- #1

paolostinz

- 25

- 1

## Homework Statement

An elevator that contains three passengers with masses of 72 kg, 84 kg, and 35 kg respectively has a combined mass of 1030 kg. The cable attached to the elevator exerts an upward force of 1.20 x 10^4 N, but friction opposing the motion of the elevator is 1.40 x 10^3 N.

- Draw a free-body diagram of all the forces acting on the elevator.

- Calculate the net acceleration of the elevator and its passengers.

- Draw a free-body diagram of all the forces acting on the 35 kg passenger

-Calculate the force normal acting on this passenger.

-Determine the velocity of the elevator 12.0 s after the passengers have entered the elevator.

## Homework Equations

F=ma

F_g=mg

F net= F_a + F_f

d=vt + 1/2at

v_2=v_1 + at

## The Attempt at a Solution

This question has me all kinds of confused, but here's my initial attempt.

My diagram has F_a (upward direction) as 1.20 x 10^4 N and F_f (downward direction) as 1.20 x 10^3 N. So then I use the formula F=mg, F=(1030 kg) (9.8 m/s^2 [down]), F=1.01 x 10^4 N [down]. I then add this force with F_a and F_f:

F net= F_a + F_f + Fg

F net= 2.07 x 10^4 [down]

F=ma

a=F/m

a=2.07 x 10^4 N[down] / 1030 kg

a=20.1 m/s^2 [down]

Now for the 35 kg person:

F_g=mg

F_g=(35 kg)(9.8 m/s^2 [down])

F_g=343 N [down]

F net= F_a + F_f + F_g

F net= 1.03 x 10^4 N [up]

This is where I start doubting myself, I feel like because the original net force had the elevator moving in the down direction that it should still be going in the same direction. I don't know if that makes sense though since that was using 3 combined masses within the calculations. Any guidance would greatly be appreciated.