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Tension on the rope

  1. Aug 21, 2015 #1
    1. The problem statement, all variables and given/known data
    An empty elevator of mass 2.7×103 kg is going down by a cable at an acceleration of 1.2 m/s2. What is the tension in the rope pulling the lift?

    2. Relevant equations
    ## T=m(g-a) ##

    3. The attempt at a solution
    ## T=m(g-a) \\ \Rightarrow T = 2.7 *10^3 (9.8-1.2) \\ \Rightarrow T= 2.7 *10^3 \times 8.6 = 23.22*10^3 N ##
    Is my solution correct. Have I used the right formula?

    Actually when an elevator goes down it moves in the direction of gravitational pull therefore why isn't the formula ## T= m(g+a)##. In stark contrast, when the elevator goes up it is moving in the direction against gravity hence shouldn't the formula then be ## T=m(g-a)##
     
    Last edited: Aug 21, 2015
  2. jcsd
  3. Aug 21, 2015 #2

    andrevdh

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    Yes. That looks fine. Truncate your answer to two significant digits though.
     
  4. Aug 21, 2015 #3
    Your working is fine . However , what formula are you referring to here ?
     
  5. Aug 21, 2015 #4

    PeroK

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    What makes you doubt your answer?
     
  6. Aug 21, 2015 #5
    Actually when an elevator goes down it moves in the direction of gravitational pull therefore why isn't the formula ## T= m(g+a)##. In stark contrast, when the elevator goes up it is moving in the direction against gravity hence shouldn't the formula then be ## T=m(g-a)##
     
  7. Aug 21, 2015 #6
    The easiest way to go about such a question is use Fnet = ma .
     
  8. Aug 21, 2015 #7
    But ## F_{net}= ma## is used when the elevator is stationary and ## a=g##
     
  9. Aug 21, 2015 #8
    Actually I am not very well able to decipher what you are trying to interpret. Do you mean that I should write the answer as ## 23220 N ##
     
  10. Aug 21, 2015 #9

    jbriggs444

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    Put yourself in the position of the elevator machinery. You are holding up the cable which is holding up the car. If the car is just sitting there, you are contending only with gravity. Clearly the tension you must put on the cable must be equal to mg.

    Now suppose that you want to allow the elevator to accelerate downward. Would you increase or decrease the tension you put on the cable?
     
  11. Aug 21, 2015 #10
    To what he meant - write it as 23 × 103 N .

    To my post - Fnet = ma is Newton's second law ( for constant mass systems ) - applicable in all cases .

    Hope this helps .
     
  12. Aug 21, 2015 #11
    I would decrease the tension I suppose. Ohh! So is that why we use the formula ## T= m(g-a) ## when the elevator accelerates downwards?
     
  13. Aug 21, 2015 #12
    I didn't get you. How can we use the ## F_{net}=ma ## here. Thanks
     
  14. Aug 21, 2015 #13

    jbriggs444

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    Bingo!
     
  15. Aug 21, 2015 #14

    andrevdh

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    Answer in post number 10 (Qwertywerty).
    The question gives values with two significant digits for both the mass
    and the acceleration, thus two significanrt digits in the final answer.
     
  16. Aug 21, 2015 #15
    Thanks a lot! :smile:
     
  17. Aug 21, 2015 #16
    Ok! Thanks
     
  18. Aug 21, 2015 #17
    Acceleration of lift is ' a ' downwards . Forces acting on the elevator are gravity , and tension .

    Therefore ,
    mg - T = ma .
     
  19. Aug 21, 2015 #18
    Ohh Ok! Thanks
     
  20. Aug 22, 2015 #19
    1) draw your Forces !

    2) equation for each force is. ##F = m*a##

    3) the two forces are gravity and acceleration

    4) tension comes from the forces pulling on the rope!

    <------ ============== ------>
    tension
    Tension = F1 + F2

    In this case, Forces are in the same direction so tension is lessened.

    ------>. ============== ------>
    F1 = ma. F2 = mg

    Tension = - F1 + F2
     
  21. Aug 22, 2015 #20
    Viraam:

    If you had drawn a free body diagram for the elevator, you would not be experiencing all this uncertainty. Draw one, and you will see what I mean.

    Chet
     
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