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    AbsoluteZero's Avatar
    AbsoluteZero Posts: 1, Reputation: 1
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    #1

    Oct 14, 2010, 09:32 PM
    AP Physics Help on Tension and Energy
    An elevator weighing 20000 N is supported by a steel cable. What is the tension in the cable when the elevator is being accelerated upaward at a rate of 3.00 m/s squared.

    I don't understand on how to find tesion and I have a test tomorrow and I have a lot of questions so I hope you guys can help me on this and other problems that involve energy :)
    Thx
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Oct 15, 2010, 05:36 AM

    ΣF=ma is at work here. You know the mass of the elevator cab, you have been told what the value of its acceleration is, so with ΣF=ma you can calculate the net force needed to make it accelerate upward. That net force equals the upward force of the tension in the cable minus the downward force of the weight of the elevator cab. Can you take it from here?
    kpg0001's Avatar
    kpg0001 Posts: 88, Reputation: 12
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    #3

    Oct 15, 2010, 06:38 AM
    Quote Originally Posted by ebaines View Post
    ΣF=ma is at work here. You know the mass of the elevator cab, you have been told what the value of its acceleration is, so with ΣF=ma you can calculate the net force needed to make it accelerate upward. That net force equals the upward force of the tension in the cable minus the downward force of the weight of the elevator cab. Can you take it from here?

    If the elevator is going up the wouldn't the forces be added together instead of subtracted? The tension is greater when the elevator is moving up than it is at rest just hanging. So, if it is going up, it had to overcome gravity making the net acceleration gravity plus 3m/s^2. This makes the force equal to the mass times gravity plus 3. Thus F=(20000/9.8)(9.8+3). Because it gives you the force while at rest, you divide that by gravity, giving you the mass, m=F/a, then multiply the mass by the total acceleration of 12.3m/s^2. Your answer would be correct if the question asked for the upward force of the elevator, not the tension in the cable. Hope this helps!
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Oct 15, 2010, 07:01 AM

    I used both of your answers, ebaines and kpg0001, and I get the same answer. So, both ways work, if you know what you are doing.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #5

    Oct 15, 2010, 07:17 AM
    Quote Originally Posted by kpg0001 View Post
    If the elevator is going up the wouldn't the forces be added together instead of subtracted?
    The math goes like this: let = tension in the cable, and = weight of the elevator, then



    So yes, the tension works out to be the sum of the weight plus the force needed to accelerate the elevator cab, but it is derived from ΣF = Force of tension MINUS the weight.

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