Action-Reaction Forces on an Elevator

[Catchingupquickly]

Homework Statement

An elevator containing three passengers (with a mass of 72 kg, 84, 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.

a) draw a free body diagram of all the forces acting on the elevator.

b) Calculate the net acceleration of the elevator and its passengers.

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

d) Calculate the force normal acting on this passenger.

e) What velocity will the elevator have 12.0 seconds after the passengers have entered the elevator?

Homework Equations

$acceleration = \frac F m$

FN = mg - ma

v = v0 + (a)(t)

The Attempt at a Solution

[/B]
So 1.20 x $10^4 N$ = 12 000 N [ up]

and 1.40 x $10^3$ = 1400 N [down]

Fnet is thus 12 000 - 1400 = 10 600 N

g = 9.80 $m/s^2$

a) I have lackluster drawing skills on this forum so a word picture: An arrow up from a central point. Top of the arrow labelled Fnet. Two arrows down from the central point on the same line, 1st arrow labelled Fg and the bottom, last labelled Ff

b) acceleration = Fnet / m

= 10600 N / 1030 kg = 10.29126213592233 $m/s^2$

c) Two arrows going up on the same line from the central point. The first arrow labelled Fnet and the top labelled FN. The arrow moving down from the central point is labelled Fg

d) $FN = mg - ma \\= 35 kg (9.80 m/s^2) - 35kg (10.29126213592233 m/s^2) \\ = 343 N - (- 360.194174757278 N) \\ = 703.194174757278 N$
or 703.2 N or just 703 N

e) $v = v0 + (a)(t) \\ = 0 + (10.29126213592233 m/s^2) (12.0 seconds) = 123.495145631068 m/s$
or 123.5 m/s or 124 m/s

I'm not feeling confident about these answers. If I'm right this is a kick @SS, potentially deadly express elevator. Which kinda makes me think I missed something somewhere.

 

[haruspex]

Fg = 9.80 m/s^2

g is that. Fg is a force, presumably, not an acceleration.

acceleration=F/m

acceleration is F_net/m. What is the net force on the elevator plus occupants?

 

[Catchingupquickly]

Yes, apologizes for being imprecise.

However, thank you very much. You pointing out that little slip caused me to see the problem(s) immediately.

Ok Round 2...

Fg = mg = 1030 kg x 9.8 m/s^2 = 10 094 [down]

Fnet = FN + Fg + Ff
= 12 000 N [up] - 10 094 N [down] - 1400N [down]
= 506 N [up]

b) a = Fnet / m
= 506 N / 1030 kg
= 0.49126213592233 m/s^2

d) FN = mg - ma
= (35 kg x 9.80 m/s^2) - (35 kg x 0.49126213592233 m/s^2)
= 325.8058252427184 N or 325.8 N

e) v = v0 + (a)(t)
= 0 + (0.49126213592233 m/s^2) (12 seconds)
= 5.895145631067961 m/s or 5.9 m/s

 

[haruspex]

FN = mg - ma

Is the acceleration up or down?

0.49126213592233

Far too many significant figures.

 

[Catchingupquickly]

Cleaning it up further:

d) FN = (35 kg) (9.80 m/s^2[down]) - (35 kg) (0.49 m/s^2 [up])
= 343 N [down] - 17.15 [up]
= 325.85 or 326 N [down]

Downward force being plausible to me via the awkward "pulling" feeling to be experienced riding up in some elevators.

e) v = 0 + (0.49 m/s^2) (12 seconds)
= 5.88 or 5.9 m/s

 

[haruspex]

FN = (35 kg) (9.80 m/s^2[down]) - (35 kg) (0.49 m/s^2 [up])

No.
The mg term is a force here, not an actual acceleration of the object.
The standard equation you should always start with is ΣF=ma. Starting with a mix of accelerations and forces on one side can confuse you.
The forces are F_N up and mg down, while the acceleration is a up.
Write the ΣF=ma equation, keeping up as positive everywhere (or down as positive everywhere, as long as you are consistent).