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Tension in a elevator cable

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data
    An elevator weighing 2.00 X10^5 N is supported by a steel cable. What is the tension in the cable when the elevator is accelerated upward at a rate of 3.00m/s^2? (g=9.81m/s^2)

    2. Relevant equations
    I realize this is probably a very simple problem, but neither my textbook, nor 2 days of searching the web have produced any help. How do I figure this problem out?


    3. The attempt at a solution
    First I drew an elevator with a cable. In side I put the weight (2.00 * 10^5 N) with an arow pointing down. On the outside I put an the speed (3.00m/s^2) with an up arrow. Then I planed a down arrow with the the amount of gravity (9.81m/s^2). The speed is a positive direction, so i placed a plus sign in front, +3.00m/s^2. For the weight and gravity, since they were negative values, I used a negative sign, -2.00 * 10^5 N, -9.81m/s^2.

    From here I wasn't sure what to do, so I added the the factors with the same units. I came up with +3.00 + (-9.81)= -6.81m/s^2. The I multiplied this value by -2.00 * 10^5 N. The answer I came up with is 1.362*10^6 N.

    I'm almost positive that it is wrong but im not sure what else to do.
     
  2. jcsd
  3. Dec 19, 2008 #2

    PhanthomJay

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    good start.
    you mean acceleration of 3m/s^2, not speed.
    No, don't do this. The acceleration is given as 3m/s^2 up. Gravity effects are already included in the weight.
    delete the word 'gravity', and the number '-9.81m/s^2', from this sentence.
    Try Newton 2, where F_net, which must be in the upward direction, is the algebraic sum of the upward tension force and the downward weight force.
     
  4. Dec 19, 2008 #3
    so i dont have to worry about the effects of gravity at all in this equation?
     
  5. Dec 19, 2008 #4

    PhanthomJay

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    the weight, which is given, is the force caused by the acceleration of gravity acting on its mass; it's not that you shouldn't be concerned with it, it's rather that its effects are already considered in the problem , since the weight is a given quantity. What you need now to know is the mass of the elevator. What formula relates weight and mass?
     
  6. Dec 20, 2008 #5
    Expanding on what Jay said a bit to help you understand:

    Using newton 2 you'll see that (using Ef=ma) you have 2 opposing forces from your force diagram, the weight (mg) and the tension in the cable (your unknown). You had the weight pointing down, which was good, now realize that the tension is pointing up.

    So using newton 2:

    Ef=ma -----> (insert your upward force) - (insert your downward force) = ma

    The accel. was given to you, as was the mass.
    Now you have everything you need to solve for the tension.

    You can approach lots of problems this way by analyzing your free body diagram.
     
  7. Dec 24, 2008 #6
    ok, what is newton 2? do you mean newton's second law? or am i missing the point completely?
     
  8. Dec 24, 2008 #7
    Yes, newtons second law ,Ef=ma
     
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