Newton's Three Laws & Free Body Diagrams

In the last lesson you were changing your mind about which frame to watch a motion from, and the velocities all shifted around depending on who was doing the looking. Accelerations did not. That is the hint this lesson is built on: acceleration is the thing that every non-accelerating observer agrees about, and forces are what produce it.

Here is the question. A crate sits on a table. The table pushes up on it, and the crate does not move. Now you put the same crate in a lift and the lift starts up, and a scale under the crate reads more than it did before, even though nothing about the crate changed. Something is deciding, moment by moment, how hard the floor pushes. What decides it? The answer is never a formula you look up. It is a bookkeeping discipline: find every force acting on one chosen object, add them as vectors, and set the sum equal to $ma$. The hard part is the finding, not the adding.

The three laws, said carefully

First law. An object with no net force on it keeps a constant velocity. Not "stays still": constant velocity, which includes staying still as the special case where that constant is zero. The law's real content is that motion needs no cause. Only change of motion needs a cause. A puck gliding across frictionless ice needs nothing pushing it forward.

Second law. The net force on an object equals its mass times its acceleration:

$\sum \vec F = m\vec a$

That is a vector equation, and a vector equation is really several scalar equations at once. Pick axes and it becomes

$\sum F_x = m a_x, \qquad \sum F_y = m a_y$

The symbol $\sum$ means "add up all of them". $m$ is the mass in kilograms, $a$ the acceleration in m/s², and force is then in newtons: one newton is one kilogram-metre per second squared. Read the equation as a statement about cause and response: forces are the input, acceleration is the output, and mass is how stubbornly the object resists. Double the force, double the acceleration. Double the mass instead, halve it.

Third law. If A pushes on B, then B pushes on A with the same magnitude in the opposite direction. The two forces in such a pair always act on different objects. That last clause is the whole point, and it is where most of the confusion about this law lives.

The procedure

Four steps, in this order, every time.

1. Choose one body and draw it alone, as a box or a dot. Nothing else on the page: no table, no rope, no other block.
2. Draw every force acting on it. On it, from outside it. Never a force it exerts on something else, and never an internal force.
3. Choose axes. Usually one axis along the known direction of the acceleration, because that makes the other component zero.
4. Write $\sum F_x = ma_x$ and $\sum F_y = ma_y$ and solve.

Step 2 needs a catalogue, so here is the short list of what can appear.

Weight, $W = mg$, always straight down, always present, magnitude $mg$ with $g = 9.8$ m/s² near the Earth's surface. Normal force $N$, from any surface in contact. The word normal here means perpendicular, not usual: it points perpendicular to the contact surface, away from it. Its magnitude is whatever is needed to stop the two objects passing through each other, which is frequently not $mg$. Tension $T$, from any rope or cable, pulling along the rope, away from the object. A rope can only pull. Applied forces, whatever the problem says someone is pushing with. And friction, which gets its own lesson next.

Nothing else. If you find yourself drawing an arrow labelled "the force of motion" or "centrifugal force", stop: neither exists in this framework.

One diagram per object

The moment there is more than one body, the discipline pays off. Two blocks of mass $m_1 = 2.0$ kg and $m_2 = 3.0$ kg sit touching on a frictionless floor, and you push the first with $F = 20$ N so the pair slides together.

Because they stay in contact they have the same acceleration. Treat the pair as one object of mass $m_1 + m_2 = 5.0$ kg. The only external horizontal force is $F$, since the contact force between them is internal to this choice of system and cancels out of the sum:

$a = \frac{F}{m_1 + m_2} = \frac{20\ \text{N}}{5.0\ \text{kg}} = 4.0\ \text{m/s}^2$

Now, to get the contact force $C$, you have to break the system apart, because $C$ was invisible while the blocks were one object. Draw the second block alone. The only horizontal force on it is $C$, pushing forward:

$C = m_2 a = (3.0\ \text{kg})(4.0\ \text{m/s}^2) = 12\ \text{N}$

Check it against the first block, which feels $F$ forward and $C$ backward by the third law: $20 - 12 = 8.0$ N, and $m_1 a = (2.0)(4.0) = 8.0$ N. It agrees. That cross-check is worth doing while you are learning, because it catches a sign error instantly.

Choosing axes: the ramp

A $12$ kg box rests on a ramp tilted at $25°$. What is the normal force?

The trap is answering $mg$. The box is not accelerating, true, but the normal force is perpendicular to the ramp and the weight is vertical, and those are no longer the same line. Tilt your axes to match the ramp: $x$ down the slope, $y$ perpendicular to it. Then $N$ lies entirely along $y$ and only the weight needs splitting. The weight makes an angle $\theta$ with the perpendicular (the same $\theta$ as the ramp, because their arms are mutually perpendicular) so its component along $y$ is $mg\cos\theta$ and along $x$ is $mg\sin\theta$.

With no acceleration perpendicular to the surface, $N - mg\cos\theta = 0$:

$N = mg\cos\theta = (12)(9.8)\cos 25° = 117.6 \times 0.906 = 106.6\ \text{N}$

Less than the $117.6$ N you would get on flat ground, which makes sense: the ramp only has to support part of the weight, and the rest goes into pulling the box down the slope. Along the slope, if nothing holds the box, $mg\sin\theta = ma$ gives $a = g\sin\theta$, independent of mass.

That second case is the one worth chasing, so work it in full. A $2$ kg block sits on a frictionless $40°$ incline, held in place by a purely horizontal force. Take $g = 10$ m/s². What is the normal force?

Tilted axes are now a poor choice, because both the weight and the applied force would need splitting. Use horizontal and vertical axes instead, and split $N$. The normal is perpendicular to a surface tilted at $40°$, so $N$ itself is tilted $40°$ from the vertical: its vertical component is $N\cos 40°$ and its horizontal component is $N\sin 40°$. The block is in equilibrium, so each direction sums to zero.

Vertical: $N\cos 40° - mg = 0$, so

$N = \frac{mg}{\cos 40°} = \frac{(2)(10)}{0.766} = 26.1\ \text{N}$

Horizontal: $F - N \sin 40° = 0$, giving $F = 26.1 \times 0.643 = 16.8$ N, which is also $mg\tan 40°$.

Notice that $N$ came out larger than $mg$, not $mg\cos\theta = 15.3$ N. The horizontal hand is squeezing the block into the ramp. Reaching for $mg\cos\theta$ here, out of habit, gets you a number that is wrong by 70%.

Apparent weight

Back to the lift. A $65$ kg student stands on a bathroom scale in a lift that accelerates upward at $1.5$ m/s². What does the scale read?

A scale reads the force it is pushed with, which by the third law equals the normal force it pushes back with. So the question is really "what is $N$?" Draw the student alone: weight $mg$ down, normal force $N$ up, nothing else. Take up as positive. The acceleration is $+1.5$ m/s², so

$N - mg = ma \quad\Longrightarrow\quad N = m(g + a)$

$N = 65(9.8 + 1.5) = 65 \times 11.3 = 734.5\ \text{N}$

Against $mg = 637$ N at rest. The same formula handles downward acceleration by making $a$ negative: at $a = -2.0$ m/s², $N = 65(9.8 - 2.0) = 65 \times 7.8 = 507$ N. And if the cable snaps, $a = -g$ and $N = 0$. That is weightlessness, not the absence of gravity, but the absence of anything pushing back.

Ropes, pulleys, and ropes with mass

A rope pulls along its own length, away from whatever it is attached to. If the rope is massless, the tension is the same everywhere in it, because a piece of rope with zero mass needs zero net force no matter how it accelerates. An ideal pulley changes the direction of the rope without changing the tension.

That gives the Atwood machine in three lines. A $3$ kg and a $2$ kg block hang over a frictionless pulley, $g = 10$ m/s². Take down as positive for the heavy block and up as positive for the light one, so a single $a$ describes both. Heavy block: $3g - T = 3a$. Light block: $T - 2g = 2a$. Add the two equations and $T$ disappears:

$3g - 2g = 5a \quad\Longrightarrow\quad a = \frac{(3-2)(10)}{5} = 2\ \text{m/s}^2$

Adding the equations is the same move as treating the two blocks plus rope as one system, where the tension is internal.

Now let the rope have mass, and "tension" stops being a single number. A $10$ kg block hangs from a uniform rope of mass $4.0$ kg, and the whole thing is hauled straight up with acceleration $2.0$ m/s². Find the tension at the midpoint of the rope.

Cut the rope at the midpoint, in your head, and ask what the upper half is holding. Everything below the cut: the block, $10$ kg, plus the lower half of the rope, $2.0$ kg. That is $12$ kg. Draw that sub-system alone. Two forces: tension $T$ at the cut, pulling up, and its own weight, $12g$ down.

$T - (12)(9.8) = (12)(2.0)$ $T = 117.6 + 24 = 141.6\ \text{N}$

The cut is the whole technique. Any point inside any extended object can be made into the boundary of a free body diagram, and the internal force there is just whatever it takes to give everything on one side of the cut its share of the acceleration. A hanging $6.0$ kg chain, one third of the way up from the bottom, supports $2.0$ kg of chain below it at rest: $T = (2.0)(9.8) = 19.6$ N. A $5.0$ kg rope on a frictionless floor pushed with $12$ N accelerates at $12/5.0 = 2.4$ m/s²; the internal force $1.0$ m from the pushed end is what drives the remaining $2.0$ m, mass $(2.0/3.0)(5.0) = 3.33$ kg, so it is $(3.33)(2.4) = 8.0$ N.

Recognising the technique

Almost every mechanics problem that is not about energy or momentum wants this method, so the useful question is not whether to draw free body diagrams but how many and around what.

Draw one per object, always, if the problem names more than one object: two blocks, a block and a wedge, a painter and a platform. Do not try to hold them in your head.

Reach for the combined system first when the question asks for an acceleration and the bodies clearly move together. Reach for the split when the question asks for a contact force, a tension, or "the force the platform exerts on the painter": any force that lives at the boundary between two objects. Those words are the signal that the boundary has to become the edge of a diagram.

Make a cut when the object is extended and the question asks about a point inside it: a rope with mass, a chain, a stack of bricks. The cut turns "inside" into "outside".

Choose tilted axes when the acceleration is along a slope and the only awkward force is gravity. Choose horizontal and vertical axes when the awkward forces point in several directions at once, or when the object is in equilibrium so neither axis is special: hanging signs on two cables, blocks held by horizontal pushes.

And when a problem says "frictionless", it is telling you the catalogue is short: weight, normal, tension, applied. Next comes the force that fills in the missing entry, and the reason the word "frictionless" was ever needed: friction.