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Homework Help: Vertical cord tension in an elevator

  1. Jun 20, 2013 #1
    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


    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
  2. jcsd
  3. Jun 20, 2013 #2
    According your conclusion, the faster the acceleration upward, the lighter the bundle seems to the passenger. Does that look right to you?
  4. Jun 20, 2013 #3
    so i have to add the acceleration?
  5. Jun 20, 2013 #4
    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?
  6. Jun 20, 2013 #5
    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.
  7. Jun 20, 2013 #6
    F=ma= (.51m/s^2)(4.8kg)= 4.448 N
  8. Jun 20, 2013 #7
    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?
  9. Jun 20, 2013 #8
    How does that help you find tension?

    F is the sum of all the forces. What forces are there?
  10. Jun 20, 2013 #9
    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
  11. Jun 20, 2013 #10
    Why do you need the velocity? The formula will give you the acceleration of the elevator.
  12. Jun 20, 2013 #11
    From the bundle's point of view, what forces are acting on it?

    Please do not use one symbol for many different things.
  13. Jun 20, 2013 #12
    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
  14. Jun 20, 2013 #13
    The bundle has no contact with the elevator. It cannot be acted upon by the elevator.
  15. Jun 20, 2013 #14
    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"
  16. Jun 20, 2013 #15
    That is correct. But what is the other force? What, except gravity, acts on the bundle?
  17. Jun 20, 2013 #16
    the tension of the rope? xD i always had difficulties with this kind of problems
  18. Jun 20, 2013 #17
    I think I understand now. The tension is one force acting on the bundle, and gravity is another right?
  19. Jun 20, 2013 #18
  20. Jun 20, 2013 #19
    ok then the total net force is F=ma and I suppose i have to substitute F by F[SUB/]b[SUB/]-T right?
  21. Jun 20, 2013 #20
    If you mean F = mg - T, then why not F = T - mg? How do you choose the right one?
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