All subjects · Mathematics · WME01 (M1)

Connected particles in a line: lifts and towing

When objects move together, you can treat them as one system or look at each separately. Using both views gives the acceleration and the internal force (tension, thrust or reaction).

Whole system

Treat the connected objects as one body of total mass. Internal forces (tow bar tension, reaction between person and lift floor) cancel out. This gives the acceleration quickly.

Separate bodies

Then write F = ma for one object alone to find the internal force. Choose the one with fewer forces acting on it.

Tension or thrust

A tow bar or rod can be in tension (pulling) or thrust (pushing). If the car brakes, the trailer pushes on the car, so the tow bar is in thrust. If your tension comes out negative, it is really a thrust of that size.

Worked example

A lift of mass 500 kg carries a person of mass 70 kg and accelerates upwards at 1.2 m/s2. Find the tension in the cable and the reaction on the person.

  1. Whole system: T - 570g = 570 × 1.2, so T = 570 × 11 = 6270 N.
  2. Person: R - 70g = 70 × 1.2, so R = 770 N.

Answer: T = 6270 N, R = 770 N

Key idea: Whole system for the acceleration, one body for the internal force. Negative tension means thrust.

Check yourself

  1. A lift of mass 400 kg contains a crate of mass 100 kg. The lift accelerates upwards at 0.6 m/s2. Find the tension in the lift cable.
    Show answer

    5200 N. T - 500g = 500 × 0.6, so T = 500 × 10.4 = 5200 N.

  2. A person of mass 60 kg stands in a lift moving downwards and decelerating at 1.5 m/s2. Find the reaction from the floor on the person.
    Show answer

    678 N. Decelerating while moving down means the acceleration is upwards: R - 60g = 60 × 1.5, so R = 678 N.

  3. A car of mass 1000 kg tows a trailer of mass 400 kg. The driving force is 2500 N, and the resistances are 300 N on the car and 200 N on the trailer. Find the acceleration, to 3 significant figures.
    Show answer

    1.43 m/s2. Whole system: 2500 - 500 = 1400a, so a = 1.43 m/s2.

  4. For the same car and trailer, find the tension in the tow bar, to 3 significant figures.
    Show answer

    771 N. Trailer: T - 200 = 400a, so T = 771 N.

  5. The same car and trailer now brake: the engine force is removed and a braking force of 2000 N acts on the car, with the same resistances. Find the magnitude of the force in the tow bar, to 3 significant figures.
    Show answer

    514 N. Whole system: -2500 = 1400a, a = -1.79. Trailer: T - 200 = 400a gives T = -514 N, so it is a thrust of 514 N.

  6. Why does the tension in the tow bar not appear when you write F = ma for the car and trailer as one system?
    • The tow bar is light
    • It is an internal force: it acts equally and oppositely on the two bodies
    • The tension is always zero
    • The system is in equilibrium
    Show answer

    It is an internal force: it acts equally and oppositely on the two bodies. By Newton's third law, the bar pulls the trailer forwards and the car backwards with equal forces, which cancel for the whole system.

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Last updated 7 November 2026. Spotted a mistake? Tell us.