Vertical cord tension in an elevator

  • #1
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1. An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.4 s. A passenger in the elevator is holding a 4.8 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates? The acceleration of gravity is 9.8 m/s^2.



Homework Equations



F=ma

The Attempt at a Solution


I did a body diagram and get to the conclusion that T=mg-a. to get the acceleration i first had to get the velocity in the first 1.4s. v=d/t= 1m/1.4s=.714m/s and then get the a=v/t=(.714m/s)/1.4s=.510m/s^2

please correct me if i'm wrong
 

Answers and Replies

  • #2
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According your conclusion, the faster the acceleration upward, the lighter the bundle seems to the passenger. Does that look right to you?
 
  • #3
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so i have to add the acceleration?
 
  • #4
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No, you do not 'have' to. But you should use Newton's second law, which you have already written down: F = ma.

You know m, and know a. What is F?
 
  • #5
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Voko is taking care of the other part, so I'll just say that your procedure of calculating the acceleration of the elevator is incorrect. What you have done is found out the average velocity by dividing total distance upon total time, and then further average acceleration.

Use one of the kinematic equations which connect distance, acceleration and time.
 
  • #6
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F=ma= (.51m/s^2)(4.8kg)= 4.448 N
 
  • #7
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so, for getting the acceleration I use x= xi+vit+(1/2)at^2 and because i know that vi=0 and xi=0 then x=1/2at^2 ? and the v=d/t?
 
  • #8
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F=ma= (.51m/s^2)(4.8kg)= 4.448 N
How does that help you find tension?

F is the sum of all the forces. What forces are there?
 
  • #9
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a downward force F=mg and the upward force being F=ma, and the tension would be F=mg? right ? because that's the force applied downwards
 
  • #10
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so, for getting the acceleration I use x= xi+vit+(1/2)at^2 and because i know that vi=0 and xi=0 then x=1/2at^2 ? and the v=d/t?
Why do you need the velocity? The formula will give you the acceleration of the elevator.
 
  • #11
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From the bundle's point of view, what forces are acting on it?

Please do not use one symbol for many different things.
 
  • #12
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Fb=mg which is the downward force acting on the bundle a.k.a. the cord and F=ma which is the upward force acting by the elevator going up
 
  • #13
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The bundle has no contact with the elevator. It cannot be acted upon by the elevator.
 
  • #14
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but obviously there has to be 2 forces acting on the bundle, and i know for sure that one of those has to be "mg"
 
  • #15
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That is correct. But what is the other force? What, except gravity, acts on the bundle?
 
  • #16
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the tension of the rope? xD i always had difficulties with this kind of problems
 
  • #17
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I think I understand now. The tension is one force acting on the bundle, and gravity is another right?
 
  • #18
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Yes.
 
  • #19
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ok then the total net force is F=ma and I suppose i have to substitute F by F[SUB/]b[SUB/]-T right?
 
  • #20
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If you mean F = mg - T, then why not F = T - mg? How do you choose the right one?
 
  • #21
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F[SUB/]b[SUB/]-T =mg tension is negative because gravity is a negative acceleration.
 
  • #22
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I do not understand. Where did ##F_b - T = mg ## come from? It makes no sense, anyway, because earlier you said ##F_b = mg##, so your equation implies T = 0.
 
  • #23
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well, a friend of mine already did the problem and he told me that he got it right, and he passed me the problem but I want to understand it it goes like this:

W=mg , F=ma so T-w=ma and to find a he uses the kinematic formula that i posted earlier.
 
  • #24
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Do you understand the part before determining a? I would hope that by now you should :)
 
  • #25
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because T and w are two opposing forces right?
 

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