1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tension in the cable of the lift

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data

    The tension in the cable supporting a lift moving upward is twice the tension when the lift moves downward. What is the acceleration of the lift?

    2. Relevant equations



    3. The attempt at a solution
    i think its only conceptual but still please show me the way how to think about it.
    Thank you.
     
  2. jcsd
  3. Apr 8, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi jatin19901! welcome to pf! :wink:

    apply good ol' Newton's second law to find the relationship between a g and T for the two cases …

    what do you get? :smile:
     
  4. Apr 8, 2012 #3
    Re: welcome to pf!


    i am doing it like this , when lift is moving downward then , Tension(T) = Mass(m) * Net Acceleration ( Acc. due to gravity(g)+ Lift acceleration(a))
    And when lift is moving upward then : 2T = M * (g-a) , and finding out the relation between a and g , i am getting , a=-g/3 , i have a option of g/3 , and i am getting -g/3 .
    please correct me where i am wrong.
     
    Last edited: Apr 8, 2012
  5. Apr 8, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    no no no no noooo :redface:

    never do that, there is only one acceleration for one body

    in other words: you can put as many forces as you like on the LHS of F = ma, but you can only put one acceleration on the RHS​

    (of course, you can subtract accelerations of different bodies, to get the relative acceleration of the two bodies, but that's not this case)

    g is not an acceleration (on the RHS), mg is a force (on the LHS)

    try again! :smile:
     
  6. Apr 8, 2012 #5
    I do not got your last line.
     
  7. Apr 8, 2012 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    you mean? …
    in F = ma, the only a on the RHS is the actual acceleration (what you, for some reason, are calling the "net acceleration" :frown:)

    on the LHS, you put all the forces, and that includes the weight, mg :smile:
     
  8. Apr 8, 2012 #7
    oh ok , yes got the answer as A = g/3 , but i don't know whether it is correct or not , can u please check it. One more thing i have some questions to ask , can you please answer them here or do i need to start a another thread?
    Thank you.
     
  9. Apr 8, 2012 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    if you want us to check your calculations, you need to show them :wink:
    always start another thread :smile:
     
  10. Apr 8, 2012 #9
    when lift is moving upward i got , 2T+mg=-ma and when lift is moving downward , T+mg=ma , and then equate the T from both equations , is it right?
     
  11. Apr 8, 2012 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    uhhh? :confused:

    how can the tension be in the same direction as the weight??​
     
  12. Apr 9, 2012 #11
    Sorry for my stupidity , now i got it. As i am touching physics after two years , so this is the result. I hope i will get my command back upon physics with the help of this forum.
    Thank you Tiny Tim for answering my questions very quickly every time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Tension in the cable of the lift
Loading...