How Do You Calculate the Net Acceleration of an Elevator with Passengers?

[paolostinz]

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.

 

[haruspex]

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]

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.

 

[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]

 

[haruspex]

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]

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

F_net=F_a + F_g

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.

F_a=1.90 x 10^3 N [up]

It's very confusing to reuse symbols to mean different things. You already have a meaning for Fa.

F_net= 500 N [up]

a_net=F_net / m
a_net=500 N [up] / 35 kg

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!

 

[HalfLostAndSearching]

I would like to commend you on your approach to this problem. Your free-body diagrams and use of equations are correct so far. However, your doubts about the direction of the net force and acceleration for the 35 kg passenger are valid.

In this scenario, the net force and acceleration for the 35 kg passenger would actually be in the downward direction, as the elevator is moving down due to the combined mass of the passengers being greater than the force of the cable pulling up. This means that the 35 kg passenger would experience a force of 1.03 x 10^4 N downward, leading to an acceleration of 29.5 m/s^2 downward.

As for the velocity of the elevator after 12 seconds, we can use the equation d=vt + 1/2at to calculate the distance traveled by the elevator in 12 seconds. Since we know that the initial velocity was 0 m/s, we can rewrite the equation as d=1/2at^2. Plugging in the values, we get d=1/2(29.5 m/s^2)(12 s)^2 = 2121.6 m. This means that the elevator would have traveled 2121.6 meters downward in 12 seconds, giving it a velocity of 176.8 m/s downward.

In conclusion, the net force and acceleration for the elevator and its passengers would be in the downward direction due to the combined mass being greater than the force pulling up. The force normal acting on the 35 kg passenger would be 343 N downward. And after 12 seconds, the elevator would have a velocity of 176.8 m/s downward.