# Homework Help: Net Force, Smhet Force

1. Dec 20, 2013

### paolostinz

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.

2. Dec 20, 2013

### haruspex

Fa acts upward, Fg downward. You seem to have taken them both downward, with only friction acting upward.
Note that you do not immediately know which way friction will act. It depends whether the net of the other forces is up or down.

3. Dec 23, 2013

### paolostinz

Wow, thank you, I can't believe I missed that part about the friction, I just assumed it was acting downwards.

Alright this is my second attempt:

F_a=1.20 x 10^4 N [up]

F_g=mg
F_g=(1030 kg)(9.8 m/s^2 [down])
F_g=1.01 x 10^4 N [down]

F_net=F_a + F_g
F_net=1.20 x 10^4 N [up] + 1.01 x 10^4 N [down]
F_net=1.90 x 10^3 N [up]

Now for the 35 kg passenger:

F_a=1.90 x 10^3 N [up]

F_f=1.40 x 10^3 N [down]

F_net=1.90 e^3 N [up] + 1.40e^3 N [down]
F_net= 500 N [up]

a_net=F_net / m
a_net=500 N [up] / 35 kg
a_net= 0.49 m/s^2 [up]

Now for the normal force on this passenger:

F_g=mg
F_g=(35 kg)(9.8 m/s^2 [down])
F_g=343 N [down]

F_net=F_a + F_g
F_net=1.20e^4 N [up] + 343 N [down]
F_net= 1.17e^4 [up]

Now for the velocity of all the passengers and the elevator:

v_2=v_1 + at
v_2=0 + (0.49 m/s ^2)(12.0s)
v_2=5.8 m/s^2 [up]

4. Dec 23, 2013

### haruspex

Rather than having to write 'up' and 'down' after each force, the usual approach is to define a positive direction (usually up) and use + and - to distinguish the actual directions of the forces.
So here you might write
F_g=(1030 kg)(-9.8 m/s^2 )
F_g=-1.01 x 10^4 N
This is not Fnet. It will only become Fnet after including friction. For now just think of it as F_a + F_g. Seeing that its value is positive (up) tells you the value of the frictional force will be negative.
It's very confusing to reuse symbols to mean different things. You already have a meaning for Fa.
500 N is the net force on the elevator plus its three passengers. If you apply that force to each passenger separately you'll have them accelerating upwards at different rates!