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Forces and elevators

  • #1

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 104 N, but friction opposing the motion of the elevator is 1.40 x 103 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 normally acting on this passenger.

Homework Equations


To be completely honest I'm not sure which of our formulas applies to this situation...

The Attempt at a Solution


I have drawn the force diagram below, but am not sure if I have done so correctly... does an elevator still have normal force? After that I am completely lost. I know f=ma, so maybe I can find the acceleration using the combined weight of the elevators and passengers, plus the net force on the elevator, but once again I am not sure about those forces. I'm even less sure about what is acting specifically on the 35 kg passenger. Any help is greatly appreciated!
Diagram.JPG
 

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Answers and Replies

  • #2
lewando
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I applaud your attempt to echo the word problem with a clear drawing as you have done. Attention to detail (and yes, penmanship!) will serve you well in your life. May I assume your avatar is also your hand-drawn creation?

The normal force, FN, looks out of place. If the elevator were resting on the surface of the Earth (and not suspended by a cable), the Earth would be providing FN in response to the weight of the elevator plus people (according to Newton's 3rd law).

This is a 2-part problem. Do the first part first (calculate the net acceleration), treating the elevator plus people as an indistinguishable mass.
 
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  • #3
I applaud your attempt to echo the word problem with a clear drawing as you have done. Attention to detail (and yes, penmanship!) will serve you well in your life. May I assume your avatar is also your hand-drawn creation?

The normal force, FN, looks out of place. If the elevator were resting on the surface of the Earth (and not suspended by a cable), the Earth would be providing FN in response to the weight of the elevator plus people (according to Newton's 3rd law).

This is a 2-part problem. Do the first part first (calculate the net acceleration), treating the elevator plus people as an indistinguishable mass.
Why thank you! Yes, the picture is mine, but I did find the original it is based off of on the internet and therefore cannot take full credit.

I eliminated the normal force, and calculated the force of gravity to find the net force, and the acceleration! Using this, I answered the last question which it appears I forgot to include. It was "what is the elevators velocity 12 seconds after the occupants entered".

49a.png
49b.png

That just leaves the questions relating to the smallest occupant... Once again confused about the forces. As a person who has ridden in a elevator I know that when travelling upwards I feel more force upon myself, and feel heavier. I am assuming there is the force of gravity on him, no normal as he's not on the ground, and the force being applied upwards as he is, in some way, attached to the elevator. However, would I also include the friction? Not sure if I'm making sense, so I tried drawing what I have now... It may be just as confusing though... :/
Diagram 2.JPG

Either way, the third question, about the forces acting upon the 35 kg passenger relate to normal forces experienced on a regular day, right? So F = mg = 35 kg x 9.8 m/s [down] = 343 N [down], correct?
 

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  • #4
lewando
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Focus only on the "forces acting on the 35 kg passenger". Fa is the force of the cable attached to the elevator, so it is not attached to the passengers head. Is there a frictional force acting on the passenger? Is there a force due to the Earth's gravity? Is there a force due to the acceleration of the elevator?

Compare the accelerating elevator scenario with a non-moving elevator, and with a passenger simply standing on the ground.

· Calculate the force normally acting on this passenger.
This is an oddly worded question. Could it mean: calculate the normal force acting on the passenger when the elevator is accelerating?
 
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  • #5
kuruman
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Could it mean: calculate the normal force acting on the passenger when the elevator is accelerating?
I interpret "normally" to mean "in the normal direction", not "under normal conditions."
 
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  • #6
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I don't know why the OP never replied, but I've been solving the same question, and my calculations for solving the force normal acting on the 35 kg passenger:
Upward net acceleration: anet=0.49m/s2
Net force on the passenger: Fnet=Fs+mg
Newton's Second Law: Fnet=manet
Fs+mg=manet
Fs=manet-mg
Fs=35 kg x (0.49m/s2[up] - 9.8m/s2[down])= 35 kg x (0.49m/s2[up] + 9.8m/s2[up])
Fs= 35 kg x 10.29 = 360 N [up]
Therefore, the passenger will feel heavier during this period of time as the elevator accelerates upwards.

And as for the free-body diagram with all the forces affecting the 35 kg passenger:
Passenger on Elevator.jpg
 

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  • #7
haruspex
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I don't know why the OP never replied, but I've been solving the same question, and my calculations for solving the force normal acting on the 35 kg passenger:
Upward net acceleration: anet=0.49m/s2
Net force on the passenger: Fnet=Fs+mg
Newton's Second Law: Fnet=manet
Fs+mg=manet
Fs=manet-mg
Fs=35 kg x (0.49m/s2[up] - 9.8m/s2[down])= 35 kg x (0.49m/s2[up] + 9.8m/s2[up])
Fs= 35 kg x 10.29 = 360 N [up]
Therefore, the passenger will feel heavier during this period of time as the elevator accelerates upwards.

And as for the free-body diagram with all the forces affecting the 35 kg passenger:
View attachment 220656
Your calculations are fine, except that there is a flaw in the question statement which invalidates the answers. The cable tension is given as 1.20x104N, so that means anything from 11950N to 12050N. Propagating that through the steps leads to an acceleration anywhere from 0.44ms-2 to 0.54ms-2.
 
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  • #8
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Therefore, if I have understood you correctly, what you are saying is that 0.50m/s2 is the most accurate acceleration of the elevator? And that 12000 N might be the most accurate upward force? Would you mind reiterating?
 
  • #9
haruspex
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0.50m/s2 is the most accurate acceleration of the elevator?
0.5, not 0.50
12000 N might be the most accurate upward force?
Which upward force? The tension in the cable is given as 1.20x104N. There is no better way to express that.
 
  • #10
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0.5, not 0.50

Which upward force? The tension in the cable is given as 1.20x104N. There is no better way to express that.
The cable exerts an upward force. I concur with you on the 0.5m/s2
After redoing the calculations I came with the results:
The velocity of the elevator 12.0s after the passengers enter it is 6.0 m/s2
The force normal acting on the 35 kg passenger will be 360.5 N

It isn't a big difference, but could very well end up being the exactly correct response to the question, due to the faults. Thank you for looking over my work on this question, I appreciate it. :smile:
 

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