Why weight of the body is not taken as negative in case of the motion in a lift

Asked by vrg.anayath | 22nd Nov, 2010, 07:28: AM

Expert Answer:

Dear student,
The apparent weight of a person in a lift or an elevator:
The actual weight of the person = mg.This acts on the weighing machine which offers  a reaction R. given by the reading of weighing machine. This reaction exerted by the surface of contact on the person is the apparent weight of the person.
When the elevator is at rest,
Accelaration of the person = 0
Net force of the person =0, (f)
R - mg =0
R =mg
Hope this helps.
Thanking you

Answered by  | 23rd Nov, 2010, 01:09: PM

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