Question

A person weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal force of the elevator's floor on the person?

Answer 1

First of all, we can find the mass of the person, since we know his weight W:
$W=mg=0.70 kN=700 N$
And so
$m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg$

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
$\sum F = ma$
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
$R_v -W = ma$
We know the acceleration of the system, $a=1.5 m/s^2$ (upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
$R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N$