vertical cord tension in an elevator


by Robertoalva
Tags: acceleration, elevator, tension
Robertoalva
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#1
Jun20-13, 08:55 AM
P: 140
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.



2. Relevant equations

F=ma

3. 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
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voko
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#2
Jun20-13, 09:03 AM
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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?
Robertoalva
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#3
Jun20-13, 09:05 AM
P: 140
so i have to add the acceleration?

voko
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#4
Jun20-13, 09:09 AM
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P: 5,489

vertical cord tension in an elevator


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?
dreamLord
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#5
Jun20-13, 09:11 AM
P: 199
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.
Robertoalva
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#6
Jun20-13, 09:11 AM
P: 140
F=ma= (.51m/s^2)(4.8kg)= 4.448 N
Robertoalva
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#7
Jun20-13, 09:18 AM
P: 140
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?
voko
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#8
Jun20-13, 09:19 AM
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Quote Quote by Robertoalva View Post
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?
Robertoalva
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#9
Jun20-13, 09:28 AM
P: 140
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
dreamLord
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#10
Jun20-13, 10:04 AM
P: 199
Quote Quote by Robertoalva View Post
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.
voko
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#11
Jun20-13, 10:09 AM
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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.
Robertoalva
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#12
Jun20-13, 10:50 AM
P: 140
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
voko
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#13
Jun20-13, 10:53 AM
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The bundle has no contact with the elevator. It cannot be acted upon by the elevator.
Robertoalva
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#14
Jun20-13, 11:01 AM
P: 140
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"
voko
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#15
Jun20-13, 11:06 AM
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That is correct. But what is the other force? What, except gravity, acts on the bundle?
Robertoalva
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#16
Jun20-13, 11:07 AM
P: 140
the tension of the rope? xD i always had difficulties with this kind of problems
Robertoalva
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#17
Jun20-13, 11:16 AM
P: 140
I think I understand now. The tension is one force acting on the bundle, and gravity is another right?
voko
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#18
Jun20-13, 11:16 AM
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P: 5,489
Yes.


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