LearnYMA01WME01 · Mechanics 1Dynamics of a particle in a straight lineConnected particles and pulleys

Dynamics of a particle in a straight line · Subtopic 3 of 4

Connected Particles and Pulleys

Particles joined by a light inextensible string share the same acceleration, and a light string over a smooth pulley has the same tension on both sides. Write $F=ma$ for each particle separately; adding the equations eliminates the tension and gives $a$ .

Edexcel YMA01 WME01≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Write an equation of motion for each connected particle.
Solve towing problems with a tow bar or string.
Solve pulley problems, including on a table or slope.
Find the force on a pulley and handle motion after a string goes slack.

Pause and think

$3$ kg and $5$ kg masses hang over a smooth pulley. Which way do they move, and what drives the motion?

Section 2 of 14

Core idea

Start hereA lift and its counterweight

A lift is connected by cable over a pulley to a counterweight. The heavier side goes down, the lighter goes up, at the same rate because the cable doesn't stretch. Each body gets its own $F=ma$ , and the equations solve together.

The big idea

Light inextensible string: same acceleration. Smooth pulley: same tension both sides. Write $F=ma$ for each particle in its own direction of motion; add to eliminate $T$ .

1. Two particles over a pulley

Two particles over a smooth pulley

The 5 kg particle goes down, the 3 kg up, both with acceleration

a

General

m_1>m_2

$m_1g-T=m_1a$
Heavier particle, down positive.
$T-m_2g=m_2a$
Lighter particle, up positive.
$a=\frac{(m_1-m_2)g}{m_1+m_2}$
Add: $T$ cancels.
$T=\frac{2m_1m_2g}{m_1+m_2}$
Substitute back.

Worked example5 kg and 3 kg

Released from rest. $a$ , $T$ , force on the pulley.

Goal: Three values.
Strategy: Formula or the two equations.

$49-T=5a,\quad T-29.4=3a$
Each particle.
$19.6=8a\Rightarrow a=2.45$
Add.
$T=29.4+7.35=36.75$
Back-substitute.
$2T=73.5\text{ N}$
Both strands pull down on the pulley.

Answer

2.45

m s

^{-2}

;

36.75

73.5

✓ Check

\frac{2(5)(3)(9.8)}8=36.75

✓;

29.4<T<49

✓.

Worked exampleString breaks after 1 s

How much higher does the 3 kg particle rise?

Goal: Extra height.
Strategy: Speed at break, then free motion under gravity.

$v=0+2.45(1)=2.45$
Speed when the string breaks.
$0=2.45^2-19.6s\Rightarrow s=0.306$
Now only gravity acts.

Answer

0.306

✓ CheckIt keeps rising briefly: a particle doesn't stop instantly ✓.

Worked exampleTable and hanging mass

4 kg on a rough table ( $\mu=0.25$ ) tied over the edge to 2 kg hanging. $a$ and $T$ ?

Goal: $a$ , $T$ .
Strategy: Friction on the table particle.

$F=0.25\times39.2=9.8$
$R=4g$ .
$T-9.8=4a,\quad19.6-T=2a$
Each particle.
$9.8=6a\Rightarrow a=1.63,\ T=16.3$
Add, then substitute.

Answer

1.63

m s

^{-2}

;

16.3

✓ CheckHanging:

19.6-16.3=3.3\approx2(1.63)

✓.

Common misconception $T=mg$ Only if the hanging mass is in equilibrium. Accelerating down means

T<mg

Pause and think

Why is the acceleration the same for both?

Summary

One equation per particle.
Add to eliminate $T$ .

Smooth pulley: Tension unchanged across it.

HistoryAtwood's machine. In 1784 George Atwood built two masses over a pulley to measure

g

. Because the acceleration

\tfrac{(m_1-m_2)g}{m_1+m_2}

is much smaller than

g

, it could be timed by hand, a clever way to study gravity in slow motion.

Fun factLifts are Atwood machines. Most lifts have a counterweight on the other side of the pulley, roughly the mass of the car plus half a full load. The motor only has to supply the difference, which saves a huge amount of energy.

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