Connected Bodies (Lifts) (AQA AS Maths): Revision Note
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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-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?
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
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.
(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.
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.
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.
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