In one of the physics book I read the following article on variation of weight in lift.I request uyou also please see it & answer my query.
It was written that
"In a lift the actual weight of the person=mg.This acts on the weighing machine which offers a reaction R given by the weighing machine & this is the apparent weight of the person".
After this the variation of R was given in different case.Here I want to ask that as R is the reaction force given by machine in response to actual weight of the person=mg,so it must be same but how it can be different.I was not able to understand mathematically so can anyone please explain it to me logically.
thank you

Since you mention a lift I assume that the question has to do with how R changes under vertical acceleration or deceleration. Keep in mind the famous equation: \sum \vec F = m \vec a. The forces acting on the person's feet are his weight pressing down (weight = mg) and the reaction force of the scale pressing up (R).

Case 1: If R = mg then the sum of forces is 0, and from F=ma that means acceleration is zero, and so either the lift is stationary or traveling at a constant rate of speed.

Case 2: If R is greater than mg then the sum of forces is positive in the upward direction and the man must therefore have positve acceleration upward. Convesely - if the man has positive acceleration then R > mg. Note that positive acceleration could mean either his upward velocity is increasing or his downward velocity is decreasing.

Case 3: If R < mg then the reaction force is not sufficient to hold the weight of the man and he accelerates downward. Conversely - when the car accelerates downward R < mg.