Pulleys and connected bodies: a light inextensible string over a smooth pulley, hanging masses, a mass on a smooth table or smooth inclined plane, equilibrium, finding acceleration and tension, and what happens when the string goes slack (new in the 2024 Extension 2 syllabus)

Syllabus dot point

“Solve problems on motion without resistance involving a single smooth pulley and a smooth horizontal or inclined plane, by applying Newton's second law to each body”

HSCMaths Extension 2Applications of Calculus to Mechanics…13 min readUpdated 7 Oct 2026

Quick answer

For a light inextensible string over a smooth pulley both bodies share the acceleration magnitude and the tension $T$ . Write Newton's second law for each body along its direction of motion, add to eliminate $T$ , and solve: Atwood $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$ , $T = \frac{2m_1m_2g}{m_1 + m_2}$ ; a mass on a smooth slope uses $mg\sin\alpha$ . The force on the pulley is $2T$ for two vertical strands, and the string goes slack when a hanging mass lands. New to Extension 2 in the 2024 syllabus.

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What this dot point is asking
The answer
Practice questions

What this dot point is asking

Pulley problems are new to Extension 2 in the Mathematics Extension 2 11-12 Syllabus (2024), first examined in the 2027 HSC. NESA's Year 12 Applications of calculus to mechanics asks you to analyse motion on a smooth inclined plane by resolving forces into components, and to "Solve motion problems involving a single smooth pulley and a smooth inclined plane where a body hangs vertically or lies on a smooth horizontal or inclined plane". That is the whole scope: one pulley, smooth surfaces, a string joining a hanging body to one on a table or a slope. Separately, the point on resisted motion says "Solve problems, excluding those with pulley systems, involving a particle moving in a straight line subject to a resistance oppositely directed to the motion and with a magnitude proportional to a power of the speed", so pulleys do not appear in motion with resistance. The inclined plane part is covered in our page on forces and Newton's laws; this page adds the string and pulley.

Note

A pulley is a wheel that only changes the direction of a rope. If you hang a heavy bucket on one end of a rope and a light one on the other, the heavy bucket falls and drags the light one upward. Both buckets move at the same speed, because the rope does not stretch. The rope pulls on both buckets with the same strength, because the wheel is smooth and does not hold any of the pull. So all you need to find is one acceleration shared by both buckets and one rope pull.

The answer

Key fact

The modelling assumptions for a "light inextensible string over a smooth pulley" are:

The string does not stretch, so both bodies have the same acceleration magnitude (and speed).
The string is light, and the pulley is smooth, so the tension $T$ is the same at both ends.
A smooth plane or table exerts only a normal reaction, with no friction.

Apply Newton's second law $\sum F = ma$ to each body separately, along the direction of its motion, then solve the two equations for $a$ and $T$ . The force of the string on the pulley is the vector sum of the two tensions: for two vertical strands it is $2T$ downward.

The method, step by step

Decide the direction of motion. Compare the "pulling" forces (for a mass on a slope this is $mg\sin\alpha$ down the slope; for a hanging mass it is $mg$ ).
Draw a force diagram for each body. Label weight, tension, and a normal reaction where there is a surface.
Choose a positive direction for each body that matches its direction of motion. For a string over a pulley, the positive direction "round the string" has both bodies accelerating positively.
Write one equation per body and solve the pair as simultaneous equations. Adding the two eliminates $T$ .
Check by substituting back into the equation you did not use.

Three standard set-ups

Both hanging (Atwood machine): Masses $m_1 < m_2$ : $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$ and $T = \frac{2m_1m_2g}{m_1 + m_2}$ .
One on a smooth horizontal table, one hanging: The table mass has $T = m_1 a$ and the hanging mass has $m_2 g - T = m_2 a$ , so $a = \frac{m_2 g}{m_1 + m_2}$ .
One on a smooth slope at angle $\alpha$ , one hanging: For the body on the slope (up the slope positive, if it is being dragged up), $T - m_1 g\sin\alpha = m_1 a$ ; for the hanging body $m_2 g - T = m_2 a$ . Then $a = \frac{(m_2 - m_1\sin\alpha)g}{m_1 + m_2}$ . If this is negative, the slope body moves down instead, and the signs must be redone.

When the string goes slack

If the hanging mass hits the floor, or the system is cut loose, the tension drops to zero. The remaining body then moves under its own weight (and the normal reaction) alone. Use the speed at that instant as the initial speed for the next stage.

How exam questions use pulley systems

"Find the acceleration and the tension." Two equations, one for each body.
"Show that the system is in equilibrium (or find the mass that gives equilibrium)." Set $a = 0$ .
"Find the speed after the hanging mass has fallen $h$ ." Use constant-acceleration formulas with the answer for $a$ .
"The string is cut / the mass lands. Find how far the other body travels." Two stages, with the speed from stage one as the initial speed for stage two.
"Find the force on the pulley." Add the two tension vectors.

Worked examples

Equal masses

Two equal masses $m$ hang from a string over a smooth pulley. Find $a$ and $T$ .

Step 1: use the formulas

With $m_1 = m_2 = m$ : $a = 0$ and $T = \frac{2m^2 g}{2m} = mg$ .

Step 2: interpret

The system stays at rest (or moves at constant speed if already moving), and each side of the string supports one weight. The force on the pulley is $2mg$ .

Marker's note: this is the equilibrium special case, and a good check on any general formula.

Common traps

Using $T = mg$ for a moving hanging mass: The tension equals the weight only when the mass does not accelerate. For a moving mass use $mg - T = ma$ .
Mixing sign conventions: Take the direction of motion as positive for each body separately, so that both bodies have acceleration $+a$ . Using "down" for one body and "up the slope" for the other without adjusting the signs gives a wrong equation.
Putting $mg\cos\alpha$ along the slope: The component of the weight along a slope at angle $\alpha$ is $mg\sin\alpha$ ; the component $mg\cos\alpha$ is balanced by the normal reaction.
Forgetting that the pulley experiences $2T$: The force of the string on a pulley with two vertical strands is twice the tension, not the tension.
Assuming the direction of motion: If you guess the wrong direction, $a$ will come out negative. Compare the pulling forces first.

Exam technique

Draw one clear diagram with every force on each body, and write "same $T$ , same $a$ " next to the string. State the positive direction in words. After solving, substitute the answer back into the equation you did not use to catch sign errors. Give answers to the accuracy of the data, with units. If the question says "smooth", ignore friction. NESA's pulley point lists smooth surfaces only; a rough surface is beyond the listed scope of this dot point.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks

Particles of mass

3

kg and

5

kg hang at the ends of a light inextensible string that passes over a smooth fixed pulley. They are released from rest. Taking

g = 9.8

m/s

^2

, find (a) the acceleration of the particles, (b) the tension in the string, and (c) the downward force that the string exerts on the pulley.

Show worked solution →

Set up. The $5$ kg mass moves down and the $3$ kg mass moves up, each with acceleration $a$ (same magnitude because the string is inextensible). Let $T$ be the tension, which is the same on both sides of a smooth pulley.

Newton's second law for each mass, taking its direction of motion as positive.

$5g - T = 5a$ and $T - 3g = 3a$ .

(a) Add the equations to eliminate $T$ : $2g = 8a$ , so $a = \frac{g}{4} = 2.45$ m/s $^2$ .

(b) From the second equation, $T = 3g + 3a = 3(9.8 + 2.45) = 36.75$ N. Check with the first: $5(9.8 - 2.45) = 36.75$ N.

(c) The string pulls down with the tension $T$ on each side of the pulley, so the downward force is $2T = 73.5$ N.

Marker's note: one mark for the two equations, one for the acceleration, one for $T$ and the force $2T$ on the pulley. Stating the positive direction for each body is what keeps the signs correct.

core3 marks

A particle of mass

4

kg lies on a smooth horizontal table. It is joined by a light inextensible string over a smooth pulley at the edge of the table to a particle of mass

6

kg hanging vertically. The system is released from rest. Take

g = 9.8

m/s

^2

. Find the acceleration and the tension, and the speed of the particles after the hanging particle has fallen

1.5

Show worked solution →

Equations. The only horizontal force on the $4$ kg particle is the tension (the table is smooth). For the hanging particle, weight $6g$ acts down and tension up.

$T = 4a$ and $6g - T = 6a$ .

Add: $6g = 10a$ , so $a = 0.6g = 5.88$ m/s $^2$ and $T = 4a = 23.52$ N.

Speed after $1.5$ m. Starting from rest with constant acceleration, $v^2 = 2as = 2(5.88)(1.5) = 17.64$ , so $v = 4.2$ m/s.

Marker's note: one mark for the two equations, one for $a$ and $T$ , one for the speed using $v^2 = 2as$ .

core4 marks

A particle

A

of mass

5

kg lies on a smooth plane inclined at

30^\circ

to the horizontal. It is joined by a light inextensible string, parallel to the line of greatest slope and passing over a smooth pulley at the top of the plane, to a particle

B

of mass

3

kg that hangs vertically. The system is released from rest. Take

g = 9.8

m/s

^2

. Find the acceleration of the system and the tension in the string, and state which way

A

moves.

Show worked solution →

Which way does it move? The component of the weight of $A$ down the slope is $5g\sin 30^\circ = 2.5g$ , while $B$ pulls with $3g$ . Since $3g > 2.5g$ , $B$ descends and $A$ moves up the slope.

Equations of motion (positive direction is the direction of motion):

For $B$ : $3g - T = 3a$ .

For $A$ (along the slope, up positive): $T - 5g\sin 30^\circ = 5a$ , that is $T - 2.5g = 5a$ .

Add: $3g - 2.5g = 8a$ , so $a = \frac{0.5g}{8} = \frac{g}{16} = 0.6125$ m/s $^2$ .

Tension. $T = 3g - 3a = 3(9.8 - 0.6125) = 27.5625 \approx 27.56$ N.

Check using $A$ : $T = 2.5g + 5a = 24.5 + 3.0625 = 27.5625$ N.

Marker's note: one mark for the direction argument, one for each equation and one for $a$ and $T$ . The normal reaction on $A$ is perpendicular to the slope and does not appear in the along-slope equation.

core2 marks

A particle of mass

10

kg rests on a smooth plane inclined at

30^\circ

to the horizontal. It is held in equilibrium by a light string, parallel to the slope, which passes over a smooth pulley at the top of the slope and carries a hanging particle of mass

M

kg. Take

g = 9.8

m/s

^2

. Find

M

, the tension in the string and the normal reaction on the

10

kg particle.

Show worked solution →

Equilibrium means zero acceleration. For the hanging particle, $T = Mg$ . For the particle on the slope, along the slope, $T = 10g\sin 30^\circ = 5g$ .

So $Mg = 5g$ and $M = 5$ kg, and $T = 5g = 49$ N.

Normal reaction. Perpendicular to the slope, $N = 10g\cos 30^\circ = 98 \times \frac{\sqrt{3}}{2} \approx 84.9$ N.

Marker's note: one mark for $M$ and $T$ together, one for $N$ . The normal reaction is $mg\cos\alpha$ , not $mg$ .

exam4 marks

A particle

A

of mass

2

kg lies on a smooth plane inclined at

30^\circ

to the horizontal. It is joined by a light inextensible string, parallel to the slope and passing over a smooth pulley at the top, to a particle

B

of mass

m

kg hanging vertically. When released,

A

accelerates up the slope at

1.4

m/s

^2

. Take

g = 9.8

m/s

^2

. Find

m

and the tension in the string.

Show worked solution →

Equations of motion. $B$ moves down with acceleration $a = 1.4$ and $A$ moves up the slope with the same magnitude.

For $B$ : $mg - T = ma$ .

For $A$ along the slope: $T - 2g\sin 30^\circ = 2a$ , that is $T - g = 2a$ .

Add the two equations: $mg - g = ma + 2a$ , so $m(g - a) = g + 2a$ .

Substitute $g = 9.8$ and $a = 1.4$ : $m(8.4) = 9.8 + 2.8 = 12.6$ , so $m = 1.5$ kg.

Tension. $T = mg - ma = 1.5(9.8 - 1.4) = 12.6$ N.

Check using $A$ : $T = g + 2a = 9.8 + 2.8 = 12.6$ N.

Marker's note: one mark for each equation, one for the value of $m$ , one for $T$ .

exam5 marks

In the system of the third question (

A

of mass

5

kg on the smooth plane at

30^\circ

B

of mass

3

kg hanging),

B

reaches the floor after falling

1.5

m from rest, and the string then goes slack. Assume

A

has not reached the pulley. How much further does

A

move up the slope before it first comes to rest? Take

g = 9.8

m/s

^2

Show worked solution →

Speed when the string goes slack. From the third question the acceleration is $a = \frac{g}{16} = 0.6125$ m/s $^2$ . After $1.5$ m,

v^2 = 2as = 2(0.6125)(1.5) = 1.8375.

Motion after the string goes slack. Now the only force on $A$ along the slope is the component of its weight, so $A$ decelerates at $g\sin 30^\circ = 4.9$ m/s $^2$ (independent of its mass).

Distance to rest. With final speed $0$ ,

0 = v^2 - 2(4.9)d \quad \Rightarrow \quad d = \frac{1.8375}{9.8} = 0.1875 \text{ m}.

$A$ moves a further $0.1875$ m up the slope (so $1.6875$ m in total), then slides back down.

Marker's note: one mark for the first acceleration, one for $v^2$ , one for recognising that the tension vanishes, one for the deceleration $g\sin 30^\circ$ , one for $d$ .

exam4 marks

Particles of masses

m_1

and

m_2

, with

m_2 > m_1

, are attached to the ends of a light inextensible string passing over a smooth fixed pulley and released from rest. Show that the acceleration is

a = \dfrac{(m_2 - m_1)g}{m_1 + m_2}

and that the tension is

T = \dfrac{2m_1m_2 g}{m_1 + m_2}

Show worked solution →

Equations. The heavier mass $m_2$ moves down and $m_1$ moves up, each with acceleration $a$ :

m_2 g - T = m_2 a \qquad \text{and} \qquad T - m_1 g = m_1 a.

Add the equations: $(m_2 - m_1)g = (m_1 + m_2)a$ , which gives

a = \frac{(m_2 - m_1)g}{m_1 + m_2}.

Tension. From the second equation, $T = m_1(g + a)$ . Substituting $a$ ,

T = m_1 g\left( 1 + \frac{m_2 - m_1}{m_1 + m_2} \right) = m_1 g \cdot \frac{2m_2}{m_1 + m_2} = \frac{2m_1m_2 g}{m_1 + m_2}.

Check with $m_1 = 3$ , $m_2 = 5$ : $a = \frac{2g}{8} = 2.45$ m/s $^2$ and $T = \frac{30g}{8} = 36.75$ N, matching direct calculation.

Marker's note: one mark for each equation, one for $a$ , one for $T$ .

Practise this

Flashcards Maths Extension 2 deck

Sources & how we know this

Mathematics Extension 2 11-12 Syllabus (2024): Year 12 Applications of calculus to mechanics — NESA (2024)
Mathematics Extension 2 11-12 Syllabus (2024): course overview and implementation — NESA (2024)
Mathematics Extension 2 Stage 6 Syllabus (2017) — NESA
Mathematics Extension 2 HSC exam papers — NESA