R2, R4–R5MechanicsStretch

Connected Particles

Two objects joined by a string, or pushing against each other, move as one — but the force between them is usually what the question wants. The technique is to treat the system as a single body to find the acceleration, then look at one object alone to find the internal force.

40 min Forces and Newton’s Laws

The big picture

This two-step method works because of the modelling assumptions, and it is worth seeing why. An $\textbf{inextensible}$ string forces both objects to have the same acceleration, which is what allows one value of $a$ to describe the whole system. A $\textbf{light}$ string has the same tension throughout, so one value of $T$ suffices. And when the system is treated as a whole, the tension appears twice with opposite signs and cancels — which is exactly why the system equation contains no unknown internal force. Take away either assumption and the method collapses.

What you'll be able to do

Apply $F = ma$ to a system as a whole to find the acceleration
Apply $F = ma$ to one object to find the tension or contact force
Solve pulley problems with objects on both sides
Handle systems combining horizontal and vertical sections

Why the two-step method works

Treating the system as one body means the internal forces — the tension pulling each object towards the other — appear as an action-reaction pair and cancel. Only $\textbf{external}$ forces remain, so the equation contains no unknown tension.

That gives the acceleration immediately, using the total mass. With $a$ known, isolating either object gives an equation containing the tension as the only unknown.

Choose the simpler object for the second step. If one has fewer forces acting on it, the equation is shorter and there is less to go wrong.

F_{\text{external}} = \left(m_{1}+m_{2}\right)a

Whole system: internal forces cancel.

\text{then apply } F = ma \text{ to one object}

To find the tension.

Example 1Blocks of 3 kg and 2 kg are connected by a light inextensible string on a smooth horizontal surface. A 15 N force pulls the 3 kg block. Find the acceleration and the tension.

1Whole system: total mass 5 kg, external force 15 N, so

a = \dfrac{15}{5} = 3\ \mathrm{m\,s^{-2}}

2Now the 2 kg block alone — the only horizontal force on it is the tension.

T = 2 \times 3

Answer

a = 3\ \mathrm{m\,s^{-2}}

;

T = 6\ \mathrm{N}

Tip — Analysing the 2 kg block gives $T$ in one line; analysing the 3 kg block gives $15 - T = 9$ and the same answer with more work. Pick the object with fewer forces.

Pulley systems

For two masses hanging over a smooth pulley, the heavier one descends and the lighter rises, with the same magnitude of acceleration. The string being light and the pulley smooth mean the tension is the same on both sides.

The system approach works if you take "positive" as the direction of motion around the pulley — clockwise or anticlockwise. Then the heavier weight drives the motion and the lighter weight opposes it, so the net external force is the $\textbf{difference}$ of the weights.

Isolating either mass then gives the tension, which always lies between the two weights: greater than the lighter weight (which is accelerating upwards) and less than the heavier (which is accelerating downwards).

a = \dfrac{\left(m_{1}-m_{2}\right)g}{m_{1}+m_{2}}

Two masses over a smooth pulley,

m_{1} > m_{2}

Example 1Masses of 5 kg and 3 kg hang from a light string over a smooth pulley. Find the acceleration and the tension. Take

g = 9.8\ \mathrm{m\,s^{-2}}

1Net driving force:

\left(5-3\right)g = 2 \times 9.8 = 19.6

2Total mass 8 kg, so

a = \dfrac{19.6}{8} = 2.45\ \mathrm{m\,s^{-2}}

3For the 3 kg mass, rising:

T - 3g = 3a

, so

T = 29.4 + 7.35

Answer

a = 2.45\ \mathrm{m\,s^{-2}}

;

T = 36.75\ \mathrm{N}

Why this works

Check the tension against both weights: $3g = 29.4$ N and $5g = 49$ N, and $36.75$ lies between them. If your tension falls outside that range, a sign has gone wrong.

Mixed horizontal and vertical systems

A common arrangement has one mass on a horizontal table connected over a pulley at the edge to a mass hanging freely. The hanging weight drives the system; the table mass contributes inertia but no driving force if the table is smooth.

For the whole system, the external driving force is the hanging weight alone. The normal reaction and the table mass’s weight are vertical and perpendicular to that mass’s motion, so they do not contribute.

If the table is rough, friction opposes the motion and is subtracted from the driving force — which is the subject of the next lesson.

Example 1A 4 kg block on a smooth table is connected over a smooth pulley to a 2 kg mass hanging freely. Find the acceleration and the tension.

1Driving force: the hanging weight,

2 \times 9.8 = 19.6

2Total mass moving:

4 + 2 = 6

kg, so

a = \dfrac{19.6}{6} = 3.267\ \mathrm{m\,s^{-2}}

3The 4 kg block: the only horizontal force is

T

, so

T = 4 \times 3.267

Answer

a = 3.27\ \mathrm{m\,s^{-2}}

;

T = 13.1\ \mathrm{N}

(3 s.f.).

Tip — The table block’s weight never enters the system equation on a smooth table — it acts vertically while the block moves horizontally. Including it is a frequent error.

Objects in contact

Two blocks pushed along in contact behave identically to a connected pair, with the $\textbf{contact force}$ between them playing the role of the tension.

The system approach gives the acceleration, and isolating the rear block gives the contact force. The key difference from a string is direction: a contact force pushes, whereas a tension pulls.

That means the order matters. Pushing a 5 kg block against a 3 kg block gives a different contact force from pushing the 3 kg against the 5 kg, even with the same applied force.

Example 1Blocks of 6 kg and 4 kg sit in contact on a smooth surface. A 30 N force pushes the 6 kg block towards the 4 kg block. Find the contact force between them.

1Whole system:

a = \dfrac{30}{10} = 3\ \mathrm{m\,s^{-2}}

2The 4 kg block is accelerated only by the contact force from the 6 kg block.

C = 4 \times 3

Answer

C = 12\ \mathrm{N}

Why this works

If the force were applied to the 4 kg block instead, the acceleration would still be $3\ \mathrm{m\,s^{-2}}$ but the contact force would be $6 \times 3 = 18$ N — the contact force must accelerate whatever mass lies ahead of it.

Think like an examiner

State the modelling assumptions you are relying on: light and inextensible string, smooth pulley.
Use the whole system for the acceleration, then one object for the internal force.
Define your positive direction, especially around a pulley where the two sides move oppositely.
Check that a tension lies between the two weights in a pulley problem.

Common misconceptions

Including the tension in the whole-system equation.

It is internal and cancels. Only external forces appear.

Adding both weights in a pulley problem.

They act in opposite directions around the pulley, so the driving force is their difference.

Including the table block’s weight as a driving force.

On a smooth horizontal table it acts perpendicular to the motion and does not drive it.

Using different accelerations for the two connected objects.

An inextensible string forces the same magnitude of acceleration throughout.

Connected systems

F_{\text{ext}} = \left(\Sigma m\right)a

Whole system.

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

Pulley with two hanging masses.

T \text{ between the two weights}

A useful check.

\text{same } a \text{ throughout}

From inextensibility.

Stretch yourself

A 5 kg block on a smooth horizontal table is connected by a light inextensible string over a smooth pulley at the table edge to a 3 kg mass hanging freely. The system is released from rest. Find the acceleration, the tension, and the speed of the hanging mass after it has descended 0.5 m. Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .

Hint — Two steps for the forces, then a suvat equation for the last part — the acceleration is constant.

Reveal solution

SolutionWhole system: the driving force is the hanging weight

3 \times 9.8 = 29.4

N, and the total mass is

8

kg, so

a = \dfrac{29.4}{8} = 3.675\ \mathrm{m\,s^{-2}}

. For the tension, take the 5 kg block on the table — the only horizontal force on it is

T

, so

T = 5 \times 3.675 = 18.375

N, about

18.4

N. Check with the hanging mass:

3g - T = 29.4 - 18.375 = 11.025 = 3 \times 3.675

✓. Note

T

is less than the hanging weight, as it must be for that mass to accelerate downwards. For the speed, the acceleration is constant so suvat applies:

v^{2} = u^{2} + 2as = 0 + 2(3.675)(0.5) = 3.675

, giving

v = 1.92\ \mathrm{m\,s^{-1}}

(3 s.f.).

\blacksquare

Questions students ask

Why does the tension cancel in the system equation?

It acts on each object with equal magnitude and opposite direction around the system, so summing the two equations eliminates it. That is precisely why the system approach gives the acceleration directly.

Which object should I use for the second step?

Whichever has fewer forces acting on it. A hanging mass or a block pulled only by the string usually gives the shortest equation.

Why must the string be light and inextensible?

Light gives a single tension throughout; inextensible gives a single acceleration. Without both, the system would need separate tensions and accelerations and could not be solved this way.

How do I check a pulley answer?

The tension must lie strictly between the two weights, and the acceleration must be less than

g

. Either check failing indicates a sign error.

Key takeaways

Treat the system as one body to find the acceleration; internal forces cancel.
Isolate one object to find the tension or contact force.
An inextensible string gives a shared acceleration; a light string gives one tension.
For a pulley, the driving force is the difference of the weights.
A contact force pushes and depends on which block is pushed first.

How this fits the course

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