Connected Particles

Modelling systems where particles move together with a common acceleration

SPEC 8.4 • FORCES AND NEWTON'S LAWS

What Are Connected Particles?

Definition

Connected particles are two or more objects joined by a string, rope, or coupling that move together as a system.

In mechanics, we model these systems using key assumptions:

Key Assumptions

Light string: The string has negligible mass (we ignore its weight)

Inextensible: The string doesn't stretch, so both particles have the same speed and acceleration

Exam Tip

Always state "the string is light and inextensible" in your assumptions when modelling connected particle problems.

Modelling Assumptions

Exams often ask: "Explain how the modelling assumptions have been used."

Light String

The string has negligible mass.

Effect: Tension is the same throughout the string. We use the same value T at both ends.

Inextensible String

The string doesn't stretch.

Effect: Both particles have the same acceleration. We use the same value a for both.

Exam Answer Template

"Light" means the tension is the same at both ends of the string (or throughout the string).

"Inextensible" means both particles have the same magnitude of acceleration.

Common Acceleration

The most important concept for connected particles:

Fundamental Principle

a_{1} = a_{2} = a

Both particles have the same magnitude of acceleration

Both particles accelerate together

Important

The acceleration has the same magnitude for both particles. For vertical systems, both particles move in the same direction (both up or both down).

The Whole System Method

When particles move in the same straight line, you can treat them as a single particle to find acceleration quickly:

Shortcut

R (\to) : Total driving force - Total resistance = (Total mass) \times a

When to Use

Use this method to find acceleration quickly - tension cancels out as it's internal to the system.

Limitation

To find tension, you must still consider particles separately using the cut method.

Worked Example01

Two particles P (2 kg) and Q (6 kg) are connected by a light inextensible string on a smooth horizontal surface. A force of 32 N is applied to Q. Find the acceleration using the whole system method.

1

Whole system - treat both particles as one:

Total mass = 2 + 6 = 8 kg

Total driving force = 32 N

Total resistance = 0 N (smooth surface)

2

Apply Newton's second law:

R (\to) : 32 = 8 a

a = 4 m s^{- 2}

Note: To find tension, we would need to consider one particle separately.

AnswerAcceleration = 4 m s⁻²

Tension vs Thrust

The internal force in a connection depends on whether it's being pulled or pushed. To find this force, we must cut the connection and show the internal forces:

Arrows point towards each other

String pulls both particles inward

Arrows point away from each other

Rod pushes both particles outward

Remember

Strings can only provide tension (pull) - they go slack if pushed

Rods can provide both tension (pull) AND thrust (push)

Always draw a red cut line to separate particles and show internal forces

The Cut Method

To solve connected particle problems, we "cut" the string and consider each particle separately using our standard resolving method:

Method: Resolving for Each Particle

Draw the system and mark where you will "cut" (red dashed line)

Draw separate force diagrams for each particle after the cut

Add tension at each cut - arrows point towards each other (tension pulls)

Resolve in direction of motion:

R (\to)

or

R (↓)

for each particle

Solve the simultaneous equations (same

a

and

T

in both)

Key Point

The tension (or thrust) is the same on both sides of the cut because the string is light (no mass).

Positive Direction

Choose positive in the direction of motion (or initial velocity). This keeps acceleration positive.

Worked Example02

Two particles A and B, of masses 3 kg and 5 kg respectively, are connected by a light inextensible string. They rest on a smooth horizontal surface. A horizontal force of 24 N is applied to B in the direction from A to B. Find the acceleration of the system and the tension in the string.

1

Whole system - to find acceleration:

R (\to) : 24 = (3 + 5) a

24 = 8 a

a = 3 m s^{- 2}

2

Particle A (3 kg) - to find tension:

R (\to) : T = 3 a

T = 3 \times 3 = 9 N

AnswerAcceleration = 3 m s⁻², Tension = 9 N

Worked Example03

Particles P (4 kg) and Q (6 kg) are connected by a light rod on a rough horizontal surface. The coefficient of friction is 0.2 for each particle. The system moves to the right but decelerates at 0.5 m s⁻². Find the force in the rod, stating whether it is a tension or thrust.

1

Find normal reactions - resolve vertically for each particle:

For P:

R (↑) : R_{P} = 4 g = 39.2 N

For Q:

R (↑) : R_{Q} = 6 g = 58.8 N

2

Find friction forces (oppose motion, so act leftwards):

F_{P} = μ R_{P} = 0.2 \times 39.2 = 7.84 N

F_{Q} = μ R_{Q} = 0.2 \times 58.8 = 11.76 N

3

Particle P (4 kg) - taking rightward as positive:

Let T = force from rod (positive if tension, pulling P to the right)

R (\to) : T - F_{P} = 4 a

T - 7.84 = 4 (- 0.5)

T = - 2 + 7.84 = 5.84 N

4

Interpret the result:

Since T is positive, the rod pulls P to the right (in direction of motion).

This means the rod is in thrust (Q is pushing P from behind).

Note: When decelerating, the front particle (P) slows down faster than friction alone would cause, so Q pushes it from behind.

AnswerThrust = 5.84 N

Car and Trailer / Train Problems

Real-world connected particle problems often involve vehicles with driving forces and resistances:

Key Points

Driving force acts only on the engine/car (not the trailer)

Each vehicle has its own resistance to motion

The tow-bar/coupling is modelled as a light rod

When accelerating: tow-bar in tension (car pulls trailer)

When braking: tow-bar in thrust (trailer pushes car)

Worked Example04

A car of mass 900 kg pulls a trailer of mass 300 kg along a horizontal road. The resistances are 200 N on the car and 100 N on the trailer. The engine provides a driving force of 1200 N. Find the acceleration and the tension in the tow-bar.

1

Whole system - to find acceleration:

R (\to) : 1200 - 200 - 100 = (900 + 300) a

900 = 1200 a

a = 0.75 m s^{- 2}

2

Trailer only (300 kg) - to find tension:

R (\to) : T - 100 = 300 a

T = 300 (0.75) + 100 = 325 N

Note: The tow-bar is in tension because the car is accelerating (pulling the trailer forward).

AnswerAcceleration = 0.75 m s⁻², Tension = 325 N

Worked Example05

A car of mass 800 kg pulls a trailer of mass 400 kg along a horizontal road. The resistances are 150 N on the car and 100 N on the trailer. The brakes are applied giving a braking force of 1000 N on the car. Find the deceleration and the force in the tow-bar, stating whether it is a tension or thrust.

1

Whole system - to find deceleration (taking positive in direction of motion):

R (\to) : - 1000 - 150 - 100 = (800 + 400) a

- 1250 = 1200 a

a = - 1.042 m s^{- 2}

(deceleration of 1.04 m s⁻²)

2

Trailer only (400 kg) - to find force in tow-bar:

Let P = force from tow-bar (positive if tension, pulling trailer back)

R (\to) : - P - 100 = 400 (- 1.042)

- P - 100 = - 416.8

P = 316.8 N

3

Interpret the result:

Since P is positive with our sign convention, the tow-bar pulls the trailer backwards.

This means the tow-bar is in thrust (the trailer is pushing into the car).

Note: When braking, the trailer pushes into the car, creating thrust in the tow-bar.

AnswerDeceleration = 1.04 m s⁻², Thrust = 317 N

Vertical Connected Particles

Particles can also be connected vertically, with one hanging directly below the other:

Both particles accelerate together

vertically with the same acceleration

Tension arrows point towards each other

String pulls both particles inward

Setup

Both particles move vertically with the same acceleration

Resolve

R (↑)

for each particle (if accelerating upward)

Tension still acts towards each particle (string pulls both)

Worked Example06

Two particles A (10 kg) and B (5 kg) are connected by a light inextensible string. B hangs directly below A. A force of 180 N is applied vertically upwards to A, causing the system to accelerate upwards. Find the acceleration and the tension in the string.

1

Whole system - to find acceleration (positive upward):

R (↑) : 180 - 10 g - 5 g = (10 + 5) a

180 - 147 = 15 a

a = 2.2 m s^{- 2}

2

Particle B (5 kg) - to find tension:

R (↑) : T - 5 g = 5 a

T = 5 (2.2) + 49 = 60 N

AnswerAcceleration = 2.2 m s⁻² upward, Tension = 60 N

Particles on Inclined Planes

When connected particles are on an inclined plane, resolve parallel to the plane in the direction of motion:

Resolving on an Incline

Resolve parallel to the plane:

R (↙)

or

R (↗)

Weight component down the plane:

m g sin θ

Normal reaction:

R = m g cos θ

Friction (if rough):

F = μ R = μ m g cos θ

07Worked Example

Two particles A (3 kg) and B (5 kg) are connected by a light inextensible string on a smooth plane inclined at 30° to the horizontal. B is higher up the plane. The system is released from rest. Find the acceleration and the tension in the string.

1

Both particles slide down the plane. Take positive direction down the plane.

Weight components: A:

3 g sin 30° = 14.7

N, B:

5 g sin 30° = 24.5

N

2

Particle A (3 kg) - resolve parallel to plane:

R (↘) : 3 g sin 30° - T = 3 a

14.7 - T = 3 a

... equation (1)

3

Particle B (5 kg) - resolve parallel to plane:

R (↘) : 5 g sin 30° + T = 5 a

24.5 + T = 5 a

... equation (2)

4

Add (1) + (2):

14.7 + 24.5 = 8 a

a = 4.9 m s^{- 2}

5

From (2):

T = 5 (4.9) - 24.5 = 0

N

Note: T = 0 because both particles have the same acceleration as free fall down the plane!

Answera = 4.9 m s⁻² (= g sin 30°), T = 0 N

08Worked Example

Particles A (2 kg) and B (4 kg) are connected by a light string on a rough plane inclined at 25° to the horizontal. The coefficient of friction is 0.3 for both particles. B is higher. A force P acts on B up the plane, causing both particles to accelerate at 1.5 m s⁻² up the plane. Find P and the tension.

1

Normal reactions - resolve perpendicular to plane for each particle:

For A:

R (⊥) : R_{A} = 2 g cos 25° = 2 \times 9.8 \times 0.906 = 17.77

N

For B:

R (⊥) : R_{B} = 4 g cos 25° = 4 \times 9.8 \times 0.906 = 35.54

N

2

Friction (opposes motion up the plane, so acts down):

F_{A} = μ R_{A} = 0.3 \times 17.77 = 5.33

N

F_{B} = μ R_{B} = 0.3 \times 35.54 = 10.66

N

3

Particle A (2 kg) - resolve parallel to plane (up positive):

R (↗) : T - 2 g sin 25° - F_{A} = 2 a

T - 8.28 - 5.33 = 2 (1.5)

T = 16.61

N

4

Particle B (4 kg) - resolve parallel to plane (up positive):

R (↗) : P - T - 4 g sin 25° - F_{B} = 4 a

P - 16.61 - 16.56 - 10.66 = 4 (1.5)

P = 49.83

N

AnswerP = 49.8 N, T = 16.6 N

Practice

Question 1: Two particles P (5 kg) and Q (7 kg) are connected by a light inextensible string on a smooth horizontal surface. A force of 36 N is applied to Q in the direction from P to Q. Find the acceleration and the tension in the string.

1

P:

R (\to) : T = 5 a

Q:

R (\to) : 36 - T = 7 a

2

Adding:

36 = 12 a

, so

a = 3

m s⁻²

3

T = 5 \times 3 = 15

N

AnswerAcceleration = 3 m s⁻², Tension = 15 N

Summary

Key Concepts

ConceptDescription

Common AccelerationConnected particles have the same magnitude of acceleration

Tension vs ThrustTension: arrows towards each other. Thrust: arrows away from each other

Inclined PlaneResolve parallel to plane: weight component

m g sin θ

down

Modelling AssumptionsLight → same T throughout. Inextensible → same a for both

Exam Strategy

Draw red cut lines and show tension arrows towards each other

Use whole system method to find acceleration quickly

On inclines: resolve

R (↗)

or

R (↘)

parallel to plane

For car/trailer: accelerating = tension, braking = thrust

Always explain modelling assumptions if asked