3.03MechanicsStretch

Connected Particles

A car towing a trailer, two masses hanging over a pulley, a block on a table pulled by a weight hanging off the edge: in each case, objects are joined so that they move together. That shared motion — the same speed and the same size of acceleration — is the key that makes these $\textbf{connected particle}$ problems solvable.

50 min Forces and Newton’s Laws

The big picture

The method for connected particles is to write $F = ma$ separately for each particle, with the tension (or thrust) in the connecting string or rod appearing in both equations with opposite effects. Adding the equations eliminates the tension and gives the acceleration; substituting back gives the tension. You can also treat the whole system as one particle to find the acceleration directly, since internal forces cancel — but you must return to an individual particle to find a tension. The standard modelling assumptions — light inextensible strings, smooth pulleys — are what guarantee the tension is the same throughout a string and the particles share an acceleration, and exam questions often ask you to state them. The lesson ends with what happens when a string goes slack.

What you'll be able to do

Model connected systems using light inextensible strings, light rods and smooth pulleys
Write equations of motion for each particle and solve simultaneously
Treat a system as a whole to find acceleration, and individual particles to find tensions
Solve multi-stage problems, including when a string becomes slack

Modelling assumptions

A $\textbf{light}$ string has negligible mass, so the tension is the same throughout it. An $\textbf{inextensible}$ string does not stretch, so connected particles have the same speed and the same magnitude of acceleration.

A $\textbf{smooth pulley}$ has no friction, so the tension is the same on both sides.

A $\textbf{light rod}$ (such as a tow bar) can push as well as pull: it can be in tension or in $\textbf{thrust}$ (compression). A string can only pull.

Tip — When asked to state an assumption, link it to its consequence: "the string is inextensible, so both particles have the same acceleration".

Towing on a straight road

For a vehicle towing a trailer, draw separate force diagrams. The tow bar pulls the trailer forwards with tension $T$ and pulls the car backwards with $T$ (Newton’s third law).

Treating the car and trailer together, the tension is internal and cancels, which gives the acceleration quickly.

Example 1A 1200 kg car tows a 400 kg trailer. The car’s driving force is 2400 N, and resistances are 500 N on the car and 200 N on the trailer. Find the acceleration and the tension in the tow bar.

1Whole system:

2400 - 500 - 200 = 1600a

, so

a = 1.0625

m s⁻².

2Trailer alone:

T - 200 = 400 \times 1.0625

, so

T = 625

3Check with the car:

2400 - 500 - 625 = 1275 = 1200 \times 1.0625

✓.

Answer

a = 1.06

m s⁻²; tension

625

Why this works

If the car brakes, the trailer can push forwards on the car and the tow bar goes into thrust. A negative value for $T$ in your working means exactly that.

Pulleys

For two masses hanging over a smooth pulley, the heavier one accelerates down and the lighter one up, with the same magnitude.

For a particle on a table connected over a pulley at the edge to a hanging particle, write the equation for each in its own direction of motion: horizontally for the table particle, vertically for the hanging one.

Example 1Particles of mass 5 kg and 3 kg hang either side of a smooth pulley and are released from rest. Find the acceleration and the tension.

15 kg (down positive):

5g - T = 5a

23 kg (up positive):

T - 3g = 3a

3Add:

2g = 8a

, so

a = 2.45

m s⁻². Then

T = 3(9.8 + 2.45)

Answer

a = 2.45

m s⁻²;

T = 36.75

Example 2A 4 kg block on a smooth horizontal table is joined by a string over a smooth pulley at the edge to a 1 kg particle hanging freely. Find the acceleration and tension.

1Block:

T = 4a

. Hanging particle:

g - T = a

2Add:

g = 5a

, so

a = 1.96

. Then

T = 4 \times 1.96

Answer

a = 1.96

m s⁻²;

T = 7.84

Tip — The pulley also experiences a force: the two tensions pull on it, so the force on a pulley with a string at right angles is $T\sqrt{2}$ , and with both strings vertical it is $2T$ .

When the string goes slack

If a hanging particle hits the ground, the string goes slack and the tension drops to zero. The other particle continues with the velocity it had at that moment, but now under different forces.

Split the motion into stages: use suvat with the connected acceleration until impact, find the velocity, then use a new acceleration for the next stage.

Example 1In the table example above, the 1 kg particle starts 0.49 m above the floor. How fast is the block moving when the particle lands, and what happens next?

1Stage 1:

v^{2} = 0 + 2 \times 1.96 \times 0.49 = 1.9208

, so

v = 1.386

m s⁻¹.

2Stage 2: the string is slack and the table is smooth, so there is no horizontal force on the block.

AnswerThe block moves at

1.39

m s⁻¹ and continues at that constant speed until it reaches the pulley.

Think like an examiner

Draw a separate force diagram for each particle.
Write one $F = ma$ equation per particle in its own direction of motion.
State the modelling assumptions and their consequences.
Treat string-goes-slack problems as separate stages.

Common misconceptions

Using the total mass with the tension included as an external force.

For the whole system tension is internal and cancels.

Giving the two particles different accelerations.

An inextensible string gives them the same magnitude of acceleration.

Assuming tension equals the hanging weight.

That is only true in equilibrium; when accelerating,

T \neq mg

Keeping the connected acceleration after the string goes slack.

Recalculate the forces on each particle once

T = 0

Connected particles

F_{\text{res}} = (m_{1} + m_{2})a

Whole system.

m_{1}g - T = m_{1}a,\ T - m_{2}g = m_{2}a

Pulley,

m_{1} > m_{2}

a = \tfrac{(m_{1} - m_{2})g}{m_{1} + m_{2}}

Pulley acceleration.

Stretch yourself

A 6 kg block rests on a rough horizontal table ( $\mu = 0.25$ ). A light inextensible string attached to it passes over a smooth pulley at the table’s edge to a 4 kg particle hanging 1.5 m above the floor. The system is released from rest. Find the speed of the block when the particle hits the floor, and how much further the block slides (assuming it does not reach the pulley).

Hint — Friction on the block is $\mu \times 6g$ . After impact, only friction acts horizontally on the block.

Reveal solution

SolutionFriction

= 0.25 \times 6g = 14.7

N. Block:

T - 14.7 = 6a

. Particle:

4g - T = 4a

. Adding:

39.2 - 14.7 = 10a

, so

a = 2.45

m s⁻² (and

T = 29.4

N). The block moves 1.5 m before impact:

v^{2} = 2 \times 2.45 \times 1.5 = 7.35

, so

v = 2.71

m s⁻¹. After impact the string is slack; friction decelerates the block at

\dfrac{14.7}{6} = 2.45

m s⁻². Distance to stop:

0 = 7.35 - 2 \times 2.45 \times s

, so

s = 1.5

m. The block slides a further 1.5 m.

\blacksquare

Questions students ask

Why is the tension the same on both sides of the pulley?

Because the pulley is smooth and the string is light. A rough pulley or heavy string would make the tensions differ.

Can I always treat the system as one particle?

Only when all parts move in the same straight line. With a pulley, directions change, so write separate equations.

What is thrust?

A pushing force in a rod. A rod in thrust pushes outwards on the objects at its ends; a string can never do this.

Why does the heavier mass not fall with acceleration

g

The tension pulls upwards on it, so the resultant force is less than its weight.

Key takeaways

Light inextensible strings give equal tension throughout and equal accelerations.
Write $F = ma$ for each particle; add equations to eliminate tension.
The whole system approach finds acceleration when everything moves in a line.
Tow bars can be in tension or thrust; strings only in tension.
When a string goes slack, $T = 0$ and a new stage begins.