Vertically moving elevators

Let us consider that a person of mass m and weight is on a scale (a model used in pharmacies) placed inside an elevator, as shown in the figure above. Suppose then that the scale display shows us the values ​​in units of force.

We know that the person on the balance plate applies a force   and, according to the Law of Action and Reaction, the balance plate exerts a force on the person of the same intensity, but in the opposite direction, so the force is – . We cannot forget that the weight force is still acting on the person .

Therefore, the scale display should show the magnitude of the force applied to the pan, that is, it should show the value of . If the elevator is at rest or moving vertically with constant velocity (up or down), the resultant of forces on the individual is zero. That way:

F _N =P → F _N =mg

That is, the marking indicated on the scale display is equal to the individual’s weight. In such a way, we can say that for an elevator at rest or vertical MRU the normal force is equal to the weight.

However, if the elevator is moving with non-zero acceleration, the resultant forces on the individual will no longer be zero. Thus, normal force is different from weight , that is, the scale will not measure the individual’s weight. In this case, the value of (which is what the scale signals) is called the apparent weight . We will then analyze the cases in which the elevator goes up or down, in accelerated or retarded motion, with an acceleration of magnitude a, remembering that:

– in an accelerated movement, the resultant force has the same direction as the movement ;

– in a delayed motion, the resultant force has the opposite direction to the motion.

In this situation the elevator is accelerating upwards.

In that case, we will have:

F _N >P
F _R = F_N-P=ma
F _N =m.g+ma
F _N =m.(g+a)

In this situation the elevator rises delayed

In that case, we will have:

P> F _N
P- F_N=m .a
F _N =mg-ma
F _N =m.(ga)
a ≤g

In this situation the elevator descends accelerated

In that case, we will have:

P> F _N
P- F_N=m .a
F _N =mg-ma
F _N =m.(ga)
a ≤g

In this situation the elevator descends delayed

In that case, we will have:

F _N >P
F _N -P= m.a
F _N =m.g+ma
F _N =m.(g+a)