Tension in strings
Understanding how strings transmit force through tension acting inwards on both ends.
What is tension?
When a string (or rope, cable, tow-bar) connects two objects, it exerts a tension force on each of them. The crucial property of tension is that it always acts inwards along the string — it pulls each attached object towards the other.
Tension pulls inwards at both ends of a string simultaneously. The magnitude is the same at each end (for a light, inextensible string).
A mass hanging from the ceiling
Consider an object of mass $m$ hanging vertically on a string attached to the ceiling. Three forces are present:
Forces on the mass
• Weight $mg$ downward
• Tension $T$ upward (string pulling mass towards ceiling)
Forces at the ceiling
This is a Newton 3 pair between the ceiling and the string.
• Tension $T$ downward (string pulling ceiling towards mass)
• Reaction, N, from ceiling structure upward on string
If the system is in equilibrium (not accelerating), applying Newton's second law to the mass gives:
If the system accelerates (e.g. the mass is being raised with acceleration $a$ upwards), then:
Tension is greater than the weight when accelerating upward, and less than the weight when accelerating downward. At free-fall ($a = -g$), tension would be zero.
Strings as force transmitters
How a string moves the point of application of a force from one location to another.
Transmitting the point of application
If a force $F$ acts directly on an object, it accelerates the object. But if the same force $F$ acts on a string attached to the object, the string transmits that force to the object — the object experiences the same force, just applied via the string.
The string effectively moves where the force acts. This is how a car tows a caravan: the engine force is applied to the car, but through the tow-rope, it pulls the caravan along too.
Car and trailer — the classic example
The car's engine creates a force, F, that pulls the car forward. The car is connected to the trailer by a rope. The car moving forward causes a tension in the rope that pulls the trailer forward. The car and trailer remain at constant distance apart because the rope is inextensible. As a result the car and trailer travel at the same speed and same acceleration (as long as the rope remains at full length, i.e. taut).
① The tension in the rope pulls the trailer forward — this is how the driving force reaches the trailer.
② The tension acts backward on the car — it resists the car's forward motion (like an internal resistance).
Two solution approaches
For connected particle problems on a horizontal surface, you have two valid approaches. Both give the same equations:
Approach 1 — Single System
Treat both objects as one combined mass. This eliminates $T$ from the equations and directly gives the acceleration $a$.
Approach 2 — Separate Objects
Apply $F = ma$ to each object individually. This gives two equations in two unknowns ($a$ and $T$).
In practice: use the single system approach first to find $a$ (tension cancels out). Then consider one object separately to find $T$. Choose whichever object has fewer unknown forces.
Horizontal connected particles with friction
Setting up and solving equations of motion for particles connected by strings on a horizontal surface.
General Setup
When two particles are connected by a light, inextensible string and move along a horizontal surface, we need to account for all horizontal forces: the applied force, any friction (on rough surfaces), and the tension in the string.
Equations of motion
Taking right as positive, and letting $F_1$, $F_2$ be friction forces on P and Q respectively:
Whole system (combined)
Note: $T$ cancels — it is an internal force for the combined system.
Particle P alone (to find T)
Particle Q alone (alternative way to find T)
① Forgetting that tension acts in opposite directions on the two particles (it can't push and pull in the same direction on both).
② Including tension in the whole-system equation — it is an internal force and must cancel.
③ Using the wrong mass in $F = ma$ (use only the mass of the object you are considering).
Smooth vs. rough surfaces
Smooth surface
No friction forces. Equations simplify to $F = (m_1+m_2)a$ and $T = m_2 a$ (for Q pulled indirectly).
Rough surface — large external force
When the external force is sufficient to overcome friction, the object moves relative to the surface. Friction is present, acts opposite in direction to the external force, and equals its maximum value $F_f = \mu R = \mu mg$. Include these separately for each particle in both approaches.
Rough surface — small external force
When the external force is not large enough to overcome friction, the object remains stationary. Friction adjusts to be exactly equal and opposite to the external force — just enough to keep the object still. Friction is less than its maximum value: $F_f = F$ and $F_f < \mu R$.
Modelling assumptions
Why strings are modelled as light and inextensible, and the consequences for our calculations.
Light and inextensible
In AS Mechanics, strings connecting particles are always modelled as light and inextensible. These two assumptions are essential simplifications that make the mathematics tractable.
| Assumption | What it means | How it is used |
|---|---|---|
| Inextensible | The string cannot stretch — it does not behave like an elastic band or spring. | The acceleration and velocity of each connected particle are identical (provided the string is taut). We use a single value $a$ for the whole system. |
| Light | The string has negligible mass — its mass is tiny compared to the particles. | ① When treating the whole system as a single particle, the mass of the string is ignored. ② In vertical systems, the tension is the same throughout the string (lower parts do not pull down on upper parts). |
Explaining the assumptions in exam questions
Part (c) questions often ask you to explain how you used one of these assumptions. Use precise language:
"Since the string is inextensible, the car and trailer have the same acceleration, so I could write a single value $a$ for both."
"Since the string is light, its mass was not included in the total mass of the system when applying $F = ma$."
Rods vs. strings
A rod (or tow-bar) can push as well as pull. The force in a rod is called a thrust (thrust pushes outward from the centre of rod on to the objects the rod is connected to. Tension pulls inward towards the centre of the rope, acting on the objects the rope is connected to.). When finding the thrust, treat the problem identically to a string problem; the sign of your answer tells you whether it is a thrust or tension.
Particles must remain in contact (or connected) to be treated as a single particle. If they separate, they must be considered independently.
Car & trailer
A classic horizontal connected particle problem with resistances.
Car pulling a trailer on a horizontal road
A car of mass 780 kg is pulling a trailer of mass 560 kg using a light, inextensible cable. The driving force on the car is constant at 1800 N. The total resistance forces on the car and the trailer are 800 N and 600 N respectively.
- a) Find the acceleration of the car.
- b) Find the tension in the cable.
- c) Explain how you used: (i) light (ii) inextensible.
Identify all forces on the combined system. The tension $T$ is internal and cancels when treating car + trailer together.
Treat car + trailer as one object. Total mass $= 780 + 560 = 1340$ kg. Tension is internal — it cancels.
Applying $F = ma$ (taking right as positive):
Consider the trailer only (mass 560 kg). Forces on trailer: tension $T$ forward, resistance 600 N backward.
Tip: use the exact fraction for $a$ to avoid rounding errors.
The mass of the cable was not included in the total mass of the two objects (1340 kg).
The car and the trailer have the same acceleration, so I used a single value $a$ for both.
Rough horizontal surface
Two particles on a rough surface — finding acceleration and tension.
Two particles on a rough horizontal plane
Two particles $P$ and $Q$ of masses 5 kg and 3 kg respectively are connected by a light inextensible string. Particle $P$ is pulled by a horizontal force of 40 N along a rough horizontal plane. $P$ experiences a frictional force of 10 N and $Q$ experiences a frictional force of 6 N.
- a) Find the acceleration of the particles.
- b) Find the tension in the string.
- c) Explain how the modelling assumptions have been used.
| Particle | Mass | Forces (→ positive) |
|---|---|---|
| $P$ | 5 kg | $+40$ N (applied), $-T$ (tension backward), $-10$ N (friction) |
| $Q$ | 3 kg | $+T$ (tension forward), $-6$ N (friction) |
Total mass $= 5 + 3 = 8$ kg. Tension cancels as an internal force.
For particle $P$ (mass 5 kg), taking right as positive:
Check using Q: $T - 6 = 3 \times 3 = 9$, so $T = 15$ N ✓
The acceleration of both masses is the same — we used a single value $a = 3$ m s$^{-2}$.
The string's mass was ignored; only $5$ kg $+ 3$ kg $= 8$ kg appears in $F = ma$.
Connected by a rod
When a rod connects two particles, the internal force can be a thrust or a tension.
A rod can push (thrust) as well as pull (tension). Solve exactly as for a string problem. If your answer for $T$ is positive, the rod is in tension. If negative, the rod is in compression (thrust).
Two particles connected by a light inextensible rod
Two particles $P$ and $Q$ of masses 20 kg and $m$ kg are connected by a light inextensible rod. The particles lie on a smooth horizontal plane. A horizontal force of 60 N is applied to $Q$ in a direction towards $P$, causing the particles to move with acceleration 2 m s$^{-2}$.
- a) Find the mass $m$ of $Q$.
- b) Find the thrust in the rod.
Note: the 60 N force acts on $Q$ in the direction from $Q$ towards $P$ (i.e. to the left). Both particles accelerate to the left at 2 m s$^{-2}$.
Smooth surface — no friction. Treating both particles as one system (thrust is internal):
Consider $P$ alone (mass 20 kg). The only horizontal force on $P$ is the thrust $T$ in the rod (pushing $P$ to the left, since the applied force pushes the whole system left).
The positive sign confirms the rod is in compression (thrust) — it pushes $P$ along, rather than pulling it.
Two strings in line
A particle is pulled by one string and connected to another — two different tensions.
Particle P pulled by a string, with Q attached behind
Particle $P$ of mass 4.5 kg is being pulled by a light inextensible string. Another light inextensible string is attached to the other side of $P$, and particle $Q$ of mass 6 kg is attached to the other end of this string. The particles move with acceleration 2.1 m s$^{-2}$ in a straight line on a smooth horizontal table.
Find the tension in each string.
Two tensions: $T_2$ is in the string pulled externally (between the external force and $P$); $T_1$ is in the string connecting $P$ and $Q$.
Treat the whole system as one object. Total mass $= 4.5 + 6 = 10.5$ kg. The external tension $T_2$ provides the net force.
Consider $Q$ alone (mass 6 kg). The only force on $Q$ is $T_1$ forward.
$T_1 < T_2$ always. The outer string carries the full load; the inner string only carries the load of particles behind it ($Q$ in this case).
Vertical system — upward force
Two vertically stacked particles accelerated upward by an applied force.
Particles A and B accelerated vertically upward
Two particles $A$ and $B$ of masses 6 kg and $m$ kg respectively are connected by a light inextensible string. Particle $B$ hangs directly below particle $A$. A force of 118 N is applied vertically upwards on $A$, causing the particles to accelerate at 2 m s$^{-2}$ upwards.
- a) Find the mass $m$ of particle $B$.
- b) Find the tension in the string.
Taking upward as positive. Both $A$ and $B$ have acceleration $a = 2$ m s$^{-2}$ upward.
For particle $A$ (mass 6 kg, taking up as positive):
For particle $B$ (mass $m$ kg):
Add equations (1) and (2) together — this eliminates $T$:
Using equation (2) with $m = 4$:
Check with equation (1): $118 - 6(9.8) - 47.2 = 118 - 58.8 - 47.2 = 12 = 6 \times 2$ ✓
Descending Lift
Finding cable tension and contact forces for boxes inside a descending lift.
Two boxes inside a descending lift
Two boxes $A$ and $B$ of masses 110 kg and 190 kg sit on the floor of a lift of mass 1700 kg. Box $A$ rests on top of box $B$. The lift is supported by a light inextensible cable and descends with constant acceleration 1.8 m s$^{-2}$.
- a) Find the tension in the cable.
- b) Find the force exerted by box $B$: (i) on box $A$ (ii) on the floor of the lift.
System masses
Total mass $= 110 + 190 + 1700 = 2000$ kg
All accelerate downward at $1.8$ m s$^{-2}$
Sign convention
Taking downward as positive (direction of acceleration). Weight acts downward (+), tension acts upward (−).
Apply $F = ma$ to the entire system (mass 2000 kg, accelerating downward):
Consider box $A$ alone (mass 110 kg). Forces: weight $110g$ downward, reaction $R_B$ from $B$ upward.
By Newton's 3rd Law, $A$ exerts 880 N downward on $B$, and $B$ exerts 880 N upward on $A$.
Consider $B + A$ together as one system (total mass $= 300$ kg). Forces: weight $300g$ down, floor reaction $R_F$ up.
All apparent weights are reduced because the lift accelerates downward. The cable tension (16 000 N) is less than the actual weight (2000 × 9.8 = 19 600 N). This is the "lighter feeling" in a descending lift.
① Always draw a clear force diagram before writing any equations.
② Use the whole system first to find acceleration (internal forces cancel).
③ Then consider one object separately to find the tension or thrust.
④ Check your answer using the other object's equation.
⑤ For vertical problems, choose your positive direction carefully and state it clearly.