An elevator accelerates upward in 2.80 s to a final speed of 6.20 m/s.

Question

An elevator accelerates upward in 2.80 s to a final speed of 6.20 m/s. A 72.0 kg person is standing on a bathroom scale in the elevator. Determine the apparent weight of the passenger.

Answer 1

First draw a free body diagram:

where "W" is the person's weight and F_N is the normal force.

An important thing to realize is that the measurement on the scale will be a reading of the normal force (or the "apparent" weight of the passenger).

The sum of the forces in the y-direction equals mass times acceleration

∑F_y = ma

F_N - W = ma

But weight (W) equals mass (m) times the acceleration due to gravity (g) or W = mg so:

F_N - mg = ma

F_N = mg + ma

F_N = m(g + a)

This is the equation we will use but we must first calculate the acceleration "a". Assuming CONSTANT acceleration and assuming the elevator starts from rest then a = Δv/Δt = (6.20 - 0)/2.80 = 2.214 m/s²

So, plugging that into the equation along with the mass and the value of "g":

F_N = 72.0(9.81 + 2.214)

F_N = 866 N