Tension in inextensible strings: Question bank — Physics deep dive

Notation used everywhere below: $T$ = tension (the pulling force along a string), $g$ = gravitational acceleration ( $\approx 9.8 m/s^{2}$ downward), $a$ = acceleration magnitude, $m$ = mass. All strings are ideal (massless + inextensible) and all pulleys ideal (massless + frictionless) unless a question explicitly says otherwise.

The three pictures every trap comes from

Before the traps, look at the three diagrams these questions keep referring to. Once you can see them, the verbal traps are obvious.

Figure 1 — the free-body diagram (FBD) of one mass. A free-body diagram means: draw only the one object, and every force arrow on it, nothing else. For a mass hanging on a string, exactly two arrows act: gravity $m g$ pulling down, tension $T$ pulling up (away from the body, into the string). The net of these two arrows is what $ma$ equals.

Figure — Tension in inextensible strings

Notice the arrow lengths: when the mass falls (accelerates down), the down-arrow $m g$ must be the longer one, so $T < m g$ . That single picture kills the "tension = weight" trap.

Figure 2 — the string element (why tension is uniform). Zoom into a tiny slice of string. Call its mass $d m$ (the " $d$ " just means "a very small piece of"). This little slice is itself a body, so it gets its own FBD: the string to its left pulls it left with $T_{1}$ , the string to its right pulls it right with $T_{2}$ . Newton's law for the slice reads $T_{2} - T_{1} = (d m) a$ .

For a massless string $d m = 0$ , so the right side is zero for any finite $a$ , forcing $T_{2} = T_{1}$ . That is why tension is the same everywhere — it is a consequence of $d m \to 0$ , not an assumption.

Figure 3 — the inextensibility constraint (why accelerations match). Set up a coordinate: measure $x_{1}$ = the length of string from the pulley down to mass 1, and $x_{2}$ = the length from the pulley down to mass 2. Positive $x$ points downward for each side (longer string on that side = mass lower down). The rope's total length is fixed: $x_{1} + x_{2} + (fixed wrap over pulley) = L = constant .$ Differentiate once in time: $\overset{x}{˙}_{1} + \overset{x}{˙}_{2} = 0$ (if one side lengthens, the other shortens equally). Differentiate again: $\overset{x}{¨}_{1} = - \overset{x}{¨}_{2}$ . Equal magnitude, opposite sign — mass 1 goes down exactly as fast as mass 2 goes up.

That is the whole content of "inextensible"; see Constraint Relations for the general rule. Now the traps.

True or false — justify

A string can push a block if you compress it fast enough.

False. A string is floppy; under compression it buckles (picture pushing a thread — it folds). It only sustains force along its length in the pulling direction, the up-arrow of Figure 1.

In an Atwood Machine with

m_{1} \neq = m_{2}

, the tension on the heavier side is larger.

False. Figure 2 shows a single

T

threaded over the pulley; for an ideal string + ideal pulley it is identical on both sides. The masses differ, so the accelerations differ — not the tension.

If two masses hang balanced (

m_{1} = m_{2}

) over a pulley, the tension in the string is zero.

False. Nothing accelerates, but in each Figure-1 diagram the up-arrow must still cancel the down-arrow, so

T = m g

, not

0

For a mass falling on a string with downward acceleration

a

, the tension equals its weight

m g

False. In Figure 1 the down-arrow is longer, so

m g - T = ma

, giving

T = m (g - a) < m g

. Tension only equals weight when

a = 0

(equal arrows).

The inextensibility of a string forces connected masses to have the same velocity vector.

False. Figure 3 gives

\overset{x}{¨}_{1} = - \overset{x}{¨}_{2}

: equal magnitudes, opposite signs. Directions differ — one goes up while the other goes down.

Cutting an ideal string makes the tension drop to zero instantly.

True. Tension is transmitted through the material (the

T_{1}, T_{2}

arrows of Figure 2); sever the material and there is nothing to pull, so

T = 0

the moment the cut completes.

If you double both masses in an Atwood machine, the acceleration doubles.

False.

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

is a ratio — doubling both leaves it unchanged. Only the tension doubles.

A massless string over a pulley with mass still has equal tension on both sides.

False. A pulley with inertia needs a net torque to spin up, which requires different

T_{1}, T_{2}

across it (Figure 2's zero-difference argument breaks). See Frictionless Pulleys vs Pulleys with Inertia.

The tension in a real (heavy) rope is the same at every point along its length.

False. In Figure 2,

d m \neq = 0

, so

T_{2} - T_{1} = (d m) a

is nonzero and

T

varies. Uniform tension is a consequence of the massless idealization only.

Spot the error

"For the falling mass I wrote

T - m_{2} g = m_{2} a

, taking up as positive."

The mass moves down, so in its Figure-1 diagram the down-arrow wins; net force is downward. Correct is

m_{2} g - T = m_{2} a

(down positive for this body).

"Both masses accelerate, so I added the two FBD equations using 'up is positive' globally."

In Figure 3 the two masses move opposite ways, so a single global sign flips one tension term and

T

no longer cancels. Fix: take direction of motion as positive for each body separately, then adding eliminates

T

cleanly.

"The hanging mass pulls with force

m_{2} g

, so the block on the table feels

T = m_{2} g

Only true if nothing accelerates. Here the system accelerates, so

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

. Use the $F=ma$ equations on each Figure-1 diagram, not the static shortcut.

"Force

F

pulls the front block

m_{2}

, dragging

m_{1}

behind, so

T = F

F

accelerates the whole system; only part of that force reaches

m_{1}

through the string. Isolating

m_{1}

's FBD gives

T = m_{1} a = \frac{m _{1} F}{m _{1} + m _{2}} < F

"The pulley changes the tension because it redirects the string."

An ideal (massless, frictionless) pulley only redirects the string's direction; Figure 2 shows the difference

T_{2} - T_{1} = (d m) a = 0

, so it neither adds nor removes tension.

"Tension is uniform, so both masses in an Atwood feel the same net force."

Same tension ≠ same net force. Each Figure-1 diagram also has its own weight arrow, so net forces (and thus acceleration directions) differ. Uniform

T

is only one of the two arrows on each body.

Why questions

Why is tension always directed away from the body it acts on?

A string can only pull, and pulling means tugging the body toward the string — so in Figure 1 the tension arrow on the body points along the string, away from the body into the string.

Why does "inextensible" translate into equal acceleration magnitudes?

Figure 3: total length

L

is constant, so

\overset{x}{˙}_{1} + \overset{x}{˙}_{2} = 0

and differentiating again gives

\overset{x}{¨}_{1} = - \overset{x}{¨}_{2}

. Equal magnitude, opposite sign — see Constraint Relations.

Why can we treat two connected blocks as one system to find

a

Internal tension appears as the equal-and-opposite

T

arrows of Figure 2, so it cancels for the whole system; only external forces set the common acceleration.

Why is the massless assumption essential for "same

T

throughout"?

The string-element equation of Figure 2,

T_{2} - T_{1} = (d m) a

, forces

T_{2} = T_{1}

only when

d m = 0

. Any mass lets tension vary along the string.

Why does the Atwood tension satisfy

T < m_{1} g

and

T > m_{2} g

when

m_{1} > m_{2}

In the heavy mass's Figure-1 diagram the weight arrow wins (

T < m_{1} g

); in the light mass's diagram the tension arrow wins (

T > m_{2} g

). The single

T

lives between the two weights.

Why doesn't the Normal Force appear in the horizontal equation of a block sliding on a frictionless table?

Normal force is vertical and cancels the block's weight in its FBD; horizontally only the tension arrow acts, so only

T = m_{1} a

survives.

Edge cases

Atwood machine with

m_{2} = 0

(one side empty): what happens to

a

and

T

a = g

(the remaining mass is in free fall) and

T = 0

(nothing on the other side to create a pull).

Atwood machine as

m_{1} \to \infty

with

m_{2}

fixed: what does

a

approach?

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

. The huge mass essentially free-falls, dragging the tiny one up at nearly

g

Block-on-table with hanging mass, as

m_{1} \to \infty

: what does

T

approach?

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

. The giant table block barely moves, so the hanging mass nearly hangs statically at

T = m_{2} g

A string goes slack (e.g. a mass thrown upward faster than the string can follow): what is

T

T = 0

. A slack string exerts no force; tension can never be negative, so it drops to zero until the string is taut again.

Two blocks pushed together on a frictionless floor by force

F

(contact, no string). Is the internal contact force a "tension"?

No — contact between blocks can push (it's a normal force). Tension only exists in something being pulled, like a connecting string.

Massless string, but the pulley has friction that grips it: is tension still equal on both sides?

No. Friction lets the pulley resist slipping, so it can support a tension difference across itself — the ideal "same

T

" rule breaks.

Recall

Recall One-line rescues for the top traps

Tension = weight? → Only when $a = 0$ ; falling mass has $T = m (g - a)$ .
Heavier side, more tension? → No — same $T$ , different accelerations.
Push a string? → Never; strings only pull.
Same $T$ over any pulley? → Only ideal (massless, frictionless) pulleys.

Connections

Newton's Second Law — every trap here is resolved by writing $F = ma$ per body.
Free Body Diagrams — the tool that draws the arrows in Figures 1–2.
Constraint Relations — the general form of Figure 3's length equation.
Atwood Machine — the canonical source of the tension traps.
Frictionless Pulleys vs Pulleys with Inertia — when "same $T$ " fails.
Normal Force — the pushing cousin, contrasted with pulling tension.