← All lessons
Introductory Physics · Feynman Vol. I · Ch. 12

Characteristics of Force

This lesson follows chapter 12 of the Feynman Lectures, which asks what a force is and where it comes from. The second law only says something when the force can be measured by some route other than the acceleration itself, and so it pays to look at forces one at a time: friction, the forces between atoms that lie behind it and behind springs, the electric and magnetic fields, and the forces that appear only to someone measuring from inside an accelerated frame. The original chapter can be read free of charge on the Caltech website.

  1. 1F = ma
  2. 2Friction
  3. 3Atoms
  4. 4Fields
  5. 5Pseudo-forces
STEP 1

Is F = ma a law or a definition?

The bench puts a 1 kg cart on a track and pulls it with a 20 N/m spring. The other end of the spring is held by someone walking ahead of the cart, who adjusts their pace so that the spring stays 0.10 m longer than at rest. The spring was calibrated beforehand by hanging known weights from it, so we know how hard it pulls at each stretch: at 0.10 m, 2 N. Press Pull and the velocity climbs along a straight line on the graph, 2 m/s every second, up to 4 m/s at the end of the 2 s run.

With the spring stretched 0.20 m, the next line comes out twice as steep, 4 m/s². With the stretch back at 0.10 m and a 2 kg cart, the slope halves, to 1 m/s². Earlier runs stay on the graph in grey, and in every one of them the mass times the measured acceleration reproduces the spring’s \(k\,x\), as the message under the bench checks after each run.

spring force   measured velocity   prediction \(k\,x/m\)   earlier runs

None of this would be news if force were only a name for the product \(m\,a\). One would measure the acceleration, multiply by the mass and call the result a force, and the second law would hold by construction, unable to go wrong about any cart. It starts to claim something once force has a measure of its own, such as the stretch of a calibrated spring, because then the prediction \(k\,x/m\) can be set beside the slope of the line, and it could fail. The test lies in using the same spring, at the same stretch, on bodies of different mass: if it always exerts the same force, \(m\,a\) has to give the same number for all of them, and a world in which that failed is perfectly conceivable.

The third law adds a requirement that no definition guarantees either. The cart pulls the spring backwards as hard as the spring pulls it forwards, and the same holds between spring and hand, so what is measured on one body tells us the size of a force acting on another. The secondary-school lesson on the second law uses \(F = m\,a\) as a rule for calculating; this bench asks what in the law could have gone wrong, and the rest of the lesson looks at where the forces on its left-hand side come from.

\(F = k\,x\)\(a = \dfrac{F}{m} = \dfrac{k\,x}{m}\)\(k\) is the spring constant, \(x\) is how far the spring is stretched beyond its rest length and \(m\) is the mass of the cart; the line on the graph has slope \(a\).

In Feynman: §12-1 What is a force? ↗

Let's discuss

  • With 20 N/m and 1 kg, make runs with 0.10 m, 0.20 m and 0.30 m of stretch. What is the acceleration divided by the stretch in each?
  • Keep 0.10 m and change the mass to 2 kg and then to 4 kg. What is \(m\,a\) in each run, and why is that the number that matters?
  • With a 40 N/m spring, what stretch gives the same acceleration as the 20 N/m spring stretched 0.20 m?
  • Is there any combination of spring and stretch that gives a 4 kg cart 2 m/s²?
Spring force (N)
Measured acceleration (m/s²)
Prediction kx/m (m/s²)
STEP 2

Why is it harder to start pushing than to keep going?

In the Pulling scenario, a 2 kg wooden block sits on a wooden table, and the Applied force control pulls it horizontally. At 4 N the block does not budge, and the table answers with 4 N of friction the other way; at 8 N, it answers with 8 N. On the graph this shows up as a line along which friction equals the applied force, and static friction only reaches 9.8 N, the static coefficient of 0.5 times the normal force of 19.6 N. A little beyond that, the block slides.

Once the block is sliding, friction drops to 5.88 N, the kinetic coefficient of 0.3 times the normal force, and stays there whatever the force and whatever the speed. With 10 N pulling, 4.12 N is left over to accelerate the 2 kg, 2.06 m/s². Lowering the force below 9.8 N does not stop the block, because while it moves the table holds it back with only 5.88 N; it slows down and stops once the force falls below that value. Getting started means beating the peak \(\mu_s N\), and keeping going needs only the plateau \(\mu_k N\).

The coefficients on the buttons are typical, approximate values. In practice they shift with the state of the surfaces, and the gap between static and kinetic depends heavily on conditions, such as how long the pieces sat still in contact and how clean they are, and is sometimes small. That friction barely depends on the contact area or on the speed is an empirical rule, which works reasonably well across a range of everyday situations, and not a fundamental law. The secondary-school lesson on friction shows the same peak with the block lying flat and standing on end.

applied force   friction   normal force   weight

The Incline scenario offers a way of measuring \(\mu_s\) with no force gauge at all. The table becomes a ramp and the angle goes up. The part of the weight along the ramp, \(mg\sin\theta\), grows, and the normal force, \(mg\cos\theta\), shrinks; the block sets off downhill when the first exceeds \(\mu_s\) times the second, that is, when \(\tan\theta\) reaches \(\mu_s\). For wood on wood this happens at 26.6°, and the mass drops out of the sum, because it appears on both sides. Once moving, at 30° the block goes down at 2.35 m/s², and it only comes to rest again if the angle falls below 16.7°, where \(\tan\theta = \mu_k\). Feynman describes the same measurement, and the secondary-school lesson on the ramp breaks the weight into parts step by step.

That leaves the question of why area hardly matters. The most widely accepted explanation is usually associated with Frank Bowden and David Tabor, in Cambridge, around the middle of the twentieth century: the true contact area between two solids is a small fraction of the apparent area and grows with the normal force, not with the size of the block. Under more weight, the points where the surfaces really touch flatten and multiply; laying the block flat spreads the same weight over more points, each pressed less hard, and the total barely changes. At those points the atoms of the two pieces come close enough to attract one another, and friction would thus be a consequence of forces between atoms, which step 3 examines closely.

\(f_s \le \mu_s N\)\(f_k = \mu_k N\)\(\tan\theta_c = \mu_s\)\(f_s\) is static friction and \(f_k\) kinetic friction, \(N\) is the normal force (\(mg\) on the table and \(mg\cos\theta\) on the ramp) and \(\theta_c\) is the angle at which the block starts to slide down; the coefficients \(\mu_s\) and \(\mu_k\) have no units.

In Feynman: §12-2 Friction ↗

Let's discuss

  • In Pulling, with wood and 2 kg, up to what force does the block stay put? What is the friction at 5 N and at 9 N?
  • With the block already sliding, bring the force down to 7 N. Does it stop? And at 4 N?
  • Set the mass to 4 kg. What happens to the peak and to the plateau on the graph? And to the critical angle on the Incline?
  • On the Incline, from what angle does the block slide with rubber on asphalt? And with steel on ice?
Friction (N)
State
Critical angle
STEP 3

Why does a spring obey Hooke’s law?

The two atoms on the bench push or pull each other with a force that depends only on the distance \(r\) between their centres. The one on the left is fixed and the one on the right can be dragged. The units are the bench’s own, with \(\sigma = 1\) for distances, \(\varepsilon = 1\) for energies and, therefore, \(\varepsilon/\sigma\) for forces. Taken to 1.05, the mobile atom is pushed outwards with 8.40; taken to 1.20, it is pulled back with 2.21. Between the two positions lies a distance, 1.122, at which the force vanishes, and there the pair can stay at rest.

The graphs show the rest of the curve. Inside 1.122 the repulsion climbs very steeply and at 1.02 is already above 16. Outside it, attraction appears, peaks at 2.40 at 1.244 and then fades, dropping below 0.1 by 2.2. The potential energy has a well of depth \(\varepsilon\) exactly at the distance where the force vanishes, and the force at each point is minus the slope of the energy curve.

Both curves come from the Lennard-Jones potential, associated with John Lennard-Jones, who worked on this form in the 1920s and 1930s, and still widely used in simulations of noble gases and simple liquids. It is a model, not the exact law of the force between two atoms. The \(r^{-6}\) term has a physical justification for the attraction between neutral atoms, but the \(r^{-12}\) term was chosen largely because it is convenient in calculations. Where these forces come from is a question for quantum mechanics, since they arise from the arrangement of the electrons and nuclei of the two atoms; the bench borrows only the general shape, repulsion up close, attraction further out and an energy minimum in between, a shape that turns up in many materials.

model force   Hooke’s prediction   potential energy

Near 1.122 the force curve barely parts from the dashed line of slope \(-k\), with \(k \approx 57.1\,\varepsilon/\sigma^2\). That is a spring. At 1.12 the line predicts 0.141 and the model gives 0.144; at 1.115, 0.426 against 0.457. The argument does not rest on the model’s exponents: any force that comes from an energy with a smooth minimum is, close to it, proportional to the displacement and opposite in direction, because there the energy curve is indistinguishable from a parabola, the dashed one in the upper graph. In a steel spring each bond between atoms strays very little from its minimum, and the linear response of the whole spring appears to come from there.

The line is only good near the bottom of the well. At 1.10 it is already off by 19%, and at 1.15 by 29%; at the peak of the attraction, at 1.244, it predicts almost three times the model’s force. In a real spring, stretching beyond the linear range leaves a permanent deformation or breaks the material. The Release button shows the difference in motion, with time measured in \(\sigma\sqrt{m/\varepsilon}\), where \(m\) is the mass of the mobile atom. Released at 1.15, it swings in to 1.099 and comes back after 0.84, close to the 0.83 of the Hooke spring; released at 1.5, it takes 1.72, barely gets past 1.015 on the inside and spends most of its time far out, in the shallow part of the well.

\(U(r) = 4\varepsilon\left[\left(\tfrac{\sigma}{r}\right)^{12} - \left(\tfrac{\sigma}{r}\right)^{6}\right]\)\(F = -\dfrac{dU}{dr}\)\(F \approx -k\,(r - r_0),\)\(k = \dfrac{d^2U}{dr^2}\Big|_{r_0}\)\(r_0 = 2^{1/6}\sigma \approx 1.122\,\sigma\) is the equilibrium distance, where \(U = -\varepsilon\); a positive \(F\) pushes the atoms apart and a negative one pulls them together.

In Feynman: §12-3 Molecular forces ↗

Let's discuss

  • At what distance is the force zero, and what is the potential energy there?
  • Between which distances does Hooke’s prediction stay within 10% of the model’s force?
  • Release the atom at 1.15 and then at 1.8. How long does each round trip take, and on which side of equilibrium does the atom spend more time?
  • Switch on damping and release the atom at 1.5. Where does it stop, and why?
Distance r (σ)
Model force (ε/σ)
Hooke, −k(r − r0)
STEP 4

How does a force act at a distance?

In the B only scenario, an electron at a million metres per second travels through a region of uniform magnetic field, 1 mT, pointing out of the screen (the dots). It does not go straight: it traces a circle of radius 5.69 mm and gets back to its starting point in 35.7 ns. At twice the speed the circle doubles, and a lap still takes 35.7 ns; with twice the field, radius and lap time both halve. The scene runs in slow motion, 12 ns of electron time for each second on screen, with the same factor in every case, so lap times can be compared by eye.

What bends the electron is the magnetic force \(q\,\vec v\times\vec B\), perpendicular to the velocity and to the field at once. Since it is always at right angles to the motion, it turns the velocity without changing the speed, and plays the part of the centripetal resultant: setting \(|q|\,v\,B\) equal to \(m v^2/r\) gives the radius, and the lap time, \(2\pi r/v\), no longer depends on \(v\). The secondary-school lesson on the magnetic force on a charge does this sum with ions. The electron’s charge is negative, so for it the force points up when it moves to the right with the field out of the screen, the opposite of what a proton would feel; that is why it circles anticlockwise.

Adding the force that an electric field \(\vec E\) exerts on any charge, moving or not, leads to the Lorentz force, \(\vec F = q\,(\vec E + \vec v\times\vec B)\). It gives the force on the charge from the two fields at the point where the charge is, without asking what produces them.

velocity   force on the electron   path

Nothing touches the electron, and yet it is deflected. The idea of a field separates whatever produces the force from whatever feels it: the sources fix the value of the field at every point, and the charge needs only the value where it happens to be. On the bench, the magnet or coil that would make this \(\vec B\) does not even appear. The force the sources produce obeys simple laws. The force between two charges at rest falls off as the inverse square of the distance, like gravitation, whose \(1/r^2\) law the lesson on gravitation discusses, and that is why electricity and gravitation are called fundamental forces, unlike friction and the force between atoms, which come from piling up a great many electrical effects.

In the E and B scenario, the electron passes between two charged plates that add a vertical electric field. The electric force does not depend on speed, the magnetic one grows with it, and the two point in opposite directions. At 1 kV/m and 1 mT they cancel only for an electron moving at \(E/B = 10^6\) m/s: that one goes straight and through the slit. Those moving much more slowly or much faster hit a plate or miss the slit; near \(E/B\) a narrow band still gets through, and with a strong \(B\) the electron can make whole loops between the plates and land in the slit by chance. An arrangement like this sorts electrons by speed, and the balance between the two deflections is the one J. J. Thomson used in 1897, in a cathode-ray tube, to measure the speed of electrons and reach the ratio of their charge to their mass.

Nuclear forces are left out. They reach no further than about \(10^{-15}\) m, roughly the width of a nucleus, and nobody has found for them a simple rule of the inverse-square kind.

\(\vec F = q\,(\vec E + \vec v\times\vec B)\)\(r = \dfrac{m v}{|q| B}\)\(T = \dfrac{2\pi m}{|q| B}\)\(v = E/B\)\(r\) and \(T\) are the radius and the lap time with \(\vec v\) perpendicular to a uniform \(\vec B\); \(v = E/B\) is the speed that goes straight through with \(\vec E\), \(\vec B\) and \(\vec v\) mutually perpendicular. For the electron, \(|q| = 1.602\cdot10^{-19}\) C and \(m = 9.109\cdot10^{-31}\) kg.

In Feynman: §12-4 Fundamental forces. Fields ↗ · §12-6 Nuclear forces ↗

Let's discuss

  • In B only, at 1 mT, take the speed from 1 to 3·10⁶ m/s. What happens to the radius and to the time of one lap?
  • What magnetic field puts an electron moving at 2·10⁶ m/s on a circle of radius 5.69 mm?
  • In E and B, at 2 mT and 3·10⁶ m/s, what electric field lets the electron through undeflected?
  • With the electric field above \(v\,B\), which way does the electron veer? And with the electric field switched off, at 1 mT and 10⁶ m/s?
Radius predicted · measured (mm)
Lap time (ns)
E/B (10⁶ m/s)
STEP 5

Where does the force that throws us back in a car pulling away come from?

A small ball on a string hangs from the roof of the bench’s car. With the car moving at constant speed the string hangs plumb, but setting the control to 3 m/s² is enough for the ball to lag behind, swing a little and settle with the string tilted 17.0° from the vertical. At 5 m/s² the angle reaches 27.0°; braking at 3 m/s², the string tilts by the same 17.0°, now forwards. Neither the mass of the ball nor the speed of the car enters the result, only the acceleration, and the string comes to rest where \(\tan\theta = a/g\). A pendulum hung like this works as an accelerometer.

With the Seen from the road button, the camera stands still on the verge and the car crosses the screen, in a scene that starts again at every pass. From there we watch the ball gain speed along with the car, and the bench’s magnifier shows the two forces acting on it, its weight and the tension in the string. With the string tilted the two do not cancel, and once the ball settles a horizontal resultant is left over, pointing forwards, worth exactly \(m\,a\). That is what accelerates the ball, and the second law works with no adjustment at all. The scene runs in slow motion, at half speed.

tension   weight   pseudo-force   resultant

With the Seen from inside the car button, the camera rides with the car. Now it is the road that moves, backwards and ever faster, and the ball is at rest, hanging askew. Weight and tension are the same as before and still add up to a forward resultant, except that, for someone in the car, the ball is not accelerating. The second law only holds in there if one more force enters the sum, \(-m\,\vec a\), pointing backwards and exerted by no body at all. It is a pseudo-force, and to someone inside it is what seems to press the passenger into the seat back as the car pulls away. On a 70 kg person, 3 m/s² amounts to 210 N, a little under a third of their weight.

The pseudo-force grows with mass exactly as weight does, and from inside the car the two combine into a tilted effective gravity of size \(\sqrt{g^2 + a^2}\), which comes to 10.25 m/s² at 3 m/s². The string lines up with it, and a stone let go inside the car would fall along the string. A passenger with the windows blacked out reads 17.0° and 10.25 m/s² on the pendulum, and the same reading would come from a car standing still where gravity were 10.25 m/s² and tilted. Albert Einstein, in 1907, turned this into a principle, the equivalence principle, which would eventually lead to general relativity, completed in 1915.

Centrifugal force is of the same kind, only in a rotating frame, such as a bus going round a bend. For someone on board, the sums balance with an outward force, proportional to mass and with no body behind it; for someone watching from the pavement it disappears, and what exists is the centripetal force, a role that the secondary-school lesson on centripetal force gives to real forces, such as friction or tension. The pseudo-force is not a mistake. It is the price of describing motion from an accelerated frame, and with it the sums done inside the car agree with those done on the road.

\(\vec F_{\text{pseudo}} = -m\,\vec a\)\(\tan\theta = \dfrac{a}{g}\)\(g_{\text{eff}} = \sqrt{g^2 + a^2}\)\(\vec a\) is the acceleration of the car, \(\theta\) is the angle of the string from the vertical once the ball stops swinging and \(g_{\text{eff}}\) is the gravity measured from inside the car, with \(g = 9.8\) m/s².

In Feynman: §12-5 Pseudo forces ↗

Let's discuss

  • At 2.5 m/s² and at 5 m/s², does the angle of the string double? And the pseudo-force on a 70 kg person?
  • Braking at 4 m/s², which way does the string tilt, and what is the effective gravity?
  • Set the acceleration to zero. The car then moves at a steady 3 m/s. From inside, can the pendulum tell you whether the car is moving or standing still?
  • Seen from inside the car, press Restart and follow the resultant arrow while the ball swings. When does it vanish?
String angle
Effective gravity (m/s²)
Pseudo-force on 70 kg (N)

Where we go next

This lesson asked what a force is and where it comes from, and left aside what happens when a force goes along with a body over a path. Feynman’s chapter 13, Work and Potential Energy (A), deals with the work the force does along that path and with potential energy, an idea that already turned up here in the energy well of step 3. It is also on the Caltech website, and the lesson in this track that goes with it comes after this one.