Connected Bodies (Lifts) (Edexcel International A Level (IAL) Maths): Revision Note

Exam code: YMA01

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Connected Bodies - The Lift Problem

What is the lift problem?

  • The lift problem involves objects (particles) that are directly in contact with each other – typically a person or crate in a lift

  • If it is not a person in the lift the object is often referred to as a load

  • There may be more than two objects involved – for example two crates stacked on top of each other on a lift floor

  • Vertical motion is involved so use g m s-2, the acceleration due to gravity, where appropriate

    • Gravity always acts vertically downwards

    • Depending on the positive direction chosen – and which other forces are acting vertically – acceleration (a m s-2) may be positive or negative

  • Remember that acceleration links F= ma (N2L) and the ‘suvat’ equations

How do I solve ‘lift problem’ type questions?

  • Lift problems will only consider motion in the vertical direction

  • As motion is involved Newton’s Laws of Motion apply so use “F = ma” (N2L)

  • The steps for solving lift problems are the same as for solving rope problems

  • As both the lift and load are travelling in the same direction the system can be treated as one particle (as well as separate particles)

    • There is no reaction force acting on the lift or load when treating the particle as one - mathematically they cancel each other out

    • You can think of the upward as counteracting the person’s weight and moving the load upwards; N3L applies so there must be an equal force acting in the opposite direction; - you can think of this as the force keeping the person in contact with the lift floor whilst it is moving

  • For constant acceleration the ‘suvat’ equations could be involved

3-2-3-the-lift-problem-diagram-1

How do we form the equations for problems involving lifts?

  • Form the equations as follows:

    • Treating the lift and person/load as one


      (↓) (M + m)g - T = (M + m)a

    • Treating the lift and person/load separately

      Lift: (↓) (Mg + R) - T = Ma

                  Person/load: (↓) mg - R = ma

  • You do not necessarily need all equations but if in doubt attempt all and it may help you make progress

Worked Example

A lift of mass 800 kg is moving vertically downwards with acceleration 0.3 m s -2. A load of mass m kg lies on the floor of the lift. The reaction force exerted on the load by the lift is 399 N.

Diagram of a block inside a square. Forces acting: tension T upwards, 800g N downwards, and acceleration a = 0.3 m/s² downwards.

(a)  Briefly explain how the force of arises in this problem.

3-2-3-fig5-we-solution-1
3-2-3-fig5-we-solution-2

(b)  Find the mass of the load, m kg . 

3-2-3-fig5-we-solution-3

(c)  Find the tension,T N, in the cable of the lift.

3-2-3-fig5-we-solution-4

Examiner Tips and Tricks

  • Sketch diagrams or add to any diagrams given in a question.

  • If in doubt of how to start a problem, draw all diagrams and try writing an equation for each.  This may help you make progress as well as picking up some marks.

  • Watch out for “hidden lift” problems – we’re not strictly talking elevators here!  For example, a load being raised by a crane; the “lift” would be a platform (such as a pallet) and the “lift cable” would be the cable connecting the crane to the load. Another common alternative is a fast rising (or falling) fairground ride.

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Paul

Author: Paul

Expertise: Maths Content Creator

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams.

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