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Work, Moving a Ladder

  1. Jun 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A uniform ladder of mass 12kg is 4.6m long. If it is lifted from a horizontal to a vertical position, how much work is done.


    2. Relevant equations



    3. The attempt at a solution

    I'm just not sure how to approach this problem. I'm guessing that I start off using centre of mass of the ladder to determine that the radius of the movement is half the length of the ladder, so 2.3m.

    Since the ladder travels in an arch as it moves to it's resting position, would the distance value be c/4=∏r/2, so c/4=3.6m ?

    Force = 12kg x 9.81m/s2 = 117.3N

    W=117.3N x 3.6m = 420J

    How does this look?
     
    Last edited: Jun 25, 2012
  2. jcsd
  3. Jun 25, 2012 #2

    PhanthomJay

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    looks good, but you started by calculating the length of the arc and then you used a different approach to arrive at your answer. Please explain how you came to this conclusion.
     
  4. Jun 25, 2012 #3
    Jay, what I did was:

    Since w=fd (I ignored cosθ here)
    I needed to figure out the distance that the object traveled to complete my equation.
    I found the arc length which was c/4= 3.6m = d and plugged it into my w=fd equation. Of course, I found my force by multiplying the mass of the ladder by gravity.

    Does that make sense?
     
    Last edited: Jun 25, 2012
  5. Jun 25, 2012 #4

    PhanthomJay

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    No. If you did it that way, you'd end up with W = 117*3.6?? wouldn't you? Which would be wrong. Please explain.
     
  6. Jun 25, 2012 #5
    Since it's a uniform mass, I used centre of mass as the length of the ladder. That's where the 3.6m comes in. I used that 3.6m as a radius to find the arc length. The arc length is the length through which the ladder travels isn't it? Which is 2.3m.

    I edited out my last comment. I misread your last post.

    Second edit, lol:

    I didn't use 3.6m like I said in my explanation. I used 2.3m, I made a typo. I'll correct that.
     
  7. Jun 25, 2012 #6

    PhanthomJay

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    117*3.6 = 420
    117*2.3 = 270

    You've got some serious :biggrin:explaining to do, Jimbo.
     
  8. Jun 25, 2012 #7
    Mathemagics, my friend :)
     
  9. Jun 25, 2012 #8
    Oh my goodness, now I see what you're talking about. This thread is a wreck lol.

    So my final work is:

    w=117.3N x 3.6m = 420J

    So 420J is correct then?

    I'll go back and make the correction!
     
  10. Jun 25, 2012 #9

    PhanthomJay

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    No dont correct your answer of 270 J which is correct but which will be incorrect if you make the incorrect correction . Just please explain why 270 J is correct. Hint: What is the work done against gravity?
     
  11. Jun 25, 2012 #10
    Hmm,

    w= 12kg x 9.8m/s2 x 2.3m
    = 270J
     
  12. Jun 25, 2012 #11

    PhanthomJay

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    yes! But why?
     
  13. Jun 25, 2012 #12
    Because of uniform mass and using centre of mass as the distance through which it has traveled from 0m to 2.3m? And, m*g = force * 2.3m = work done or potential energy at that height.

    Is that right?
     
  14. Jun 25, 2012 #13

    PhanthomJay

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    Well not exactly, but close. Work done by gravity (-270 in this example) is independent of the path taken and only depends on its potential energy change between its start and end points. So the work done by the lifting force assuming the lift occurs at constant speed and angular speed must be ________
     
  15. Jun 25, 2012 #14
    I'm a little confused, sorry Jay. When I see constant anything, I think conservation of energy. If gravity is -270J then I would think the work in getting it up is 270J (for a straight line at least), the arc length is throwing me off.
     
  16. Jun 26, 2012 #15

    PhanthomJay

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    Oh, yes, conservation of energy....that is a great way to solve this problem; the arc length path of the center of mass does tend to confuse, even more than I do. You are probably familiar with the equation
    [tex] W_{nc} = \Delta KE + \Delta PE [/tex] where [tex]W_{nc}[/tex]represents the work done by all non conservative forces , which in this example is the work done by the lifting force. And since there is no change in speed during the motion (this is assumed in the problem, although not stated), then there is no KE change and thus the work done by the lifting force is just the change in PE of the ladder, mgh, or 270 J.
    The arc path of the center of mass doesn't matter, see my last post.
     
  17. Jun 26, 2012 #16
    That's great! Thanks for your help Jay!
     
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