Free body diagrams — systematic drawing technique: Visual walkthrough

We use only four ideas, each defined before we lean on it: a force (a push or pull, drawn as an arrow), mass $m$ (how hard something is to speed up), weight $m g$ (gravity's pull), and acceleration $a$ (how fast the speed is changing). If any of those feel shaky, that's fine — we re-earn them below.

Step 1 — Meet the machine

WHAT. Two blocks. Block $A$ (mass $m_{1}$ ) sits on a flat, frictionless table. A string runs from $A$ horizontally to the edge, bends over a pulley (a little wheel), and drops down to block $B$ (mass $m_{2}$ ) hanging in the air.

WHY draw the whole scene first? Before we shrink anything to a dot, we must see who touches whom. Forces only appear where objects touch or where gravity reaches. This picture is the map of every contact.

PICTURE. The red string is the messenger between the two blocks. Block $B$ (the accent object) hangs, ready to fall and drag $A$ sideways.

Figure — Free body diagrams — systematic drawing technique

Step 2 — Free body diagram of the hanging block $B$

WHAT. Shrink $B$ to a dot. Two forces touch it: gravity pulling down with strength $m_{2} g$ , and the string pulling up with strength $T$ .

WHY only two? Scan $B$ 's boundary. The only thing touching it is the string (→ tension $T$ , up). Nothing else touches it, so nothing else can push it — except gravity, which reaches without touching (→ $m_{2} g$ , down). That's the complete list.

PICTURE. Up-arrow $T$ (red string) fights down-arrow $m_{2} g$ . Because $B$ actually falls, gravity wins — the down arrow is longer.

Now apply Newton's second law, Newton's Second Law, along the vertical. We choose down as positive because that's the direction $B$ actually moves:

$down pull, wins m_{2} g - up pull, loses T = B’s inertia m_{2} shared accel a$

Step 3 — Free body diagram of the table block $A$

WHAT. Shrink $A$ to a dot. Three forces touch it: gravity $m_{1} g$ down, the table's normal force $N$ up, and the string's tension $T$ pulling sideways toward the pulley.

WHY sideways, not down? The string on $A$ 's side runs horizontally, so it can only pull horizontally. The table can only push perpendicular to itself — straight up — because a frictionless surface has nothing to grip with (see Normal Force and Friction).

PICTURE. Vertically, $N$ (up) exactly balances $m_{1} g$ (down): $A$ never leaves the table. Horizontally, only $T$ (red) acts — so all of $A$ 's motion is horizontal.

Split into two axes (this is why the parent note insists on axes — vectors become plain arithmetic):

$y : table push up N - weight down m_{1} g = 0 \Rightarrow N = m_{1} g$

WHY $= 0$ ? $A$ stays flat on the table — no vertical acceleration — so the vertical forces must cancel exactly.

$x : string pull T = A’s inertia m_{1} shared accel a$

Step 4 — Why the tension is the same on both ends

WHAT. We wrote the same letter $T$ in Step 2 and Step 3. That's a claim, not an accident. An ideal string is massless and inextensible (cannot stretch).

WHY same $T$ ? Newton's second law on any tiny piece of string is $\sum F = (mass of piece) \times a$ . For a massless string that right side is $0$ , so the pull entering a piece equals the pull leaving it — the tension can't change as you walk along the rope. The pulley just redirects that unchanged pull around the corner. (See Tension in Strings and Pulleys.)

PICTURE. Follow the red string from $A$ , over the wheel, down to $B$ : the tension arrows are the same length everywhere. The pulley bends the direction, not the strength.

Step 5 — Why the acceleration is the same for both blocks

WHAT. We also wrote the same letter $a$ in both equations. This is the inextensible part of "ideal string."

WHY same $a$ ? The string cannot stretch or go slack, so its total length is fixed. If $B$ drops $1$ cm, the string on $A$ 's side must shorten by exactly $1$ cm, dragging $A$ exactly $1$ cm toward the pulley. Same distance in the same time $\Rightarrow$ same speed $\Rightarrow$ same acceleration.

PICTURE. The red length that leaves $A$ 's side is exactly the red length that arrives on $B$ 's side. Distances marked equal $\Rightarrow$ accelerations marked equal.

Step 6 — Add the equations, kill the tension

WHAT. Line up the two equations and add them. $(B) (A) m_{2} g - T = m_{2} a m_{2} g - T = m_{1} a$

WHY add? $T$ is an internal force — it lives inside the system, pulling one block via the other. In equation (B) it appears as $- T$ ; in equation (A) as $+ T$ . Adding makes them cancel: $- T + T = 0$ . Internal forces always cancel this way, leaving only the external driver.

PICTURE. Stack the two arrows: the $+ T$ and $- T$ annihilate, and what survives is the driving weight $m_{2} g$ on one side against the combined inertia $m_{1} + m_{2}$ on the other.

Adding term by term:

$external driver m_{2} g = total inertia moved (m_{1} + m_{2}) a$

Solve for $a$ by dividing both sides by $(m_{1} + m_{2})$ :

$a = \frac{m _{2}}{m _{1} + m _{2}} g$

Step 7 — Edge and degenerate cases (never let a limit surprise you)

WHAT & WHY. A formula you trust is one you've stress-tested at its extremes. We push each input to $0$ or $\infty$ and check the answer stays sane.

PICTURE. Three sliders, three sanity checks, each with the block cartoon that makes it obvious.

Case	Formula gives	Does it make sense?
$m_{2} \to 0$ (nothing hanging)	$a \to \frac{0}{m _{1}} g = 0$	✓ No driver, no motion.
$m_{1} \to 0$ (table block weightless)	$a \to \frac{m _{2}}{m _{2}} g = g$	✓ $B$ falls freely; string drags a massless $A$ for free.
$m_{1} = m_{2} = m$	$a = \frac{m}{2 m} g = \frac{g}{2}$	✓ Equal partners split gravity's pull.
$m_{2} \to \infty$ (huge hanging block)	$a \to g$	✓ $A$ 's inertia becomes negligible; near free-fall.

The one-picture summary

The entire derivation on one canvas: two FBDs (left), the two equations they produce (middle), and the single boxed answer after tension cancels (right). Follow the red string — it ties every panel together.

Recall Feynman retelling — the whole story in plain words

Picture a block on a slippery table with a rope running off the edge, over a little wheel, down to a second block dangling in the air. The dangling block wants to fall, but it's tied to its lazy friend on the table, so as it drops it drags the table block sideways. First I look at the hanging block alone: gravity pulls it down hard, the rope pulls it up a bit, and it loses, so it falls — that gives me one equation. Then I look at the table block alone: gravity down is exactly cancelled by the table pushing up, and the only thing left to move it is the rope pulling sideways — that gives me a second equation. The clever bit: the rope can't stretch and weighs nothing, so both blocks feel the same rope-pull and speed up at the same rate. That means my two equations share the same two unknowns. When I add the equations, the rope-pull cancels itself out (it's an inside force — it pulls one block by pulling the other), and I'm left with a beautifully simple story: gravity on the hanging block does the driving, and both blocks' masses share the load. So the acceleration is the hanging weight divided by the total weight-to-move, times gravity — always a bit less than a real free-fall, because the lazy table block is holding things back.

Visual walkthrough — Free body diagrams — systematic drawing technique