Tension in inextensible strings: Visual walkthrough

Step 1 — The scene, with nothing assumed

WHAT. Two objects hang from the two ends of a single rope. The rope goes up and over a wheel (a pulley) that lets it change direction. We call the objects' amounts of "stuff" (their masses) $m_{1}$ and $m_{2}$ , and we agree $m_{1}$ is the heavier one ( $m_{1} > m_{2}$ ).

WHY. Before any maths, we must see the players. A symbol like $m_{1}$ is meaningless until it is pinned to a picture. Look at the left weight — that is $m_{1}$ . Look at the right — that is $m_{2}$ .

PICTURE. The heavier block sits on the left; both hang straight down because the only thing tugging them sideways would be the rope, and the rope here pulls straight up.

Figure — Tension in inextensible strings

Step 2 — Draw every force as an arrow (the Free Body Diagram)

WHAT. We separate the two blocks and, for each one alone, draw every push or pull acting on it as an arrow. This picture is a Free Body Diagram (FBD).

WHY. Newton's Second Law — the law $F = ma$ we are about to use — applies to one body at a time. If we mix forces from two bodies, the law breaks. So we isolate each block and ask: what touches it, what pulls it?

Two forces act on each block:

Gravity, pulling straight down, of size $m_{1} g$ on the left block and $m_{2} g$ on the right.
Tension $T$ , the rope's pull, going straight up into the rope on both blocks.

PICTURE. Down-arrows are weight; up-arrows are tension. Notice both up-arrows have the same label $T$ — the parent note proved this: a massless rope over an ideal pulley pulls with equal strength everywhere.

WHAT. The rope cannot stretch. So if the left block drops by a distance $d$ , exactly that much rope must appear on the right, lifting the right block up by the same $d$ . Same distance in the same time ⟹ same speed ⟹ same acceleration magnitude, which we will call $a$ .

WHY. This is the constraint that turns two unknown motions into one. Without it we would have two accelerations and could not solve. "Inextensible" is the hidden gift.

PICTURE. Watch the coloured markers: the left one slides down one grid square exactly as the right one slides up one grid square. They are locked together.

Formula The constraint — with an explicit coordinate choice

To turn "same distance" into signs, we must first choose axes. Measure each block by a single coordinate that grows in a chosen direction:

Let $x_{1}$ = how far the left block has moved downward from its start (so $x_{1}$ grows as $m_{1}$ descends, i.e. down is positive for $x_{1}$ ).
Let $x_{2}$ = the rope length from the pulley down to the right block (so $x_{2}$ grows as $m_{2}$ descends, i.e. down is positive for $x_{2}$ too).

The total rope length is fixed. As the left block drops by $x_{1}$ , exactly $x_{1}$ of rope is pulled over the pulley, so the right side shortens by the same amount: $x_{1} + x_{2} = constant .$

$x_{1} + x_{2}$ :: total rope hanging on the two sides, which cannot change (inextensible).

Differentiate twice with respect to time (velocity, then acceleration): $\overset{x}{¨}_{1} + \overset{x}{¨}_{2} = 0 ⟹ \overset{x}{¨}_{1} = - \overset{x}{¨}_{2} .$

$\overset{x}{¨}_{1}$ :: acceleration of the left block measured with down positive;
$\overset{x}{¨}_{2}$ :: acceleration of the right block measured with down positive;
the minus sign :: is forced by our axis choice — since both coordinates count down as positive, one block's downward gain must be the other's downward loss (i.e. an upward move). One descends, the other rises.

Taking sizes, both accelerations have the same magnitude: $∣ \overset{x}{¨}_{1} ∣ = ∣ \overset{x}{¨}_{2} ∣ = a .$ That single number $a$ , defined above, is what the next steps solve for.

Step 4 — Newton's law on the heavy block ( $m_{1}$ )

WHAT. $m_{1}$ is heavier, so it wins and moves down. We choose down as the positive direction for this block, and write $F = ma$ .

WHY. We pick "positive = the way it actually moves" so the acceleration comes out positive — cleaner bookkeeping. Newton's Second Law says the net force (all arrows added with signs) equals mass times acceleration.

$weight, down (+) m_{1} g - tension, up (-) T = mass \times accel m_{1} a$

$m_{1} g$ :: the downward gravity pull, positive because down is our chosen positive here;
$- T$ :: the rope's upward pull, negative because it opposes the motion;
$m_{1} a$ :: the leftover unbalanced force, driving it downward.

PICTURE. The down-arrow ( $m_{1} g$ ) is drawn longer than the up-arrow ( $T$ ): the difference between them is what accelerates the block.

Step 5 — Newton's law on the light block ( $m_{2}$ )

WHAT. $m_{2}$ is dragged up. We choose up as positive for this block, and write $F = ma$ again.

WHY. Same law, but a separate sign choice — each body gets its own "positive = its direction of motion". The parent note flagged mixing global signs as a classic mistake; local conventions add up cleanly.

$tension, up (+) T - weight, down (-) m_{2} g = mass \times accel m_{2} a$

$T$ :: rope pull, now positive because up is this block's positive direction;
$- m_{2} g$ :: gravity, negative because it fights the upward motion;
$m_{2} a$ :: the net upward force lifting the light block.

PICTURE. Here the up-arrow ( $T$ ) is the longer one: tension out-pulls this block's weight, so it rises.

Step 6 — Add the two equations to kill $T$

WHAT. We have two equations sharing the same $T$ and same $a$ . Adding them makes $- T$ and $+ T$ cancel.

WHY. $T$ is an unknown we don't yet want. Because we chose each block's positive direction along its own motion, the tension terms have opposite signs and vanish when added — leaving one equation in one unknown, $a$ .

m_{1} g - T = m_{1} a T - m_{2} g = m_{2} a m_{1} g - m_{2} g = (m_{1} + m_{2}) a

Solve for $a$ :

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

$m_{1} - m_{2}$ :: the imbalance in weight — the "extra" pull the heavy side has;
$m_{1} + m_{2}$ :: the total mass being pushed around (both blocks move);
so $a$ is "unbalanced pull ÷ total mass moved" — exactly Newton's law in disguise.

PICTURE. The bar chart shows the two weights; only the overhang (the burnt-orange slice) drives the system, and it must accelerate both masses.

Step 7 — Put $a$ back to find the tension $T$

WHAT. Now use $a$ in the $m_{2}$ equation ( $T - m_{2} g = m_{2} a$ ) to get $T$ .

WHY. $T$ was the other thing we wanted. Any of the two equations works; we substitute and simplify.

$T = m_{2} g + m_{2} a = m_{2} (g + \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}) = m_{2} \cdot \frac{( m _{1} + m _{2} ) + ( m _{1} - m _{2} )}{m _{1} + m _{2}} g$

The top becomes $2 m_{1}$ , so:

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

$2 m_{1} m_{2}$ :: symmetric in the two masses — swap them and $T$ is unchanged (the rope doesn't care which side is heavier);
$m_{1} + m_{2}$ :: total mass again;
the whole thing sits between $m_{2} g$ and $m_{1} g$ — the rope tension is a compromise, never as big as the heavy weight, never as small as the light one.

PICTURE. A number line shows $T$ landing strictly between the two weights $m_{2} g$ and $m_{1} g$ .

Step 8 — All the edge cases (never leave a scenario unshown)

WHAT. We test the formulas at the extremes to be sure they behave.

WHY. A formula you cannot sanity-check is a formula you cannot trust. Each limit is a picture in itself.

Case	$a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$	$T = \frac{2 m _{1} m _{2}}{m _{1} + m _{2}} g$	Meaning
$m_{1} = m_{2} = m$	$0$	$m g$	Balanced — nothing moves, each side just holds its own weight.
$m_{2} = 0$	$g$	$0$	Left block in free fall; nothing on the right, no pull.
$m_{1} ≫ m_{2}$	$\to g$	$\to 2 m_{2} g$	Heavy side almost free-falls; light side gets yanked — tension near twice its own weight!

PICTURE. Three mini-scenes: the balanced still-life, the free-fall, and the violent yank — each with its arrows sized to match.

The one-picture summary

Everything above, compressed into a single annotated map: the two FBDs (Steps 4–5) feed two $F = ma$ equations; the constraint (Step 3) forces one shared $a$ ; adding cancels $T$ to give $a$ (Step 6); back-substitution gives $T$ (Step 7). The figure lays the algebra beside the arrows so you can trace each symbol back to a picture.

Recall Feynman retelling — say it to a friend

Two blocks dangle from the ends of a rope slung over a wheel. I look at each block by itself and draw its arrows: gravity yanks it down, the rope tugs it up. The rope can't stretch, so both blocks are forced to move the same distance in the same time — one shared acceleration $a$ . I write "net force = mass × a" for each block, being careful to call its own direction of motion positive. When I add the two little equations, the rope's tug ( $T$ ) cancels out and I'm left with: the difference of the weights drives the sum of the masses. That's $a = (m_{1} - m_{2}) g / (m_{1} + m_{2})$ . Then I slide $a$ back into one equation and out pops the tension $T = 2 m_{1} m_{2} g / (m_{1} + m_{2})$ — a value that always sits politely between the two weights. Finally I test silly extremes: equal masses → nothing moves; empty right side → left block free-falls; giant left mass → the little block gets whipped upward with nearly twice its weight in rope-pull. Every extreme makes sense, so I trust the formulas.

Connections

Newton's Second Law — each boxed step is $F = ma$ on one block.
Free Body Diagrams — Steps 2, 4, 5 are FBDs.
Constraint Relations — Step 3, the "same $a$ " lock.
Atwood Machine — this whole page is its derivation.
Frictionless Pulleys vs Pulleys with Inertia — where "equal $T$ " stops holding.
Normal Force — the pushing cousin of tension.