1. The problem statement, all variables and given/known data 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. 2. Relevant equations F=ma F_g=mg F net= F_a + F_f d=vt + 1/2at v_2=v_1 + at 3. 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.