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Homework Help: Tension+ centripetal force

  1. Oct 19, 2007 #1

    klm

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    Two wires are tied to the 400 g sphere shown in figure. The sphere revolves in a horizontal circle at a constant speed of 8.30 m/s

    1) What is the tension in the upper wire?
    2) What is the tension in the lower wire?

    28w1ipz.jpg

    okay i know that m=.4 and v=8.3
    i dont know r, which i thought that maybe i could find by doing the square root (1^2-.5^2) =.866
    and then Force= m V^2 / R sooo.. (.4)(8.3^2)/(.866) = 31.819 . but i dont know where to go from there?
     
  2. jcsd
  3. Oct 19, 2007 #2

    klm

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    okay i also think..
    Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
    Fnety= Tsin 26 + -T2 sin 26 -Fg= mv^2/ r
     
  4. Oct 19, 2007 #3

    klm

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    so my free body diagram looks like 1 wire pulling in the 2nd quad, one in the 3 quad, and Fg going straight down on the y axis
     
  5. Oct 19, 2007 #4
    yep. No motion in y-dir., so forces sum up to zero. In X- dir. they will contribute to centripetal force.
     
    Last edited: Oct 19, 2007
  6. Oct 19, 2007 #5

    klm

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    but i dont understand how the tension in the 2 wires are going to be different
     
  7. Oct 19, 2007 #6

    klm

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    Fnety= Tsin 26 + -T2 sin 26 -Fg= 0
    T1 sin 26 - T2 sin 26 = Fg
    T1 sin 26 - T2 sin 26 = 3.92

    Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
    -T1 cos 26 + -T2 cos 26 = 31.82
     
  8. Oct 19, 2007 #7

    klm

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    ^ i am not sure how to solve for T1 or T2 . can you please help me with that
     
  9. Oct 19, 2007 #8

    klm

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    please can someone help me!
     
  10. Oct 19, 2007 #9

    klm

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    i figured out the T2, but now i cant figure out T1. shouldnt it just be plug T2 into one of the equations and it should give you T1??
     
  11. Oct 19, 2007 #10

    Doc Al

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    Staff: Mentor

    Where does the 26 degrees come from?

    Which way does the centripetal acceleration act?

    Once you correct your equations, isolate the T1 term in each, like this:

    [tex]T_1 \sin\theta = mg + T_2 \sin\theta[/tex]
     
  12. Oct 19, 2007 #11

    klm

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    26 came from inverse tan (.5/1)
     
  13. Oct 19, 2007 #12

    klm

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    does centripetal acc act on the Horizontal force?
     
  14. Oct 19, 2007 #13

    Doc Al

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    Staff: Mentor

    But the 1 m side is the hypotenuse. That should be inverse sin (.5/1).
     
  15. Oct 19, 2007 #14

    Doc Al

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    Staff: Mentor

    My question is: What's the direction of the centripetal acceleration?
     
  16. Oct 19, 2007 #15

    klm

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    umm this may be wrong, but i thought centripetal acc just means that the direction is toward the center
     
  17. Oct 19, 2007 #16

    klm

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    ohh wait i know what i did. it is just the angle that is messing me up! thank you for pointing that out. i dont know why, but the angle i used still let me get the tension of the first string??!
     
  18. Oct 19, 2007 #17

    Doc Al

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    Staff: Mentor

    That's exactly right. So, what direction is that? Should it be positive or negative?
     
  19. Oct 19, 2007 #18

    klm

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    negative
     
  20. Oct 19, 2007 #19

    klm

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    would that be right?
     
  21. Oct 19, 2007 #20

    Doc Al

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    Staff: Mentor

    Good. So fix your equation for horizontal forces.
     
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