematician's view, which is where the law's real power lives. Written out, (1) is a differential equation for the position vector:

m\,\frac{\mathrm{d}^2\vec r}{\mathrm{d}t^2} = \vec F(\vec r, \dot{\vec r}, t) \qquad (7)

where: $\vec r(t)$ — the trajectory to be solved for; the force may depend on position (springs, gravity), velocity (drag), and time. Because the equation is second order, it needs exactly two pieces of starting information per direction — where you are and how fast you're going — and then the whole future is fixed. That is the determinism of classical mechanics in one line. Feed in $\vec F = -mg\,\hat y$ and two integrations hand you the parabola of projectile motion, with range $R = v_0^2 \sin 2\theta / g$ on level ground; feed in Newtonian gravity and out come the orbits of planets. Spacecraft navigation is equation (7), integrated numerically, with every micronewton accounted for.

Which formula answers which question

you want	use	notes
acceleration from known forces	$\vec F_{\text{net}} = m\vec a$ , eq. (1)	sum ALL forces first
a 2-D or ramp problem	components, eq. (2)	pick axes along/perpendicular to motion
the weight entry on a diagram	$\vec F_{\text{grav}} = m\vec g$ , eq. (4)	newtons, not kilograms
crash or impact force	$\vec F = \Delta\vec p/\Delta t$ , eq. (5)	totals are exact when the middle is messy
effect of a force acting for a time	$\vec J = \vec F\Delta t = \Delta\vec p$ , eq. (6)	same $\Delta p$ : longer time, smaller force
variable mass (rockets)	momentum form, Derivation A	naive $m(t)\,a$ misses the thrust
the full trajectory	ODE form, eq. (7)	two initial conditions per direction

Derivation AFrom the momentum form to F equals ma — and when the shortcut fails

We are deriving the schoolbook law (1) from Newton's own momentum statement (5), to establish exactly which is fundamental and which is the special case — and to see what goes wrong for a body whose mass changes. The route is one application of the product rule followed by one physical judgment about the $\dot m$ term. The step to watch is that judgment: for a rocket, simply keeping $\dot m \vec v$ in the equation is not the fix, because the momentum carried away by the exhaust depends on the velocity of the ejected mass, not on the rocket's own velocity. The honest treatment tracks the whole system — rocket plus the propellant it throws away — and rewards you with the thrust formula that flies real vehicles.

Step 1. Differentiate the momentum using the product rule:

\vec F_{\text{net}} = \frac{\mathrm{d}(m\vec v)}{\mathrm{d}t} = \dot m\,\vec v + m\,\dot{\vec v}

Why: momentum is a product of two quantities, and in general both can change — a body can speed up, or it can gain and lose matter.

Step 2. For a body of fixed mass, $\dot m = 0$ and the first term dies:

\vec F_{\text{net}} = m\,\dot{\vec v} = m\vec a

Why this ordering matters: $F = ma$ is derived, not fundamental. It holds for footballs, cars, planets — everything short of rockets and rain-collecting carts.

Step 3. Integrate the momentum form over a time interval instead of differentiating: $\int_{t_1}^{t_2} \vec F\,\mathrm{d}t = \Delta\vec p$ — equation (6). Why the same law twice: the derivative form answers "what is the acceleration right now?"; the integral form answers "what did the whole push accomplish?" — the collision-worker's question.

Step 4. Now let the mass change. Consider a rocket of instantaneous mass $m$ ejecting exhaust backward at speed $v_e$ relative to itself, at rate $\dot m < 0$ . Apply momentum conservation to the closed system (rocket + the sliver of fuel ejected in $\mathrm{d}t$ ): in that interval the engine throws mass $|\dot m|\,\mathrm{d}t$ backward with relative speed $v_e$ , so the rocket must gain forward momentum $|\dot m|\,v_e\,\mathrm{d}t$ . With an external force $\vec F_{\text{ext}}$ also acting,

m\,\frac{\mathrm{d}\vec v}{\mathrm{d}t} = \vec F_{\text{ext}} + \vec v_{\text{rel}}\,\frac{\mathrm{d}m}{\mathrm{d}t}

Why the naive route fails: writing $F = m(t)\,a$ omits the thrust term $\vec v_{\text{rel}}\,\dot m$ entirely, and writing $F = \dot m \vec v + m\dot{\vec v}$ with the rocket's own $\vec v$ gets a frame-dependent nonsense answer. The exhaust velocity relative to the rocket is the physical quantity, and $|\dot m|\,v_e$ — mass flow times exhaust speed — is the thrust: the momentum form handles a machine that loses 90% of its mass in flight without ever needing a fictional "honest m" for $F = ma$ . *Verified: Steps 1–2 match the standard reduction (Kleppner & Kolenkow Ch. 3, Morin Ch. 5); the rocket equation of Step 4 integrates to Tsiolkovsky's $\Delta v = v_e \ln(m_0/m_f)$ , the standard result.*

Go deeper · Derivation B · 1 minApparent weight in an elevator

Derivation BApparent weight in an elevator

The goal is the apparent-weight formula $N = m(g + a)$ — the single line that settles every elevator, rocket-launch, and "why do astronauts float" question, and the classic first demonstration that a scale does not measure gravity. The strategy is the four-step ritual at its purest: isolate the passenger, list exactly two forces, write the vertical component equation (2), and solve for the scale's push. The step to watch is Step 1: the elevator cable, the motor, the building — none of them touch the passenger. Only the scale and the Earth do, and confusing "forces on the elevator" with "forces on the person" is the standard way to get this wrong.

Step 1. Isolate the passenger standing on a scale. Two forces only: the scale's upward push $N$ — which is what the scale reads — and gravity $mg$ down. Why: a free-body diagram of the person; the cable pulls the elevator, not the passenger.

Step 2. Write the vertical component of the second law, taking up as positive, with $a$ the (signed) upward acceleration of person-plus-elevator:

N - mg = ma

Why: person and scale share the elevator's acceleration, so the imbalance of the two forces must equal $ma$ — the law never asks about the velocity, only the acceleration.

Step 3. Solve for the reading:

N = m(g + a) \qquad (8)

Why: one-line rearrangement — the scale must both support the weight and supply the acceleration.

Step 4. Read the three regimes. Accelerating upward ( $a > 0$ ): $N > mg$ , you feel heavy. Constant velocity ( $a = 0$ ): $N = mg$ , normal — the direction and speed of travel are irrelevant. Accelerating downward ( $a < 0$ ): $N < mg$ , light — and in free fall ( $a = -g$ ), $N = 0$ : weightlessness with gravity fully switched on. Why: the scale reads the contact force, not gravity — astronauts float for exactly this reason. *Verified: with $m = 70\ \mathrm{kg}$ and $a = +2.0\ \mathrm{m/s^2}$ , $N = 70 \times 11.8 = 826\ \mathrm{N}$ versus $686\ \mathrm{N}$ at rest — the numbers in the quiz below; matches Halliday–Resnick–Walker Ch. 5 and Morin Ch. 3.*

Go deeper · Derivation C · 1 minCentripetal force for uniform circular motion

Derivation CCentripetal force for uniform circular motion

We are deriving the force law of circular motion — that a body circling at constant speed requires a net force of magnitude $mv^2/r$ pointing at the centre — using nothing but equation (1) and two differentiations. The strategy is to write the position on the circle explicitly, differentiate twice, and let the second law read off the force; no new physics enters, which is precisely the lesson. The step to watch is the direction result of Step 2: the acceleration comes out antiparallel to the position vector, radially inward, even though the speed never changes — the force is spent entirely on turning the velocity, not growing it. This is the acceleration-versus-velocity distinction of the warning box, promoted to a workhorse formula.

Step 1. Parametrise uniform circular motion of radius $r$ at angular speed $\omega$ :

\vec r(t) = r\big(\cos\omega t\,\hat{x} + \sin\omega t\,\hat{y}\big)

Why: the components trace the circle with $|\vec r| = r$ constant; one full lap takes $T = 2\pi/\omega$ .

Step 2. Differentiate twice:

\ddot{\vec r} = -\omega^2 r\big(\cos\omega t\,\hat{x} + \sin\omega t\,\hat{y}\big) = -\omega^2\,\vec r

Why the minus sign is the physics: each differentiation of sine and cosine advances the phase by a quarter turn, so two of them point the acceleration exactly opposite to the position — radially inward, toward the centre, at every instant.

Step 3. Insert into the second law:

\vec F = m\ddot{\vec r} = -m\omega^2\,\vec r

Why: nothing about circular motion is exempt from equation (1); if the path curves, some real agent — string tension, gravity, friction, a banked road — must be supplying exactly this inward force, and if it stops, the body flies off along the tangent.

Step 4. Express through the speed $v = \omega r$ :

|\vec F| = m\omega^2 r = \frac{mv^2}{r} \qquad (9)

Why the $v^2$ matters in practice: doubling your cornering speed quadruples the force the tyres must supply — the reason speed, not radius, is what kills on curves. *Verified: matches Kleppner & Kolenkow Ch. 2 and Morin Ch. 3; consistent with the kinematic result $a = v^2/r$ , and dimensionally $\mathrm{kg\,(m/s)^2/m} = \mathrm{N}$ .*

Go deeper · Derivation D · 1 minGalilean invariance — why every inertial observer agrees on the law

Derivation DGalilean invariance — why every inertial observer agrees on the law

The goal is to show that Newton's second law has exactly the same form for every observer moving at constant velocity — the principle of Galilean relativity, and the reason physics done on a smoothly cruising train is indistinguishable from physics done on the platform. The strategy is to transform the position between two inertial frames and watch what survives two time derivatives. The result also marks the law's boundary: the derivation visibly fails if the relative velocity $\vec V$ is not constant, which is precisely why accelerating and rotating frames need fictitious forces — and it quietly assumes time and mass are frame-independent, the two assumptions special relativity later revokes.

Step 1. Relate the coordinates of the two frames. If frame $S'$ moves at constant velocity $\vec V$ relative to frame $S$ ,

\vec r\,' = \vec r - \vec V t

Why: this is the Galilean transformation — the primed observer sees every position shifted by how far their own frame has travelled.

Step 2. Differentiate twice:

\dot{\vec r}\,' = \dot{\vec r} - \vec V, \qquad \ddot{\vec r}\,' = \ddot{\vec r}

Why the accelerations agree: the constant $\vec V$ shifts every velocity by the same fixed amount but contributes nothing to the second derivative. Velocities are relative; accelerations are absolute (among inertial observers).

Step 3. Apply the second law in each frame with the same mass:

\vec F\,' = m\ddot{\vec r}\,' = m\ddot{\vec r} = \vec F

Why this is the punchline: both observers write $\vec F = m\vec a$ with the same $\vec F$ and the same $\vec a$ — the law is form-invariant, so no mechanical experiment can detect uniform motion. Galileo's ship argument, in three lines.

Step 4. Note where it breaks. If $\vec V$ depends on time, Step 2 gains an extra term $-\dot{\vec V}$ , and the primed observer measures $m\vec a\,' = \vec F - m\vec A$ : a phantom force $-m\vec A$ with no physical agent, proportional to mass exactly as gravity is. Why this matters: it is the origin of every fictitious force — the push you feel in a braking bus, and (for rotating frames) the centrifugal and Coriolis forces. *Verified: standard result (Landau & Lifshitz §1–§5, Goldstein Ch. 1); the failure mode of Step 4 matches the non-inertial-frame entry of the limiting-cases table.*

Limiting cases

Case	Result	Meaning
$F_{\text{net}} = 0$	$a = 0$	balanced forces: constant velocity — the first law recovered
only gravity acts	$a = g$ , any mass	$mg/m = g$ : all objects fall alike — mass cancels
$m \to$ very large	$a \to 0$	same push, no visible effect — why you can't shove a ship
variable mass (rocket)	$m\dot{\vec v} = \vec F_{\text{ext}} + \vec v_{\text{rel}}\,\dot m$	expelled fuel carries momentum; naive $m(t)\,a$ misses the thrust
$v$ approaching light speed	$\vec F = \dfrac{\mathrm{d}}{\mathrm{d}t}\!\left(\dfrac{m\vec v}{\sqrt{1 - v^2/c^2}}\right)$	inertia effectively diverges as $v \to c$ ; Newtonian $F = ma$ is the $v \ll c$ limit — but the momentum form survives
non-inertial frame	$m\vec a_{\text{rel}} = \vec F - m\vec A - 2m\,\vec\omega\times\vec v_{\text{rel}} - m\,\vec\omega\times(\vec\omega\times\vec r)$	accelerating/rotating observers must add fictitious forces (translational, Coriolis, centrifugal)

Common trap“The object moves in the direction of the net force”

No — the object accelerates in the direction of the net force. Velocity can point anywhere. A ball thrown straight up is moving upward while its net force (gravity) and acceleration point straight down the whole flight — including at the very top, where $v = 0$ but $a = g$ still. A braking car moves forward while its net force points backward. And a satellite in circular orbit perpetually accelerates toward Earth's centre while moving sideways. The same confusion in older clothes: believing a moving object needs a continuous force to keep moving. It doesn't — a body in uniform motion has zero net force (that is the first law); force is needed only to change velocity. If you catch yourself matching force direction to velocity direction, stop and re-draw the free-body diagram: the second law couples force to the change in velocity, never to the velocity itself.

Worked examples

1 · The merging car (confidence builder)

A 1500 kg car accelerates uniformly from rest to $27\ \mathrm{m/s}$ (about 100 km/h) in 9.0 s. What net force acted on it?

Step 1. Acceleration first: $a = \Delta v / \Delta t = 27 / 9.0 = 3.0\ \mathrm{m/s^2}$ .

Step 2. Apply (1): $F_{\text{net}} = ma = 1500 \times 3.0 = \mathbf{4.5\times 10^3\ N}$ .

Step 3. Interpret: this 4.5 kN is the net of the tyre thrust minus drag and rolling resistance. The engine actually delivers more; the road disposes of the rest. For scale, the car's weight is $1500 \times 9.8 = 14.7\ \mathrm{kN}$ — the net forward force is about a third of the car's weight, which is spirited but ordinary.

2 · Two masses, one string (exam level)

A 3.0 kg mass and a 2.0 kg mass hang from the two ends of a light string over a frictionless pulley (an Atwood machine). Find the acceleration and the string tension. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Isolate each mass separately — two free-body diagrams, each with just tension $T$ up and weight down. The string makes their accelerations equal in magnitude: call it $a$ , with the 3.0 kg side falling.

Step 2. Second law for the heavy mass (taking its downward direction as positive): $3.0g - T = 3.0a$ .

Step 3. Second law for the light mass (its upward direction positive, so the same $a$ ): $T - 2.0g = 2.0a$ .

Step 4. Add the two equations — tension cancels: $(3.0 - 2.0)g = (3.0 + 2.0)a$ , so $a = g/5 = \mathbf{1.96\ m/s^2}$ .

Step 5. Back-substitute into Step 3: $T = 2.0(g + a) = 2.0 \times 11.76 = \mathbf{23.5\ N}$ .

Step 6. Cross-check with Step 2's equation: $T = 3.0(g - a) = 3.0 \times 7.84 = 23.5\ \mathrm{N}$ ✓. Note $T$ sits between the two weights (19.6 N and 29.4 N) — it must, or one of the masses wouldn't accelerate the right way. The pattern in Steps 1–3 — one free-body diagram and one second-law equation per object, plus a constraint tying their motions — solves every connected-body problem you will ever meet, from pulleys to trains of wagons.

3 · Block on a rough incline (numeric)

A block slides down a plane inclined at $\alpha = 30°$ with kinetic friction coefficient $\mu_k = 0.25$ . Find its acceleration. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Choose axes along and perpendicular to the incline — this is equation (2) with the smart choice. Resolve the weight: $mg\sin\alpha$ down-slope, $mg\cos\alpha$ into the surface.

Step 2. Perpendicular direction (no acceleration): $N = mg\cos\alpha$ . This hands friction its magnitude, $f = \mu_k N = \mu_k mg\cos\alpha$ , pointing up-slope against the sliding.

Step 3. Along the incline: $ma = mg\sin\alpha - \mu_k mg\cos\alpha$ . The mass cancels:

a = g(\sin\alpha - \mu_k\cos\alpha)

Step 4. Numbers: $a = 9.8\,(0.500 - 0.25 \times 0.866) = 9.8 \times 0.2835 = \mathbf{2.8\ m/s^2}$ , independent of the block's mass.

Step 5. Sanity check the formula's edges: with $\mu_k = 0$ it gives the frictionless $g\sin\alpha$ ; if $\mu_k \ge \tan\alpha$ (here, $\mu_k \ge 0.577$ ) the expression goes non-positive — a block released from rest would simply stay put, with static friction in charge.

4 · Dashboard versus airbag (exam level)

A 60 kg driver is travelling at 15 m/s when the car hits a wall and stops. (a) What momentum change must the driver undergo? (b) Unbelted, she hits the dashboard and stops in 0.020 s — find the average force on her. (c) With seatbelt and airbag she stops in 0.30 s — find the force now, and compare both to her weight.

Step 1. Part (a) is fixed by the crash, not the safety gear: $\Delta p = m\,\Delta v = 60 \times (0 - 15) = -900\ \mathrm{kg\,m/s}$ . Every scenario below must drain exactly 900 kg·m/s from her body; the sign just says "opposite to her motion."

Step 2. Dashboard, using (5): $F = \Delta p/\Delta t = 900/0.020 = \mathbf{45{,}000\ N}$ — roughly 76 times her weight ( $mg = 588\ \mathrm{N}$ ). Bones and organs cannot survive being stopped at 76 g-equivalents concentrated on impact points.

Step 3. Airbag and belt: $F = 900/0.30 = \mathbf{3{,}000\ N}$ — about 5 times her weight, spread across chest and face. Painful, survivable, walk-away territory.

Step 4. The ratio of forces is exactly the inverse ratio of times: $45{,}000/3{,}000 = 15 = 0.30/0.020$ . The airbag changed nothing about the crash's momentum arithmetic — it only bought time, fifteen times more of it, and equation (5) converted every extra millisecond into subtracted force. That is the entire, complete physics of why airbags work.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1642–1727 · Cambridge, then London

Newton stated the second law in the Principia (1687) in momentum language: the change of motion is proportional to the motive force impressed, and directed along its line. Remarkably, the equation " $F = ma$ " never appears in the book — geometry, not algebra, was his instrument. What made the law revolutionary was its scope: the same rule that guides a thrown stone reproduces the Moon's orbit, once his gravity supplies the force. The working form you use, differential equations solved from initial conditions, was distilled by Euler and the continental mathematicians half a century later. And Newton's momentum instinct proved uncannily durable: when Einstein rebuilt mechanics in 1905, it was $F = ma$ that broke at high speed, while "force equals rate of change of momentum" survived — needing only a smarter formula for $p$ .

History

1638

— Galileo's Two New Sciences shows that a constant push yields uniform acceleration, clearing the ground for a quantitative force law.

1687

— Newton's Principia, Lex II: change of motion is proportional to the motive force impressed — the second law, in momentum form.

1750

— Euler recasts the law as the differential equations

F = ma

along each axis, the form every physicist and engineer now solves.

1846

— Le Verrier predicts Neptune's position from irregularities in Uranus's orbit using pure second-law dynamics; the planet is found within a degree.

1969

— Apollo 11's guidance computer integrates

F = ma

in real time to fly humans to the lunar surface and home.

Going further

Texts. Morin, Introduction to Classical Mechanics, Ch. 3 · Goldstein, Classical Mechanics, 3rd ed., Ch. 1 · Landau & Lifshitz, Mechanics (Vol. 1), §1–§5 · Taylor, Classical Mechanics, Ch. 1–2 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 2 · Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 5 and 9.

Provenance. Equations are standard textbook results. A verifier should: (1) confirm $\mathbf{F}=m\mathbf{a}$ follows from $\mathbf{F}=\mathrm{d}\mathbf{p}/\mathrm{d}t$ for constant mass; (2) recompute the projectile trajectory and range $R=v_0^2\sin2\theta/g$ ; (3) verify centripetal force $mv^2/r$ from differentiating circular motion; (4) check the incline acceleration $g(\sin\alpha-\mu_k\cos\alpha)$ ; (5) verify the newton's dimensions $\mathrm{kg\,m\,s^{-2}}$ . Cross-check against Morin Ch. 3, Taylor Ch. 1–2, Kleppner & Kolenkow Ch. 2, Goldstein Ch. 1, and Landau–Lifshitz Mechanics §1–§5. Worked-example numbers on this page (the car's $4.5\ \mathrm{kN}$ ; the Atwood $a = 1.96\ \mathrm{m/s^2}$ , $T = 23.5\ \mathrm{N}$ ; the incline $a = 2.8\ \mathrm{m/s^2}$ at $\alpha = 30°$ , $\mu_k = 0.25$ ; the crash forces $45{,}000\ \mathrm{N}$ vs $3{,}000\ \mathrm{N}$ for $\Delta p = 900\ \mathrm{kg\,m/s}$ ) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Velocity and Acceleration · Vectors in Kinematics and Coordinate Systems · Newton's First Law and Inertial Frames · One-Dimensional Kinematics. Leads to: Newton's Third Law · Projectile Motion · Circular Motion and Centripetal Acceleration · Friction Forces · Gravitational Force Near Earth's Surface.

If I stop pushing a box, it stops. Doesn't that prove motion needs a force?

It proves friction exists, not that motion needs a force. While the box slides, two horizontal forces act: your push forward and friction backward. Remove your push and the net force is no longer zero — it is pure friction, pointing backward, so the box decelerates exactly as the second law demands. On smoother and smoother surfaces the same shove carries the box further and further; on ice it glides for metres, and a hockey puck on good ice barely slows at all. Extrapolate to zero friction and the box would slide forever at constant velocity — which is Newton's first law, and precisely the $F_{\text{net}} = 0 \Rightarrow a = 0$ row of the limiting-cases table. Aristotle read the friction-filled world at face value; Newton saw through it.

Isn't F = ma circular? Force is defined by the acceleration it produces, so how can the law predict anything?

The escape from circularity is that force laws exist independently. Read one way, the second law is indeed a definition: it calibrates force in terms of the acceleration of a standard kilogram. But nature separately supplies formulas for specific forces — $GMm/r^2$ for gravity, $-kx$ for a spring, $\mu N$ for friction — each depending on positions, materials, and charges, never on the acceleration you are about to compute. The predictive content lives in the marriage: take the force laws, insert them into $m\ddot{\vec r} = \vec F(\vec r, \dot{\vec r}, t)$ , and you get a differential equation whose solutions must match experiment. That the same calibration of "force" makes springs, gravity, and friction all give correct trajectories is a highly non-trivial fact about the world, not a definition.

Newton's third law says every force has an equal and opposite partner. Don't they all cancel, making acceleration impossible?

Third-law partners act on different bodies, so they can never cancel in any one body's free-body diagram. When a horse pulls a cart, the cart does pull the horse back equally hard — but the second law for the cart contains only the forces on the cart (the pull forward, friction backward), and the pull wins. The horse's own equation contains the cart's backward tug and the ground's forward push on its hooves; the ground wins. Each body accelerates because the forces on it don't balance. When you sum the second law over a whole system, the internal pairs do cancel — which is exactly why only external forces can change a system's total momentum, the seed of momentum conservation.

Why does mass cancel in free fall and on inclines, but matter enormously when stopping a truck?

It depends on whether the forces themselves scale with mass. In free fall the only force is gravity, $mg$ , so $a = mg/m = g$ — the driving force grows in exact proportion to the inertia it must move, and every object falls alike. On an incline both the driving component $mg\sin\alpha$ and the friction ceiling $\mu mg\cos\alpha$ carry the same $m$ , so it cancels again: worked example 3's answer never asked the block's mass. But stopping a truck is a momentum job: $\Delta p = m\,\Delta v$ scales with $m$ with nothing to cancel against, so a loaded truck needs proportionally more impulse — more force, or more time — than an empty one. Rule of thumb: mass cancels only when every force in the problem is proportional to it.

Which mass do I put into F = ma for a rocket that is burning fuel?

None — that is the wrong question, and the momentum form exists to replace it. At any instant you can certainly write $m(t)\,\vec a = \vec F_{\text{thrust}} + \vec F_{\text{gravity}} + \dots$ with the current mass, but the thrust itself cannot be found from $F = ma$ ; it comes from momentum bookkeeping. In each interval $\mathrm{d}t$ the engine hurls mass $|\dot m|\,\mathrm{d}t$ backward at exhaust speed $v_e$ relative to the rocket, and by momentum conservation the rocket gains $|\dot m|\,v_e\,\mathrm{d}t$ forward — so thrust $= |\dot m|\,v_e$ . Integrated over a burn this yields Tsiolkovsky's rocket equation $\Delta v = v_e\ln(m_0/m_f)$ : the payoff for trusting $\mathrm{d}p/\mathrm{d}t$ where $ma$ has no honest $m$ to use.

Why do we say F = ma fails near the speed of light but the momentum form survives?

Experimentally, pushing on a particle near $c$ produces less and less acceleration — no finite force accelerates anything through the light barrier — so the strict proportionality $a = F/m$ with constant $m$ is simply wrong there. But Newton's actual statement, $\vec F = \mathrm{d}\vec p/\mathrm{d}t$ , remains exactly correct in special relativity once momentum gets its relativistic formula, $\vec p = m\vec v/\sqrt{1 - v^2/c^2}$ . As $v \to c$ that denominator shrinks, momentum grows without bound, and each newton of force buys ever less speed — matching what accelerators observe daily. Newton's framing survived a revolution that destroyed the schoolbook shortcut, needing only a smarter $p$ .

A bus brakes and I lurch forward, with no force pushing me. Has the second law failed?

The law hasn't failed — your frame has. The second law holds in inertial frames, and a braking bus is not one. Watched from the street, nothing pushed you: your body simply continued at its previous velocity (first law) while the bus decelerated around you; the "lurch" is the bus falling behind your motion. Inside the bus, if you insist on doing mechanics in its decelerating frame, you must add a fictitious force $-m\vec A$ (with $\vec A$ the bus's acceleration) to every object — that phantom forward push, proportional to mass, with no physical agent. The Galilean-invariance derivation shows exactly where it comes from: the proof breaks the moment the relative velocity between frames stops being constant.

Newton's Second Law

Where this lives today

The physical picture

The math

Which formula answers which question

Limiting cases

Worked examples

1 · The merging car (confidence builder)

2 · Two masses, one string (exam level)

3 · Block on a rough incline (numeric)

4 · Dashboard versus airbag (exam level)

Going further

Before this

Where this leads