# Action-Reaction Forces on an Elevator

• Catchingupquickly
In summary: You will have two force terms and one acceleration term on the left side. Solve for FN.In summary, the elevator and its passengers have a combined mass of 1030 kg and experience an upward force of 12,000 N from the cable and a downward force of 1,400 N from friction. The net acceleration of the elevator and its passengers is 0.49 m/s^2 and the force normal acting on the 35 kg passenger is 326 N downward. After 12 seconds, the elevator will have a velocity of 5.9 m/s.
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.

Last edited:
Catchingupquickly said:
Fg = 9.80 m/s^2
g is that. Fg is a force, presumably, not an acceleration.
Catchingupquickly said:
acceleration=F/m
acceleration is Fnet/m. What is the net force on the elevator plus occupants?

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

Catchingupquickly said:
FN = mg - ma
Is the acceleration up or down?
Catchingupquickly said:
0.49126213592233
Far too many significant figures.

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

Catchingupquickly said:
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 FN 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).

## 1. What is an action-reaction force on an elevator?

An action-reaction force on an elevator refers to the equal and opposite forces that are exerted on the elevator and its occupants when the elevator is in motion. This is due to Newton's Third Law of Motion which states that for every action, there is an equal and opposite reaction.

## 2. How does an action-reaction force affect the movement of an elevator?

The action-reaction forces on an elevator cancel each other out, meaning that the elevator and its occupants experience a net force of zero. This allows the elevator to move smoothly and without any acceleration or deceleration while in motion.

## 3. Can an elevator experience action-reaction forces even when it is not moving?

Yes, an elevator can still experience action-reaction forces even when it is not moving. When the elevator is stationary, the force of gravity pulling down on the elevator is balanced by an equal and opposite force exerted by the ground, preventing the elevator from falling.

## 4. How does the weight of the elevator and its occupants affect the action-reaction forces?

The weight of the elevator and its occupants does not affect the action-reaction forces. According to Newton's Second Law of Motion, the acceleration of an object is dependent on its mass and the applied force. Since the action-reaction forces on an elevator cancel each other out, the weight of the elevator and its occupants do not have an impact on its movement.

## 5. Are there any safety implications of action-reaction forces in elevators?

There are no significant safety implications of action-reaction forces in elevators. Elevators are designed and built to withstand these forces and ensure the safety of its occupants. However, regular maintenance and safety inspections are necessary to ensure that the elevator is functioning properly and can withstand the forces exerted on it.

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