Newton's laws

Kinematics can describe any motion whatsoever and predict none of them, because the acceleration has to be supplied from outside, and the three laws published by Newton in 1687 are the supply.

This lesson assumes only that acceleration is the second derivative of position, which the first two lessons built. It is the least computational lesson in the course and the most important, because every calculation after it is an application of one equation stated here.

The wrong answer that lasted two thousand years

Aristotle's physics held that a body's natural state is rest, and that continued motion requires a continued cause. It is a reasonable summary of the evidence available to anyone pushing a cart on a road: stop pushing and the cart stops. It also has an obvious embarrassment, the thrown stone, which keeps moving after the hand has let go. The standard repair was that the air closes behind the stone and drives it forward, which is worse than the problem it solves.

The medieval alternative, developed by Jean Buridan in Paris around 1350, was impetus: the thrower imparts to the stone a quantity that keeps it going and gradually runs out. This is closer, and it survived because it makes roughly the right predictions for real projectiles in real air. What it gets wrong is the running out. Impetus is spent by the motion itself; momentum is not spent at all, and is only changed by something else pushing.

Galileo got the decisive result by a thought experiment about inclined planes. A ball rolling down one incline and up another rises to nearly the height it started from, whatever the second incline's slope; make the second incline gentler and the ball travels further to reach the same height; make it horizontal and there is no height to reach, so the ball should travel forever. He could not test this, because there is always friction, and the argument works by extrapolating friction away. That extrapolation is the birth of theoretical physics: the claim is not about any ball anyone has rolled, but about the limit that no experiment can reach.

The first law: inertial frames exist

Newton's first law, in the Principia of 1687, states that a body continues in its state of rest, or of uniform motion in a straight line, unless compelled to change that state by forces impressed upon it.

Read carelessly, this is a special case of the second law with $\vec{F} = 0$ , and therefore redundant. Read properly it is a separate and stronger claim, because both laws are silently about some frame of reference, and the first law is what says a suitable frame exists.

Consider a ball resting on the floor of a train. If the train brakes, the ball rolls forward with no one touching it. In the frame of the train, a body with no force on it has accelerated, so the first law is false in that frame. The law is therefore not a universal truth about bodies; it is a test that selects frames. A frame in which an isolated body stays at uniform velocity is called an inertial frame, and the first law asserts that such frames exist. Once one is found, any frame moving at constant velocity with respect to it is another, since a constant added to a velocity vanishes on differentiation.

This matters because the second law is only true in an inertial frame. Everything else in this course is a calculation done in one, and the honest question is whether the ground under our feet qualifies. It does not exactly. The Earth spins, so a point on the equator is accelerating towards the axis at $ω^{2} R$ , with $ω = 2 π / 86164 = 7.292 \times 10^{- 5}$ radians per second and $R = 6.378 \times 10^{6}$ m, giving $0.0339$ m s⁻², about 0.35 per cent of $g$ . The Earth also orbits the Sun, adding $5.9 \times 10^{- 3}$ m s⁻². Both are far too small to notice while dropping a stone, and far too large to ignore in a Foucault pendulum, in ballistics at long range, or in the circulation of the atmosphere.

The second law: force and mass, defined together

The second law is the working equation of the whole subject:

\sum \vec{F} = m \vec{a}

the vector sum of all forces on a body equals its mass times its acceleration. Being a vector equation, it is really one equation per axis, $\sum F_{x} = m a_{x}$ and $\sum F_{y} = m a_{y}$ , which is what makes the component method of the previous lesson pay off. The unit of force follows: one newton is the force that gives one kilogram an acceleration of one metre per second squared, so 1 N = 1 kg m s⁻².

There is a circularity here that textbooks often slide past, and it is worth facing. What is a force? Something that causes acceleration. What is mass? The resistance to acceleration by a force. Each term is defined by the other, so as it stands the law says nothing falsifiable.

The way out is experimental, and it comes in two steps. First, mass can be compared without knowing anything about forces at all: let two bodies interact with each other alone, on an air track, and measure the two accelerations. Experiment says the ratio $a_{1} / a_{2}$ is always the same for the same pair of bodies, whatever the interaction is, whether they collide, or repel by magnets, or are joined by a spring. Define the mass ratio as the inverse of the acceleration ratio, $m_{2} / m_{1} = a_{1} / a_{2}$ , pick one body as the standard, and every mass in the world follows by comparison. Since May 2019 the standard is not a metal cylinder in Sèvres but the fixed value of the Planck constant, $6.62607015 \times 10^{- 34}$ J s, which fixes the kilogram through measurement rather than through an object that could be scratched.

Second, with masses known, the law becomes a real claim about forces: measure the acceleration a spring stretched by 2 cm gives to a known mass, and the law predicts what it gives to any other. That prediction can fail, and does not. The content of the second law is not the algebra but the discovery that the same number $m$ works for every kind of force applied to a given body, and that the same force gives a given body the same acceleration regardless of what else is happening to it. Forces add as vectors: that is a physical finding, not a definition.

Example. A 3.0 kg block on a frictionless horizontal surface is pulled by a 12 N force along the positive $x$ axis and simultaneously by a 9.0 N force at 60° to it. Find the acceleration.

Resolve into components. The 12 N force is $(12, 0)$ . The 9.0 N force is $(9 \cos 60^{\circ}, 9 \sin 60^{\circ}) = (4.50, 7.79)$ N. The sum is $(16.50, 7.79)$ N, whose magnitude is $\sqrt{{16.50}^{2} + {7.79}^{2}} = 18.25$ N at $\arctan (7.79 / 16.50) = {25.3}^{\circ}$ from the $x$ axis. Dividing by the mass, $a = 18.25 / 3.0 = 6.08$ m s⁻² in that same direction. The direction of the acceleration is the direction of the net force, always, and never the direction of the largest single force.

Now you. A 2.5 kg block is pulled by 15 N along $x$ and 8.0 N along $y$ , with no friction. Find the magnitude and direction of its acceleration.

Answer

The net force is $(15, 8)$ N, of magnitude $\sqrt{225 + 64} = 17.0$ N at $\arctan (8 / 15) = {28.1}^{\circ}$ from the $x$ axis. The acceleration is $17.0 / 2.5 = 6.80$ m s⁻² in that direction.

Mass, weight, and what a scale measures

Mass is a property of a body and is the same everywhere. Weight is the gravitational force on it, $W = m g$ , and depends on where it is. A 70 kg astronaut weighs 687 N on Earth, 114 N on the Moon where $g = 1.62$ m s⁻², and has a mass of 70 kg in both places. Confusing the two is harmless in ordinary speech and fatal in a calculation, because it is $m$ and not $W$ that appears on the right of the second law.

A bathroom scale does not measure either. It measures the normal contact force it exerts on your feet, and reports that force divided by 9.81 as though you were not accelerating. When you are, the reading is wrong in an informative way, and the whole of the sensation of a lift is contained in this.

Example. A 70 kg person stands on a scale in a lift accelerating upward at 2.0 m s⁻². What does the scale read, in newtons?

Two forces act on the person: gravity $m g = 687$ N down, and the normal force $N$ up. Take up as positive, so the second law gives $N - m g = m a$ , and $N = m (g + a) = 70 (9.81 + 2.0) = 827$ N. The scale reads 827 N, twenty per cent above the standing value, which is the pressed-into-the-floor feeling on starting to rise. If instead the cable snapped and the lift fell freely at $a = - g$ , then $N = 0$ : the person floats, weightless, while the gravitational force on them has not changed at all. Weightlessness in orbit is exactly this, and not the absence of gravity.

Now you. The same person is in a lift accelerating downward at 1.5 m s⁻². What does the scale read?

Answer

$N = m (g - a) = 70 (9.81 - 1.5) = 582$ N, about fifteen per cent light.

The third law and the pair it refers to

If body A exerts a force on body B, then B exerts on A a force equal in magnitude and opposite in direction, along the same line. The two forces are of the same kind, they act at the same instant, and, decisively, they act on different bodies.

That last clause is what makes the law useful and what makes it constantly misapplied. The stock objection is that a horse cannot pull a cart, since the cart pulls back equally hard and the two cancel. They do not cancel, because they never appear in the same equation: the horse's pull acts on the cart, and the cart's pull acts on the horse. Ask what accelerates the cart and the answer involves only forces on the cart. Ask what accelerates the horse and the answer involves the cart's backward pull and the ground's forward push on its hooves, which is a different pair.

The reliable test is to name both bodies for every force: not "the weight of the book" but "the Earth pulls the book down". Its partner is then automatic: the book pulls the Earth up with an equal force. The normal force of the table on the book is not that partner, despite being equal and opposite in the common case of a book at rest, and the giveaway is that it stops being equal the moment you press down on the book, while the true third law partner remains equal always.

The Earth really does accelerate upward towards a dropped stone. For a 1 kg stone the force on the Earth is 9.81 N, and dividing by $5.97 \times 10^{24}$ kg gives $1.6 \times 10^{- 24}$ m s⁻², which is why nobody notices.

Momentum, and the form Newton actually used

Newton did not write $F = m a$ . He defined the quantity of motion as mass times velocity, what we call the momentum $\vec{p} = m \vec{v}$ , and stated the second law as the claim that the change of motion is proportional to the impressed force:

\sum \vec{F} = \frac{d \vec{p}}{d t}

For constant mass this gives $m d \vec{v} / d t = m \vec{a}$ and the two forms agree. When mass is not constant, as for a rocket burning fuel or a rope being lifted onto a table link by link, the momentum form is the one to reason with, though even then it must be applied to a fixed collection of matter rather than to a shrinking body, a subtlety a later lesson handles properly.

The momentum form also makes the third law say something startling. If A and B interact and nothing else acts, then $d {\vec{p}}_{A} / d t = - d {\vec{p}}_{B} / d t$ , so the total momentum ${\vec{p}}_{A} + {\vec{p}}_{B}$ has zero derivative and never changes. Conservation of momentum is not an extra law; it is the third law rewritten. Two lessons from now that observation carries the whole of collision theory.

Example. A rifle of mass 4.0 kg fires a 12 g bullet at 850 m s⁻¹. What is the recoil speed of the rifle, and what average force acts on the shoulder if the recoil is stopped in 0.15 s?

Total momentum starts at zero and the third law keeps it there, so $m_{b} v_{b} = m_{r} v_{r}$ , giving $v_{r} = (0.012) (850) / 4.0 = 2.55$ m s⁻¹ backward. Stopping that in 0.15 s requires a force of $Δ p / Δ t = (4.0) (2.55) / 0.15 = 68$ N, which is comfortable. The bullet leaves with the same 10.2 kg m s⁻¹ of momentum, but with vastly more energy, and the reason for that asymmetry is the subject of a later lesson.

Now you. A 60 kg skater standing at rest on frictionless ice throws a 3.0 kg ball forward at 8.0 m s⁻¹. How fast does the skater move backward?

Answer

Momentum starts at zero, so $60 v = (3.0) (8.0) = 24$ kg m s⁻¹ and $v = 0.40$ m s⁻¹ backward.

An equation with an empty right hand side

Write the second law out as what it is, a differential equation:

m \frac{d^{2} \vec{r}}{d t^{2}} = \sum \vec{F}

Given the forces as functions of position, velocity and time, plus the initial position and velocity, the solution is unique and the entire future of the body follows. That is the claim of determinism that made mechanics the model for every science that came after, and it survived intact until quantum mechanics and, in a different way, until the discovery that some solutions depend so sensitively on the initial conditions that predicting them requires knowing the start to impossible precision.

None of that power is available yet, because the right hand side is empty. The laws say what force does, not what forces there are. Gravity near the ground, the push of a surface, the pull of a rope, the resistance of friction and the drag of air are separate empirical discoveries, each with its own formula and its own range of validity, and assembling that catalogue is the next lesson's work. It is also where the method that makes mechanics tractable appears: draw one body, name every force on it, and turn the picture into two equations.