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

  1. the tension in the cable supporting the lift;
  2. the normal reaction force the floor exerts on the woman.

How are these results changed if the cable snaps?

[Take g=10 ms-2]

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Hint

You’ll need to choose two different ‘systems’ to which to apply the 2nd Law.
 

Solution

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Solution

Denote by M and m the masses of the lift and woman, respectively.

  1. Consider first the system comprising the lift and the woman. The forces acting on this system from outside are

    1. a force exerted by the cable, which has magnitude T (the tension in the cable) and is directed upwards
    2. 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)a
    The 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 that
    T=(M+m)(g+a)=11550 N

  2. Now consider the system comprising the woman alone. The forces acting on this system from outside are

    1. the normal reaction (contact) force exerted by the floor, which has magnitude FN (say) and which points upwards.
    2. 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=ma
    implying
    FN=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.