IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

This lesson develops lesson 2-2 of the dynamics unit "Applications on Newton’s Laws of Motion".
It goes beyond the vertical two-mass pulley and treats rough and inclined cases and the force on the pulley.
After it you should be able to:
write one equation of motion for each body and solve for the common acceleration and the tension;
handle a string that is cut, a mass that is added or removed, and a body that hits the ground;
find the force on the pulley and solve problems with friction and inclined planes.

🌳 Concept tree

Connected Bodies→Pulleys, Tables, Inclined Planes and Pulley Reaction

├── 2-2 (a) General method for connected bodies

│ ├── Assumptions

│ └── Steps

├── 2-2 (b) Two masses hanging over a smooth pulley

│ ├── Results

│ └── The pulley reaction

├── 2-2 (c) Adding, removing and cutting

│ ├── Extra mass on one side

│ ├── When the extra mass is removed

│ └── When the string is cut

├── 2-2 (d) A body on a smooth table joined to a hanging body

│ ├── Set-up

│ ├── Pulley reaction at a corner

│ └── With friction on the table

├── 2-2 (e) Inclined planes

│ ├── Body on an incline joined to a hanging body

│ └── Equilibrium conditions

├── 2-2 (f) The force on an inclined pulley

│ └── Rule

├── 2-2 (g) Lifts and apparent weight (connected systems)

│ └── Forces in a lift

└── 2-2 (h) Summary

├── Comparison

└── Check list

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 1 / 15

2-2 (a) General method for connected bodies

Assumptions

The string is light (negligible mass) and inextensible, so connected bodies move with the same speed and the same magnitude of acceleration.
The pulley is small and smooth, so the tension is the same on both sides of it.
The pulley itself is fixed, light, and frictionless unless stated otherwise.

Draw a separate diagram for each body with all the forces on it (weight, tension, normal reaction, friction).
Choose for each body the positive direction as the direction of its motion.
Write the equation $F = m a$ for each body; the unknowns are usually $a$ and $T$ .
Solve the equations together (adding them eliminates $T$ ).
Use the constant-acceleration formulas $v = u + a t$ , $s = u t + \frac{1}{2} a t^{2}$ , $v^{2} = u^{2} + 2 a s$ for the questions about time, distance and speed.
If the string is cut, or a mass is removed, start a new stage of motion with the velocity reached as the initial velocity.

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 2 / 15

2-2 (b) Two masses hanging over a smooth pulley

Results

Let $m_{1} > m_{2}$ hang at the ends of the string. Equations: $m_{1} a = m_{1} g - T$ and $m_{2} a = T - m_{2} g$ .
$a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g, T = \frac{2 m_{1} m_{2}}{m_{1} + m_{2}} g$
If the masses start level, after time $t$ each has moved $\frac{1}{2} a t^{2}$ in opposite directions, so they are $a t^{2}$ apart vertically.

Connected Bodies: a smooth pulley carrying masses of 3 kg and 5 kg shows connected bodies: acceleration 2.45 m/s² and tension 36.75 N.

The pulley reaction

The string pulls the pulley with $T$ on each side, both downward, so the force on the pulley’s axle is
$P = 2 T$
This is the pressure the pulley exerts on its support (the support’s reaction is equal and opposite).
A light pulley’s own weight is ignored.

✍️ Worked examples

Example 1:
Masses 105 g and 70 g hang over a smooth pulley, level and at rest.
Find the acceleration.
If the heavier body hits the ground after falling 50 cm, find the total time until the lighter body first comes to rest instantaneously.
Solution:
$a = \frac{105 - 70}{105 + 70} \times 980 = 0.2 \times 980 = 196$ cm/s $^{2}$ .
When the heavier mass lands v2=0+2(196)(50)=19 600, so v=140 cm/s, after t1=140196=57 s.
The lighter mass then rises freely with initial speed 140 cm/s and deceleration $g$ : it stops after $t_{2} = \frac{140}{980} = \frac{1}{7}$ s.
Total time $= \frac{5}{7} + \frac{1}{7} = \frac{6}{7}$ s.
Example 2:
A 90 g mass and a 70 g mass hang from a string over a smooth pulley.
The 90 g mass starts from rest 245 cm above the ground.
Find (a) the time for it to reach the ground and (b) the time until the string becomes taut again.
Solution:
$a = \frac{90 - 70}{160} \times 980 = 122.5$ cm/s $^{2}$ .
(a) $245 = \frac{1}{2} (122.5) t^{2}$ , so $t^{2} = 4$ and $t = 2$ s. Its speed on landing is $122.5 \times 2 = 245$ cm/s.
(b) The 70 g mass is then 245 cm above its start, moving up at 245 cm/s. The string goes slack and the mass moves freely under gravity.
The string becomes taut when the 70 g mass has returned to the height of 245 cm, i.e. after $\frac{2 \times 245}{980} = 0.5$ s.
That is 2.5 s from the beginning of the motion.
Example 3 A 5 kg and a 3 kg mass hang level over a smooth pulley. Find $a$ , $T$ and the force on the pulley.
Solution:
$a = \frac{5 - 3}{8} (9.8) = 2.45$ m/s $^{2}$ .
$T = 3 (9.8 + 2.45) = 36.75$ N.
Force on the pulley P=2T=73.5 N.
Example 4 Two masses $m_{1} > m_{2}$ start level. After 1 s the vertical distance between them is 20 cm. Find $m_{1} : m_{2}$ .
Solution:
Each has moved 10 cm 10=12a(1)2, so a=20 cm/s2.
$\frac{20}{980} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}$ , so $m_{1} + m_{2} = 49 (m_{1} - m_{2})$ .
m2=48m1, so m1 m2=25:24.

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 3 / 15

2-2 (c) Adding, removing and cutting

Extra mass on one side

If a mass m3 rests on m1 and m1+m3>m2 (m1+m3)a=(m1+m3)g−T and m2a=T−m2g.
The tension in a string joining $m_{3}$ to $m_{1}$ (if tied rather than resting) is found from the equation of $m_{3}$ alone: $m_{3} a = m_{3} g - T^{'}$ .
When equal masses $m$ hang on both sides and a small mass $m^{'}$ is added to one, $a = \frac{m^{'} g}{2 m + m^{'}}$ .

After removal the velocity is the velocity at that moment; the acceleration changes to the new value and may be negative (a deceleration).
The system stops instantaneously if the new acceleration is a deceleration, then reverses.
If the remaining masses are equal, the system continues at a constant velocity.

Each body then moves under gravity only→a=g downward, whatever its direction of motion.

✍️ Worked examples

Example 5:
A 30 g mass hangs on one side of a smooth pulley; two 20 g masses hang tied together on the other side.
After 2 s one of the 20 g masses is removed.
Find the total distance travelled by the 30 g mass until it first comes to rest.
Solution:
Stage 1 a=40−3070×980=140 cm/s2 (the 40 g side goes down).
After 2 s v=280 cm/s and s1=12(140)(4)=280 cm.
Stage 2 masses 30 g and 20 g with 30 g now heavier: a′=20−3050×980=−196 cm/s2.
It comes to rest after $s_{2} = \frac{28 0^{2}}{2 \times 196} = \frac{78 400}{392} = 200$ cm.
The 30 g mass rises $s_{1} + s_{2} = 480$ cm before coming to rest, then reverses.
Example 6:
Masses 5 kg and 3 kg start level at 2.45 m above the ground and the string is cut after 1 s.
Find the speed with which each strikes the ground.
Solution:
$a = 2.45$ m/s $^{2}$ ; at the cut $v = 2.45$ m/s and each has moved $1.225$ m.
The 5 kg mass is $2.45 - 1.225 = 1.225$ m above the ground: $v^{2} = 2.4 5^{2} + 2 (9.8) (1.225) = 6.0025 + 24.01 = 30.0125$ , so $v \approx 5.48$ m/s.
The 3 kg mass is $2.45 + 1.225 = 3.675$ m above the ground and moving upward: take downward positive, $u = - 2.45$ , $s = 3.675$ : $v^{2} = 6.0025 + 2 (9.8) (3.675) = 78.0325$ , so $v \approx 8.83$ m/s.

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 4 / 15

2-2 (d) A body on a smooth table joined to a hanging body

Set-up

A mass $m_{1}$ on a smooth horizontal table is joined by a string over a pulley at the edge to a hanging mass $m_{2}$ .
Equations T=m1a and m2g−T=m2a. a=m2m1+m2 g,T=m1m2m1+m2 g
The table does not change the result because the normal reaction cancels the weight $m_{1} g$ .

At the table edge one part of the string is horizontal and one vertical, both with tension $T$ , so the resultant force on the pulley has magnitude
$P = \sqrt{T^{2} + T^{2}} = \sqrt{2} T$
directed at $4 5^{\circ}$ below the horizontal (towards the line between the strings).
In general, if the two parts of the string make an angle $φ$ with each other, $P = 2 T \cos \frac{φ}{2}$ .

✍️ Worked examples

Example 7 A 3 kg mass on a smooth table is joined over a pulley to a hanging 2 kg mass. Find $a$ , $T$ and the force on the pulley.
Solution:
$a = \frac{2}{5} (9.8) = 3.92$ m/s $^{2}$ .
$T = 3 (3.92) = 11.76$ N.
$P = \sqrt{2} \times 11.76 \approx 16.63$ N.

Friction on the table body is $F = μ m_{1} g$ , opposing motion. The equations become $T - μ m_{1} g = m_{1} a$ and $m_{2} g - T = m_{2} a$ .
$a = \frac{m_{2} - μ m_{1}}{m_{1} + m_{2}} g$
The system moves only if $m_{2} g > μ m_{1} g$ (the hanging weight exceeds limiting friction).

✍️ Worked examples

Example 8:
A 4 kg block on a rough table ( $μ = 0.2$ ) is joined to a hanging 3 kg mass.
Find $a$ and $T$ .
After the hanging mass has fallen 1 m, how far does the block slide further if the 3 kg mass lands and the table is long enough?
Solution:
$a = \frac{3 - 0.2 (4)}{7} (9.8) = \frac{2.2}{7} (9.8) = 3.08$ m/s $^{2}$ .
$T = 3 (9.8 - 3.08) = 20.16$ N. Check on the block: $20.16 - 0.2 (4) (9.8) = 20.16 - 7.84 = 12.32 = 4 (3.08)$ .
After 1 m v2=2(3.08)(1)=6.16.
The string goes slack when the hanging mass lands; the block then decelerates with $a = - μ g = - 1.96$ m/s $^{2}$ .
Further distance $= \frac{v^{2}}{2 μ g} = \frac{6.16}{3.92} \approx 1.57$ m.

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 5 / 15

2-2 (e) Inclined planes

Body on an incline joined to a hanging body

Comparison aspect	Main idea	Keywords
•	A mass $m_{1}$ on a smooth incline of angle $θ$ is connected by a string over a pulley at the top to a hanging mass $m_{2}$ .	masssmoothinclineangle
Component of weight along the plane	m1gsin⁡θ normal reaction: m1gcos⁡θ	m1gsinnormalreactionm1gcos
If m2 moves down	m2g−T=m2a and T−m1gsin⁡θ=m1a. a=m2−m1sin⁡θm1+m2 g	m1gsinm1sin
With friction	μ times the normal reaction opposes motion so the numerator becomes m2−m1sin⁡θ−μm1cos⁡θ	timesnormalreactionopposes

✍️ Worked examples

Example 9 A 6 kg block on a smooth $3 0^{\circ}$ incline is joined over a pulley to a 4 kg hanging mass. Find $a$ and $T$ .
Solution:
$a = \frac{4 - 6 (0.5)}{10} (9.8) = \frac{1}{10} (9.8) = 0.98$ m/s $^{2}$ .
$T = 4 (9.8 - 0.98) = 35.28$ N.
Check on the block 35.28−6(9.8)(0.5)=35.28−29.4=5.88=6(0.98).
Example 10 A 5 kg block on a rough incline with $\sin θ = \frac{3}{5}$ , $\cos θ = \frac{4}{5}$ and $μ = \frac{1}{4}$ is joined over a pulley to a hanging 5 kg mass. Find $a$ and $T$ .
Solution:
Normal reaction 5g×45=4g; friction: 14×4g=g.
Weight component along the plane 5g×35=3g.
Net driving force 5g−3g−g=g, total mass 10 kg, so a=g10=0.98 m/s2.
$T = 5 (g - a) = 5 (8.82) = 44.1$ N.
Check on the block T−3g−g=44.1−39.2=4.9=5(0.98).
Example 11:
Two masses on two smooth inclined planes meeting at a top edge carry a string over a pulley: a 10 kg mass on a plane with $\sin α = 0.6$ and a 5 kg mass on a plane with $\sin β = 0.8$ .
Find $a$ and $T$ .
Solution:
Driving components 10g(0.6)=6g down the first plane and 5g(0.8)=4g down the second plane.
The first body goes down a=6g−4g15=2g15=1.307 m/s2.
$T = 5 (a + 4 g / 5) = 5 a + 4 g = 6.53 + 39.2 = 45.73$ N.
Check on the first body 6g−T=58.8−45.73=13.07=10(1.307).

Equilibrium conditions

A body on a rough incline is in limiting equilibrium with a hanging mass when $m_{2} g = m_{1} g \sin θ + μ m_{1} g \cos θ$ (about to move up) or $m_{2} g = m_{1} g \sin θ - μ m_{1} g \cos θ$ (about to move down).

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 6 / 15

2-2 (f) The force on an inclined pulley

Rule

For a pulley at the top of an incline with one part of the string along the incline and the other vertical, the angle between the two parts is $9 0^{\circ} - θ$ (measured on the side where they pull).
Then $P = 2 T \cos \frac{9 0^{\circ} - θ}{2}$ .
For θ=30∘ P=2Tcos⁡30∘=3 T; with T=35.28 N from Example 9, P≈61.1 N.
Check by components along the plane T and vertical T; the horizontal component of the pulley load is Tcos⁡30∘ and the vertical component is T+Tsin⁡30∘=1.5T, giving (0.866T)2+(1.5T)2=3 T.

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 7 / 15

2-2 (g) Lifts and apparent weight (connected systems)

Forces in a lift

Comparison aspect	Main idea	Keywords
•	A person of mass $m$ in a lift pressing on the floor with force $N$ :	personmassliftpressing
lift accelerating upward with a	N−mg=ma so N=m(g+a)
lift accelerating downward with a	mg−N=ma so N=m(g−a)
constant velocity or at rest	N=mg
For the lift plus person as one system with total mass M	the wire tension is T=M(g+a) going up (and M(g−a) going down)	wiretensiongoingdown
•	The tension in the wire is the sum of the weight-type forces, so it is also $T = N + m^{'} (g + a)$ where $m^{'}$ is the lift’s own mass.	tensionwireweighttype

✍️ Worked examples

Example 12:
A man of 70 kg is in a lift of mass 420 kg that accelerates upward at 0.7 m/s $^{2}$ .
Find the tension in the cable and the man’s pressure on the floor, in kg·wt.
Solution:
Man N=70(9.8+0.7)=735 N =75 kg·wt.
Whole system T=490(9.8+0.7)=5145 N =525 kg·wt.
Check with the lift alone T−420g−N=420(0.7) gives T=4116+735+294=5145 N.
Example 13:
A body is hung from a spring scale fixed to the roof of a lift.
The scale reads 8 kg·wt when the lift rises with acceleration $a$ , and 5 kg·wt when the lift descends with acceleration $2 a$ .
Find the mass and $a$ .
Solution:
Rising T−mg=ma, so 8g−mg=ma . . . (1).
Descending mg−T=m(2a), so mg−5g=2ma . . . (2).
Divide (2) by (1) m−58−m=2, so m−5=16−2m and m=7 kg.
From (1) 7a=(8−7)g, so a=g7=1.4 m/s2.

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 8 / 15

Comparison

Point of comparison	Vertical pulley	Table and hanging body	Incline and hanging body
Acceleration	$\frac{m_{1} - m_{2}}{m_{1} + m_{2}} g$	$\frac{m_{2}}{m_{1} + m_{2}} g$	$\frac{m_{2} - m_{1} \sin θ}{m_{1} + m_{2}} g$
Tension	$\frac{2 m_{1} m_{2}}{m_{1} + m_{2}} g$	$\frac{m_{1} m_{2}}{m_{1} + m_{2}} g$	$m_{2} (g - a)$
Force on pulley	$2 T$	$\sqrt{2} T$	$2 T \cos \frac{9 0^{\circ} - θ}{2}$
With friction	not applicable	subtract $μ m_{1}$ in the numerator	subtract $μ m_{1} \cos θ$ in the numerator

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 9 / 15

Check list

One equation per body, signs consistent with the direction of motion.
The tension is the same throughout a light string over a smooth pulley, and the accelerations are equal in magnitude.
When the string is cut or goes slack, change to $a = \pm g$ (or $- μ g$ on a rough table).
Convert units before substituting (g with cm/s $^{2}$ when masses are in grams and distances in cm).

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 10 / 15

📌 Comprehensive Scientific Terminology & Definitions:

Comparison aspect	Main idea	Keywords
Atwood Machine / Vertical Pulley System (البكرات الملساء الرأسية)	Two masses m₁ and m₂ (m₁ > m₂) connected by a light inextensible string passing over a smooth fixed pulley: both move with common acceleration a = [(m₁ – m₂)g] / (m₁ + m₂)	massesconnectedlightinextensible
Smooth Horizontal Table System (حركة جسم على نضد أفقي أملس)	Mass m₁ on table connected by string over pulley at edge to hanging mass m₂: acceleration a = [m₂ g] / (m₁ + m₂) String tension T = (m₁ m₂ g) / (m₁ + m₂)	Massconnectedstringover
Rough Horizontal Table System (حركة جسم على نضد خشن)	Friction force f_k = μ_k · N = μ_k · m₁ g opposes motion: a = [(m₂ – μ_k m₁)g] / (m₁ + m₂).	Frictionforceopposesmotion
Inclined Plane Pulley System (حركة على مستوى مائل)	Mass m₁ on inclined plane of angle α connected to hanging mass m₂: equation depends on comparing m₂ g with component of weight m₁ g sin α	Massangleconnectedhanging

Pressure / Reaction on Pulley Axle (الضغط على محور البكرة)→Resultant force R exerted by string on the pulley: 1. Two vertical strings: R = 2T; 2. Horizontal and vertical strings: R = T√2; 3. Angle θ between strings: R = 2T cos(θ/2).

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 11 / 15

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

(‘Vertical Pulley Systema = [(m₁ – m₂) / (m₁ + m₂)] · g; T = [2 m₁ m₂ / (m₁ + m₂)] · g; Pressure on Pulley R = 2T.’,)

(‘Smooth Table Pulley Systema = [m₂ / (m₁ + m₂)] · g; T = [m₁ m₂ / (m₁ + m₂)] · g; Pressure on Pulley R = T√2.’,)

(‘Inclined Plane (Angle α) with Vertical Hanging Mass m₂a = [(m₂ – m₁ sin α) / (m₁ + m₂)] · g (for smooth plane).’,)

(‘General Pulley Pressure FormulaR = 2T cos(θ / 2), where θ is the angle between the two taut string segments.’,)

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 12 / 15

🔍 In-Depth Explanations, Deductions & Core Concepts:

Comparison aspect	Main idea	Keywords
Why acceleration is identical for both bodies	Because the string is inextensible (length is strictly constant L = y₁ + y₂ = const → a₁ = a₂ = a)	Becausestringinextensiblelength
What happens when the string is cut after time t (قطع الخيط)	The bodies continue moving from their acquired common velocity v = u + at. The hanging mass falls freely under gravity (acceleration = g downwards) the mass on an inclined plane decelerates under -g sin α (or -μ_k g on a rough table) until momentarily coming to rest	bodiescontinuemovingacquired
•	Distance between masses on a vertical pulley after time t from release: If each mass moves a distance s = (1/2) a t² the vertical distance between them becomes 2s	Distancebetweenmassesvertical

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 13 / 15

📊 Summary of Pulley Systems: Equations of Motion and Pressure on Axle:

System Configuration	Equation of Common Acceleration (a)	String Tension (T)	Resultant Pressure on Pulley (R)
Two Vertical Hanging Masses (m₁ > m₂)	a = (m₁ – m₂)g / (m₁ + m₂)	T = (2 m₁ m₂ g) / (m₁ + m₂)	R = 2T
Smooth Horizontal Table + Hanging m₂	a = m₂ g / (m₁ + m₂)	T = (m₁ m₂ g) / (m₁ + m₂)	R = T √2 (acts at 45°)
Rough Horizontal Table (μ_k) + Hanging m₂	a = (m₂ – μ_k m₁)g / (m₁ + m₂)	T = m₁ m₂ g (1 + μ_k) / (m₁ + m₂)	R = T √2
Smooth Plane Inclined at α + Hanging m₂	a = (m₂ – m₁ sin α)g / (m₁ + m₂)	T = [m₁ m₂ (1 + sin α)g] / (m₁ + m₂)	R = T √(2 + 2 sin α) = 2T cos(45° – α/2)

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 14 / 15

💡 High-Yield Exam Traps & Key Takeaways:

Comparison aspect	Main idea	Keywords
Exam Trap	When finding the vertical separation between two hanging masses on a vertical pulley remember it is 2s (one goes down by s the other goes up by s)	findingverticalseparationbetween
String Cut Question	The moment the string is severed tension T vanishes (T = 0). Write Newton’s 2nd Law anew for each body independently using the acquired velocity as initial velocity v₀	momentseveredtensionvanishes

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction · 15 / 15

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Connected Bodies: Pulleys, Tables, Inclined Planes and Pulley Reaction

🌳 Concept tree

├── 2-2 (a) General method for connected bodies

│ ├── Assumptions

│ └── Steps

│ ├── Results

│ └── The pulley reaction

├── 2-2 (c) Adding, removing and cutting

│ ├── Extra mass on one side

│ ├── When the extra mass is removed

│ └── When the string is cut

│ ├── Set-up

│ ├── Pulley reaction at a corner

│ └── With friction on the table

├── 2-2 (e) Inclined planes

│ └── Equilibrium conditions

├── 2-2 (f) The force on an inclined pulley

│ └── Rule

│ └── Forces in a lift

└── 2-2 (h) Summary

├── Comparison

└── Check list

Assumptions

Results

The pulley reaction

✍️ Worked examples

Extra mass on one side

✍️ Worked examples

Set-up

✍️ Worked examples

✍️ Worked examples

Body on an incline joined to a hanging body

✍️ Worked examples

Equilibrium conditions

Rule

Forces in a lift

✍️ Worked examples

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