Newton's Third Law in an Elevator

  • Thread starter Sammy101
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  • #1
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Main Question or Discussion Point

Hi,

I was in an elevator the other day and I began to think about the forces acting on me and the elevator during my ascent and I became very confused.
Let's say for example that my mass and the elevator's mass combined is 100kg so this weighs 98N. When the elevator is at rest I figured that we would weigh 98N, but if the elevator began to acclerate, I think I might weigh more because there is now a net force on me and the elevator that is causing us to accelerate up at let's say 3m/s^2.
Fnet = ma
= 100kg (3m/s^2) = 300N of net force

If the opposing force is the weight of the elevator and my weight combined which is -98N, then the applied force is 398N right?

Is this 398N the tension in the rope? Or is 300N the tension in the rope since 98N is already the reaction force of the pull of weight? My main question is I understand that one pair of action reaction forces is the -98N of weight and the normal force of 98N but if the rope is pulling the elevator up for a net force, would the elevator not pull the rope down and cancel this net force? Why does the object accelerate?

Thank you!!
 

Answers and Replies

  • #2
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Firstly the elevator and you have a downwards force of 980N not 98N.
So the combined force of 980+300, i.e. 1280N is the tension in the rope.

1280N upwards force is partially cancelled by 980N downwards force giving a net upwards force of 300N so it accelerates upwards at 3m/s2
 
  • #3
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If the elevator is sitting still, the opposing force is -980N or the weight of the object and the applied force is 980N which is the normal force of the floor pushing up on the object. The net force is zero. If the elevator is accelerating at 3m/s^2 the net force is 300N because the mass of the elevator and person is 100kg.

So if the opposing force is -980N, the applied force must be 1280N. I understand that, but if the floor is already pushing up with a normal force of 980N, then would a external force only need to apply and extra 300N to cause a net force? why is the tension in the rope not just 300N? Why is it adding the normal force of the object 980N + 300N to get 1280N?
 
  • #4
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So if the opposing force is -980N, the applied force must be 1280N. I understand that, but if the floor is already pushing up with a normal force of 980N, then would a external force only need to apply and extra 300N to cause a net force? why is the tension in the rope not just 300N? Why is it adding the normal force of the object 980N + 300N to get 1280N?
So what is the tension in the rope when the elevator is standing still (but suspended from the rope)?
 
  • #5
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Oh I think I get it it would be 980 N. So when accelerating at 3m/s^2, the tension is 1280N I get that, but does that mean that the motor pulling the elevator up is only applying 300N?
 
  • #6
rcgldr
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does that mean that the motor pulling the elevator up is only applying 300N?
Most cable driven elevators have a counter weight that is close to the weight of the elevator, so the motor only needs to compensate for the difference in mass between elevator and load versus the counter weight and the rate of acceleration.
 
  • #7
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I'm sorry it sometimes takes me awhile to understand things. So, is the motor only applying 300N of force to accelerate the elevator since a force of 980N is already applied by the normal force? I am confused by the role of the opposite force of weight, the normal force, in this situation. Is it an applied force?
 
  • #8
rcgldr
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Here's a link to an elevator article with an animated diagram showing a counterweight:

elevator3.htm
 

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