Newton's Second Law
You already know two big ideas. First, that when the forces on an object don't cancel, there is a
leftover resultant force.
Second, that accelerating
means changing velocity — speeding up, slowing down, or turning. Newton's Second Law is the bridge
between them: it says exactly how big an acceleration a given resultant force produces.
Push a shopping trolley and it speeds up. Push harder and it speeds up faster. Load it with bricks and
the same push barely gets it going. Everyone knows this in their bones — Newton's genius was to write
it down as a clean, exact rule you can put numbers into. Two things decide how quickly something
accelerates: how hard you push it (the resultant force) and how much stuff it
is made of (its mass).
The law itself
Newton found that the acceleration of an object is set by the resultant force acting on it and by its
mass, tied together in one short equation. In words: force equals mass times acceleration.
F = m\,a
Every symbol has a unit that must be used for the numbers to come out right:
- F — the resultant force, measured in newtons (N).
- m — the mass, measured in kilograms (kg).
- a — the acceleration, measured in
metres per second squared (m/s²).
In fact a newton is defined by this law: 1 N is exactly the resultant force
that gives a 1 kg mass an acceleration of 1 m/s². So the units lock
together perfectly — 1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2 — and if you
feed in kilograms and metres-per-second-squared, the answer is guaranteed to be in newtons.
- The resultant force on an object equals its mass multiplied by its acceleration:
F = m\,a.
- The acceleration is in the same direction as the resultant force.
- F is in newtons (N), m in kilograms (kg),
and a in metres per second squared (m/s²).
Reading the two stories inside F = m\,a
A single equation, but it carries two everyday truths at once. Keep one of the three quantities fixed
and watch what the other two do.
Same mass, bigger force → bigger acceleration. If the mass stays put, then
a is proportional to F: double the
resultant force and you double the acceleration; treble it and you treble the acceleration. Push a
2 kg mass with 4 N and it accelerates at 2\ \text{m/s}^2; push the same mass
with 8 N and it accelerates at 4\ \text{m/s}^2. Twice the shove, twice the
pick-up.
Same force, bigger mass → smaller acceleration. If instead the force stays put, then
a is inversely proportional to m: double
the mass and you halve the acceleration. A 12 N force on a 2 kg mass gives
6\ \text{m/s}^2; the very same 12 N on a 6 kg mass gives only
2\ \text{m/s}^2. More stuff to shift, so the same push achieves less.
These four slips catch out almost everyone meeting the law:
-
The F in F = m\,a is the RESULTANT
force, not just one push. If a 5 kg box is shoved forwards with 30 N while friction drags
back 10 N, the force you put in the formula is the leftover
30 - 10 = 20\ \text{N}, giving
a = 20/5 = 4\ \text{m/s}^2 — not 30/5. Always
find the resultant first.
-
Force causes acceleration, not velocity. A resultant force does
not set the speed — it sets the change of speed. A spacecraft coasting at
20 000 m/s with no resultant force keeps that speed forever; it doesn't need a force to stay
fast, only to change.
-
No resultant force means no acceleration — but it can still be moving. Zero
resultant force gives a = 0, which means steady velocity: either sitting
still or gliding at a constant speed in a straight line, not necessarily stopped.
-
Doubling the mass halves the acceleration. Because a = F/m,
mass sits underneath. It's easy to assume "twice the mass, twice the acceleration" — but
it's the exact opposite.
Rearranging the law
F = m\,a is a triangle of three quantities: give any two and you can always
find the third. Rearranged, the same law reads:
a = \dfrac{F}{m}, \qquad m = \dfrac{F}{a}.
Use whichever version fits the question. If you know the force and the mass and want the acceleration,
reach for a = F/m. If you know the force and the acceleration and want the
mass, use m = F/a. And if you already have the mass and the acceleration and
want the force, the original F = m\,a is ready to go.
A quick tip for getting them right: cover the quantity you want in the words "F over m-a". Cover
F and you see m \times a; cover
m and you see F over a;
cover a and you see F over
m.
Worked examples
Example 1 — find the force. A 2\ \text{kg} ball is
accelerated at 3\ \text{m/s}^2. What resultant force is needed? Use
F = m\,a directly:
F = m\,a = 2 \times 3 = 6\ \text{N}.
Example 2 — find the acceleration. A 1200\ \text{N} resultant
force acts on a 800\ \text{kg} car. Rearrange to
a = F/m:
a = \dfrac{F}{m} = \dfrac{1200}{800} = 1.5\ \text{m/s}^2.
Example 3 — find the mass. A resultant force of 50\ \text{N}
gives a trolley an acceleration of 2.5\ \text{m/s}^2. Rearrange to
m = F/a:
m = \dfrac{F}{a} = \dfrac{50}{2.5} = 20\ \text{kg}.
Example 4 — friction first, then the law. A 5\ \text{kg}
sledge is pulled with 40\ \text{N} while friction resists with
15\ \text{N}. First find the resultant:
F_{\text{resultant}} = 40 - 15 = 25\ \text{N},
and only then apply the law:
a = \dfrac{F}{m} = \dfrac{25}{5} = 5\ \text{m/s}^2.
Example 5 — the proportion at work. A 10\ \text{N} force on
a 4\ \text{kg} mass gives
a = 10/4 = 2.5\ \text{m/s}^2. Double the force to
20\ \text{N} and the acceleration doubles to
5\ \text{m/s}^2; instead double the mass to
8\ \text{kg} and the acceleration halves to
1.25\ \text{m/s}^2. Two ways to change a, pulling in
opposite directions.
Try it: push a box
Here is a box on the ground with a single resultant force arrow pushing it to the
right. Set the force F in newtons and the mass m in
kilograms with the sliders, and the figure works out the acceleration
a = F/m live, drawing an acceleration arrow whose length grows with
a. Turn the force up and watch a climb; then pile on
mass with the force held fixed and watch the very same push produce less and less acceleration.
Weight is Newton's Second Law in disguise
Drop anything near the Earth and it accelerates downwards at about
g = 10\ \text{m/s}^2 (more precisely 9.8). What resultant force produces that
acceleration? By F = m\,a, a mass m falling at
g must feel a force
W = m\,g.
That downward force is exactly the object's weight.
So W = m\,g is not a separate rule to memorise — it is just
F = m\,a with the acceleration set to gravity's value. A
2\ \text{kg} bag weighs
W = 2 \times 10 = 20\ \text{N} on Earth; take it to the Moon where
g \approx 1.6\ \text{m/s}^2 and its weight drops to about
3.2\ \text{N}, even though its mass — the amount of stuff — hasn't changed at
all.
Inertial mass: mass is resistance to acceleration
Look again at a = F/m. The mass is the thing that resists being
accelerated: for a given force, the bigger the mass, the smaller the acceleration you get. Seen this
way, mass is a measure of an object's reluctance to change its motion. Physicists call
this the inertial mass, and Newton's Second Law is really its definition — mass is
precisely F/a, the force needed per unit of acceleration.
This is why a fully loaded lorry pulls away so ponderously from the lights while an empty one leaps
forward on the same engine, and why it is so much harder to get a heavy shopping trolley rolling — or to
stop it once it is. The engine or your arms supply roughly the same force either way; it is the mass that
decides how much acceleration that force can buy.
A fully laden articulated lorry can mass 40 000 kg. Even a mighty engine pushing with, say,
20 000 N of resultant force only manages
a = 20\,000 / 40\,000 = 0.5\ \text{m/s}^2 — a gentle crawl, which is why big
trucks take so long to reach motorway speed and need such enormous room to stop. Empty the same truck
to 10 000 kg and that identical push now gives
a = 20\,000 / 10\,000 = 2\ \text{m/s}^2, four times livelier. Nothing about
the engine changed — only the mass being shifted.
The law also runs backwards, and that is what crumple zones exploit. In a crash a car
must lose its velocity, so it must decelerate — an acceleration, and by
F = m\,a a big deceleration means a big force on the passengers. A crumple
zone deliberately folds and collapses to stretch out the time the stopping takes, which makes
the deceleration a gentler, and so shrinks the force
F felt by the people inside. Same change in velocity, spread over more time,
equals a smaller — survivable — force.