Why is tension the same throughout a massless string?

Because a massless string has zero mass, any net force on a segment would give infinite acceleration. So Newton's second law forces the tensions at the two ends of every segment to be equal. If the string runs over a massless frictionless pulley, the pulley needs no torque to spin and tension transfers across it unchanged — the same T appears on both sides.

Why are accelerations of connected bodies equal in magnitude?

Because the string is inextensible. If the heavier block moves down by x, the lighter one moves up by the same x — the total string length is fixed. Differentiating twice, the magnitudes of the two accelerations must match. Signs differ when they move in opposite directions, which is why a sign convention is essential.

What if the pulley has mass?

Then the tensions on the two sides differ. The net torque (T1 − T2)R must equal Iα, where I is the pulley's moment of inertia and α is its angular acceleration. NEET typically assumes a massless frictionless pulley, but a question that mentions moment of inertia I is signalling that rotational dynamics is in play — covered fully in System of Particles and Rotational Motion.

Does friction affect the formula for Atwood's machine?

If the pulley is frictionless and the two masses hang freely, the classical Atwood formulas a = (m2 − m1)g/(m1 + m2) and T = 2 m1 m2 g/(m1 + m2) hold. If one mass slides on a rough surface, you add the kinetic friction term to that block's equation. If the pulley itself has friction or rotational inertia, the two tensions differ and the formula no longer applies as-is.

How do I find tension in a multi-pulley system?

Write the constraint that the total length of the string is constant, then differentiate twice to relate the accelerations. In a single movable-pulley system, if the free end moves with acceleration a, the load moves with a/2. Draw one FBD per body, write F = ma for each, and use the constraint to eliminate one unknown. The result for the ideal lifting case is that the applied force is half the load.

What happens if the string goes slack?

When the string goes slack, tension drops to zero and the connection is broken for that instant. Each body then becomes a free particle under whatever other forces act on it. This is exactly the NEET 2017 spring-cut scenario — once the string is severed, block B has only gravity acting and falls freely with a = g, while A keeps the spring force it had a moment earlier.

Pulley & Connected-Body Problems — NEET Physics

What "connected-body" means

A connected-body system is two or more particles linked by an inextensible massless string, optionally routed over one or more pulleys. The string imposes a kinematic constraint: if block A moves, block B must move by a related amount. Three canonical setups cover almost every NEET question — two blocks on a smooth surface pulled by an external force, a hanging block tied to a block on a table, and the Atwood machine (two masses hanging on either side of a single pulley).

The four golden rules

Massless string ⇒ tension is the same everywhere along that string. A massless segment cannot have a net force, so the tensions at its two ends must be equal.
Massless, frictionless pulley ⇒ tension transfers across unchanged. The pulley redirects the string without needing torque, so $T$ on the left equals $T$ on the right.
Inextensible string ⇒ equal magnitudes of acceleration. If A goes down by $x$, B comes up by $x$. Differentiating twice: $|a_A| = |a_B|$.
One FBD per body, then $F = ma$ on each. Use the constraint to eliminate one unknown.

NEET Trap 1 · Pulley with mass

"Same tension on both sides" requires BOTH conditions: massless string AND massless frictionless pulley.

If the stem specifies a moment of inertia $I$ for the pulley, tensions on the two sides differ: $(T_1 - T_2)R = I\alpha$ with $\alpha = a/R$. NEET sometimes plants this single phrase to disqualify the standard Atwood formula.

Default for NEET: assume massless string and massless frictionless pulley unless the stem says otherwise.

Two blocks on a smooth horizontal surface

Blocks $m_1$ and $m_2$ sit on a smooth surface, joined by a light string. A horizontal force $F$ pulls $m_1$ (NCERT Exercise 4.15: $m_1 = 10$ kg, $m_2 = 20$ kg, $F = 600$ N). Both share the same acceleration.

Treat them as a single body of mass $m_1 + m_2$ — tension is internal:

\[ a = \frac{F}{m_1 + m_2} = \frac{600}{30} = 20\ \text{m/s}^2. \]

Isolate $m_2$ — only $T$ acts horizontally:

\[ T = m_2 \, a = 20 \times 20 = 400\ \text{N}. \]

General results:

\[ \boxed{\ a = \frac{F}{m_1 + m_2},\qquad T = \frac{F\, m_2}{m_1 + m_2}.\ } \]

NEET Trap 2 · Which mass goes in the tension numerator?

For two blocks pulled by F from one end, the TRAILING block's mass goes in the numerator of T.

$F$ acts on $m_1$; only $T$ pulls the trailing $m_2$. Hence $T = m_2 a = F m_2 /(m_1+m_2)$ — with $m_2$ on top, not $m_1$. NEET sets this up by drawing $F$ at the left versus right and switching the labels.

Rule: the tension equals "force × (mass on the OTHER side of the string) / total mass".

Hanging block + horizontal trolley (with friction)

NCERT Example 4.9: a $3$ kg block hangs over a frictionless edge-pulley, connected to a $20$ kg trolley on a horizontal surface with $\mu_k = 0.04$ ($g = 10$ m/s²).

Take the hanging block's descent as positive. FBD of $3$ kg (weight $30$ N down, $T$ up):

\[ 30 - T = 3a. \]

FBD of $20$ kg trolley: $N = 200$ N, $f_k = 0.04 \times 200 = 8$ N opposing motion; $T$ forward:

\[ T - 8 = 20a. \]

Add:

\[ 30 - 8 = 23a \;\Rightarrow\; a = \tfrac{22}{23} \approx 0.96\ \text{m/s}^2,\qquad T \approx 27.1\ \text{N}. \]

Sanity check: $T < 30$ N — if $T = mg$, the hanging block would not accelerate.

Friction primer

Confused about $f_k = \mu_k N$ vs the static maximum? Read the Friction (static and kinetic) deep-dive before tackling rough-surface pulley problems.

Atwood machine — the canonical two-mass pulley

Two masses hanging on either side of a single frictionless massless pulley. Take $m_2 > m_1$, so $m_2$ descends and $m_1$ rises with the same magnitude of acceleration $a$.

FBD of $m_2$ (down positive): $m_2 g - T = m_2 a$. FBD of $m_1$ (up positive): $T - m_1 g = m_1 a$. Add and solve:

\[ \boxed{\ a = \frac{(m_2 - m_1)\,g}{m_1 + m_2},\qquad T = \frac{2\, m_1 \, m_2 \, g}{m_1 + m_2}.\ } \]

Verify with NCERT Exercise 4.16 ($m_1 = 8$ kg, $m_2 = 12$ kg):

\[ a = \frac{4\times 10}{20} = 2\ \text{m/s}^2,\qquad T = \frac{2\times 8\times 12\times 10}{20} = 96\ \text{N}. \]

Sanity check: $T$ should lie between $m_1 g = 80$ N and $m_2 g = 120$ N. If your answer falls outside, you have a sign error.

NEET Trap 3 · Sign convention

Pick ONE direction for the system's positive acceleration and stick to it.

For Atwood, the heavier mass descends. Choose "down for the heavier, up for the lighter" as positive — both equations then use the same $a$. Mixing conventions gives nonsense like $T < 0$.

Rule: follow the string. The direction the string moves is the positive direction for the system — "down" on one side, "up" on the other.

Connected blocks on an inclined plane

Mass $M$ on a smooth incline of angle $\theta$ is connected over a top pulley to a hanging mass $m$. Assume $m$ descends. FBD of $m$: $mg - T = ma$. FBD of $M$ along the slope: $T - Mg\sin\theta = Ma$. Add:

\[ a = \frac{m - M\sin\theta}{m + M}\,g,\qquad T = \frac{Mm(1 + \sin\theta)}{m + M}\,g. \]

The motion goes in the assumed direction only if $m > M\sin\theta$; equality gives equilibrium with $T = mg$.

The constraint equation

Behind "equal-magnitude accelerations" lies a general principle: the total length of the string is constant. Differentiate twice and you relate accelerations. For two blocks across a fixed pulley, $x_A = x_B \Rightarrow a_A = a_B$ — that is rule (iii).

For a movable pulley — one string end fixed to the ceiling, the string loops under the pulley supporting a load, the free end is pulled — the load rises by $x/2$ when the free end is pulled by $x$, because the same shortening splits across two supporting segments:

\[ a_{\text{load}} = \tfrac{1}{2}\, a_{\text{free end}}. \]

This 2:1 ratio is the basis of the simple pulley's mechanical advantage: pull with half the load's weight, but pull twice as much rope.

NEET Trap 4 · Movable pulleys halve velocity

In a movable-pulley system, the load moves at HALF the free end's velocity (and acceleration).

Students assume equal accelerations because that's the fixed-pulley default. With a movable pulley, two string segments support the load — both shorten by $x$ as the load rises by $x$, so the free end descends by $2x$. Ratio 2:1, not 1:1.

Rule: count the segments supporting the load. If $n$ segments support it, the load's velocity is $1/n$ of the free end's, and the lifting force is the load's weight divided by $n$.

Effect of pulley mass and string mass

Massive pulley. Tensions on the two sides differ; $(T_2 - T_1)R = I\alpha$ with $\alpha = a/R$ (rotational dynamics, Chapter 6). Massive string. Tension varies along the string; for a uniform vertical string of linear density $\mu$, $T(y) = \mu g y$ above the bottom. NEET defaults to a massless string and massless frictionless pulley unless explicitly stated.

Quick recap

Pulley playbook in five lines

Default: massless string, massless frictionless pulley ⇒ one tension $T$ everywhere on the string.
Inextensible string ⇒ $|a|$ is the same for both connected bodies (single fixed pulley).
Atwood machine: $a = (m_2 - m_1)g/(m_1 + m_2)$, $T = 2 m_1 m_2 g/(m_1 + m_2)$.
Two blocks pulled by $F$ on smooth surface: $a = F/(m_1+m_2)$, $T = F m_{\text{trailing}}/(m_1+m_2)$.
Movable pulley: load's acceleration is half the free end's acceleration. Mechanical advantage 2:1.

Worked examples

Worked example 1 · NCERT Ex 4.16 (Atwood)

Two masses $8\text{ kg}$ and $12\text{ kg}$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.

Solution. Take the heavier mass ($m_2 = 12$ kg) descending. FBD of $12$ kg: $12g - T = 12a$. FBD of $8$ kg: $T - 8g = 8a$. Adding: $4g = 20a \Rightarrow a = g/5 = 2\text{ m/s}^2$. Substituting back: $T = 8(g+a) = 8\times 12 = 96\text{ N}$. Answer: $a = 2$ m/s², $T = 96$ N.

Worked example 2 · NCERT Ex 4.15 (two blocks + force)

Two blocks of masses $10\text{ kg}$ and $20\text{ kg}$ are connected by a light string and placed on a smooth horizontal surface. A horizontal force of $600\text{ N}$ is applied to the $10\text{ kg}$ block. Find (i) the acceleration of the system and (ii) the tension in the string.

Solution. Treat as one body for acceleration: $a = 600/(10+20) = 20\text{ m/s}^2$. For tension, isolate the $20$ kg block (only $T$ pulls it horizontally): $T = 20\times 20 = 400\text{ N}$. Answer: $a = 20$ m/s², $T = 400$ N.

Worked example 3 · NCERT Ex 4.9 (hanging + trolley + friction)

A $3\text{ kg}$ block hangs over a pulley and is connected to a $20\text{ kg}$ trolley on a horizontal surface. Coefficient of kinetic friction between trolley and surface $= 0.04$, $g = 10$ m/s². Find the acceleration and tension.

Solution. FBD of $3$ kg: $30 - T = 3a$. FBD of $20$ kg trolley: $N = 200\text{ N}$, $f_k = 0.04\times 200 = 8\text{ N}$, equation $T - 8 = 20a$. Add: $22 = 23a \Rightarrow a = 22/23 \approx 0.96\text{ m/s}^2$. Then $T = 30 - 3(0.96) \approx 27.1\text{ N}$. Answer: $a \approx 0.96$ m/s², $T \approx 27.1$ N.

Worked example 4 · Table block + hanging mass (smooth)

A $5\text{ kg}$ block sits on a smooth horizontal table, connected over a frictionless edge-pulley to a $3\text{ kg}$ block hanging off the side. Find the acceleration and tension. ($g = 10$ m/s².)

Solution. FBD of $3$ kg hanging: $3g - T = 3a$. FBD of $5$ kg on table (no friction): $T = 5a$. Add: $30 = 8a \Rightarrow a = 30/8 = 3.75\text{ m/s}^2$. Then $T = 5\times 3.75 = 18.75\text{ N}$. Answer: $a = 3.75$ m/s², $T = 18.75$ N. Note the general formula $a = mg/(M+m)$ for this geometry.

Pulley and Connected-Body Problems