Mechanics

# Vertically moving elevators

In situation (a) there is a person of mass m on a scale inside the elevator; in situation (b) there are forces acting on the situation

We’ve all been on an elevator. At first we were a little apprehensive: what would happen if he stopped in the middle of his journey? If the supporting cables broke, would he fall at full speed, as we see in some movies? Well, it’s not as dangerous as we sometimes see in the movies. In fact, the elevator is a very interesting place to study some concepts about Physics. For example, for most of its path it describes a uniform motion, we will only notice the change in velocity in its initial or final motion. Let us then analyze the behavior of a body inside an elevator.

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:

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.

– 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:

N >P
R = F_N-P=ma
N =m.g+ma
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
N =mg-ma
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
N =mg-ma
N =m.(ga)
a ≤g

In this situation the elevator descends delayed

In that case, we will have:

N >P
N -P= m.a
N =m.g+ma
N =m.(g+a)