JEE Mains Physics · Laws of Motion

Connected Bodies and Pulleys

Bodies tied by a taut string move together, so treat them as one system to find the acceleration, then cut the string at one body to find the tension.

All Laws of Motion notes JEE Mains Physics strategy

Why this matters

Twenty PYQs, sixteen of them multiple choice, and two from 2026. Twelve move as one system with one acceleration: Atwood machines (four of them ask only for a mass ratio), blocks pulled in a line, masses hanging in series, a chain over a pulley. Four add a movable pulley, where the bodies' accelerations differ; four add friction on a table or an incline.

Concept 1 of 3: One acceleration, then one tension at a time

If the string stays taut, every body on it moves with the same size of acceleration. So first take all of them together: the tensions inside the system cancel, and the acceleration is the net driving force over the total mass. Then cut the string next to one body and apply F = ma to that body alone to get the tension.

Definition

System: $a = \dfrac{\text{net driving force}}{\text{total mass}}$ . Tensions inside the system cancel.
Atwood machine: $a = \dfrac{(m_2 - m_1)g}{m_1 + m_2}$ , $T = \dfrac{2m_1m_2g}{m_1 + m_2}$ .
Blocks pulled in a line on a smooth floor: the tension in a string is the mass BEHIND it times a.
A hanging body: $T = m(g - a)$ if it accelerates down, $m(g + a)$ if up. Masses hanging in series: the top string holds everything below it.
Double incline: each block's weight is replaced by $mg\sin\theta$ on its own face.
Chain of length L over a pulley, l on the short side: $a = \dfrac{(L - 2l)g}{L}$ . Balls or chain over a table edge: $a = \dfrac{\text{hanging mass}}{\text{total mass}}\,g$ .
A spring between two bodies pushes or pulls both with the same force, in opposite directions.

Atwood machine

a = \frac{(m_2 - m_1)g}{m_1 + m_2}, \qquad T = \frac{2m_1m_2\,g}{m_1 + m_2}

Worked example

Masses of 3 kg and 7 kg hang over a smooth pulley. A 2 kg mass hangs below the 7 kg mass on a second string. Find the acceleration and both tensions. (g = 10 m/s²)

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The same idea in a real exam question:

JEE Mains · 2022 · 28 July 2022 · Q94Moderate

Example 1 · Laws of Motion · Connected Bodies and Pulleys

A uniform metal chain of mass

m

and length '

L

' passes over a massless and frictionless pulley. It is released from rest with a part of its length '

l

' is hanging on one side and rest of its length '

L - l

' is hanging on the other side of the pulley. At a certain point of time, when

l =\frac{L}{x}

, the acceleration of the chain is

\frac{g}{2}

. The value of

x

The tension is not the hanging weight

A hanging mass that accelerates down pulls with m(g − a), less than mg. Only when nothing moves is the tension equal to the weight.

Divide by the total mass

The driving force accelerates every body on the string. Dividing by the hanging mass alone gives an acceleration that is far too large.

Read which ratio is asked

Atwood questions ask for m₁/m₂ in one paper and m₂/m₁ in another, and the stem may name either mass as the heavier one. Decide which is heavier before writing the ratio: the heavier over the lighter is always above 1.

Drill 11 more on one acceleration, then one tension at a time

Concept 2 of 3: Movable pulleys and string constraints

A string does not stretch, so its total length is fixed. When a pulley hangs in a loop of the string, the load rises or falls by half as much as the free end moves, so its acceleration is half. Write this link between the accelerations first; then the force equations have only one unknown acceleration.

Definition

Write the string's length in terms of the positions of the bodies, keep it constant, and differentiate twice.
A movable pulley held by two segments of one string moves half as far as the free end: $a_{\text{load}} = a_{\text{end}}/2$ .
The movable pulley is pulled by 2T (both segments), the free end's body by T.
General rule: over every point a string pulls, $\sum T_i a_i = 0$ , with each $a_i$ measured along the string.
A body resting on, or sliding along, a moving block shares that block's acceleration along the direction it cannot move relative to it.

Constraint

\sum T_i\,a_i = 0, \qquad a_{\text{load}} = \frac{a_{\text{end}}}{2}

Worked example

A 6 kg block on a smooth table is tied to a string that runs over a pulley at the edge, down round a movable pulley carrying a 4 kg mass, and up to a fixed hook. Find the accelerations and the tension. (g = 10 m/s²)

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The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q2Moderate

Example 2 · Laws of Motion · Connected Bodies and Pulleys

All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass

2\text{ }kg

is:

Equal accelerations do not hold with a movable pulley

Only bodies on the same straight run of string share an acceleration. The load on a movable pulley moves at half the rate of the free end; giving both the same a makes every option wrong.

The movable pulley feels 2T

Two segments of the string pull the movable pulley up, so its force equation has 2T. The body on the free end feels only T.

Drill 3 more on movable pulleys and string constraints

Concept 3 of 3: Friction inside a pulley system

Friction is one more resisting force in the system equation. First check that the system moves at all: if the driving force does not beat the largest static friction, nothing moves and the friction is only as large as the driving force. Friction acts only on surfaces that slide over each other.

Definition

Check first: does the driving force exceed $\mu_s N$ ? If not, a = 0 and the friction equals the driving force.
If it moves: $a = \dfrac{\text{driving force} - \mu_k N}{\text{total mass}}$ . On a table, N is the weight of the block on it.
Block on an incline tied to a hanging mass: the resisting forces are $mg\sin\theta$ and $\mu mg\cos\theta$ if the block is dragged up.
Stacked blocks: friction acts between two surfaces only if they slide or tend to slide. A free block riding on another at constant velocity feels none.

Table block and hanging block

a = \frac{m_h g - \mu_k m_t g}{m_h + m_t}

Worked example

A 30 kg trolley on a table (

\mu_k = 0.1

) is tied over a pulley at the edge to a hanging 10 kg block. Find the acceleration and the tension. (g = 10 m/s²)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q105Moderate

Example 3 · Laws of Motion · Connected Bodies and Pulleys

A block of mass

40\text{ }kg

slides over a surface, when a mass of

4\text{ }kg

is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is 0.02 . The acceleration of block is. (Given

g = 10{\text{ }ms}^{- 2}

Check that it moves before using kinetic friction

If the driving force is smaller than the largest static friction, the answer is a = 0. Subtracting μN anyway gives a negative acceleration, which means the check was skipped.

No sliding, no friction

A block lying freely on another block that moves at constant velocity feels no friction, because nothing tries to make it slide. Adding μmg there adds a force that is not present.

Drill 3 more on friction inside a pulley system

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

One acceleration, then one tension at a time
Atwood machine
$a = \frac{(m_2 - m_1)g}{m_1 + m_2}, \qquad T = \frac{2m_1m_2\,g}{m_1 + m_2}$
Movable pulleys and string constraints
Constraint
$\sum T_i\,a_i = 0, \qquad a_{\text{load}} = \frac{a_{\text{end}}}{2}$
Friction inside a pulley system
Table block and hanging block
$a = \frac{m_h g - \mu_k m_t g}{m_h + m_t}$

Watch out for (7)

The tension is not the hanging weight
→ One acceleration, then one tension at a time
Divide by the total mass
→ One acceleration, then one tension at a time
Read which ratio is asked
→ One acceleration, then one tension at a time
Equal accelerations do not hold with a movable pulley
→ Movable pulleys and string constraints
The movable pulley feels 2T
→ Movable pulleys and string constraints
Check that it moves before using kinetic friction
→ Friction inside a pulley system
No sliding, no friction
→ Friction inside a pulley system

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