Q3.7 Tension and normal forces (K)
A woman of mass 50 kg stands in a lift of mass 1000 kg. The lift accelerates upwards at an acceleration a=g/10. Find
- the tension in the cable supporting the lift;
- the normal reaction force the floor exerts on the woman.
How are these results changed if the cable snaps?
[Take g=10 ms-2]
Solution
Reveal

Solution
Denote by M and m the masses of the lift and woman, respectively.
Consider first the system comprising the lift and the woman. The forces acting on this system from outside are
- a force exerted by the cable, which has magnitude T (the tension in the cable) and is directed upwards
- the weight of the lift and its contents, which has magnitude (M+m)g and is directed downwards.
Newton’s 2nd law applied to this system (taking the upward direction as positive) gives
T-(M+m)g=(M+m)aThe left hand side gives the net force acting vertically upwards; the right hand side is the mass of the relevant system times the upwards acceleration. It follows thatT=(M+m)(g+a)=11550 NNow consider the system comprising the woman alone. The forces acting on this system from outside are
- the normal reaction (contact) force exerted by the floor, which has magnitude FN (say) and which points upwards.
- her weight, which has magnitude mg and is directed downwards.
Newton’s 2nd law applied to this system (still taking the upward direction as positive) gives
FN-mg=maimplyingFN=m(g+a)=550 N
If the cable snaps the acceleration of woman and lift changes immediately to a=-g (the minus sign signifying downwards). Setting a=-g in our results for T and FN we see that both vanish. The latter expresses the feeling of ‘weightlessness’ which the passenger would experience, briefly.