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-1) may be positive or negative
- Round final answers based on the number of significant figures used for g
- if g = 10 m s-2, round to 1 s.f.
- if g = 9.8 m s-2, round to 2 s.f.
- Remember that acceleration links F = ma(N2L) and the ‘suvat’ equations
How do I solve ‘lift problem’ type questions?
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- 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
How do we form the equations for problems involving tow bars and ropes?
- Form the equations as follows:
- Treating the lift and person/load as one
(↓) (M + m)g - T = (M + m)a
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- 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) Briefly explain how the force of 800g N arises in this problem.
(b) Find the mass of the load, m kg .
(c) Find the tension, T N, in the cable of the lift.
Exam Tip
- 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.
- Unless told otherwise, use g = 9.8 m s-2 and round your final answer to two significant figures.
- Some questions may direct you to use g = 10 m s-2 in which case round your final answer to one significant figure.
