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Lowering a Piano with 2 ropes.

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data
    The two ropes are used to lower a 244.37 kg piano 9.47 m from a second-story window to the ground. How much work is done by gravity force? (T1=1805 N, q1=64°, T2=1112 N, q2=44°.)
    C11P111.jpg

    How much work is done by T1 force?
    How much work is done by T2 force?
    2. Relevant equations
    Wt1= mgdTcos(theata)


    3. The attempt at a solution

    Wt1= mdTcos(theata)=RIDICULOUSLY HUGE and wrong lol. What is wrong with the way I am approaching this problem?
     
  2. jcsd
  3. May 20, 2009 #2

    dx

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    The force on the piano due to gravity is just the weight. So the work done by gravity is (force)(distance moved in direction of force) = weight times the distance lowered.
     
  4. May 20, 2009 #3
    mmhmmm. I understand that part. But the main problem is the tension parts
     
  5. May 20, 2009 #4

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    Oh sorry, I didn't see there were two more parts to the question.
     
  6. May 20, 2009 #5
    I am sorry, I mean the work done by the first rope. I was thinking tension in my head. I thought the equation i came up with is right? What is wrong with it?
     
  7. May 20, 2009 #6

    dx

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    The vertical component of T1 is T1sin(θ), not T1cos(θ). You will also need to add a minus sign because the direction of displacement is opposite to this vertical component. So the work done by T1 will be -dT1sin(θ) (force times displacement in direction of force).
     
  8. May 20, 2009 #7
    so the equation i should be using is Wt1=-d*sin(theata)*T

    isn't T the force?
     
  9. May 20, 2009 #8

    dx

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    What is d? Isn't it the displacement?
     
  10. May 20, 2009 #9
    Yep! I got it thanks for the help I got it right but more importantly undersand where the anwser came from.
     
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