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ksaimouli

A swing ride at a carnival consists of chairs that are swung in a horizontal circle by 12.0 m cables that are attached to a vertical rotating pole, as the drawing shows. Suppose that the total mass of the chair and it occupant is 220 kg. i need just the free body diagram

  • one year ago
  • one year ago

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  1. ksaimouli
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    |dw:1350335809666:dw|

    • one year ago
  2. imron07
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    |dw:1350336122902:dw|

    • one year ago
  3. ksaimouli
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    where is fn then

    • one year ago
  4. imron07
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    What is fn

    • one year ago
  5. ksaimouli
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    normal force

    • one year ago
  6. ksaimouli
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    and how did u get t cos 65 because that is y coordinate right so it would be t sin 65 ?

    • one year ago
  7. ksaimouli
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    @imron07

    • one year ago
  8. ksaimouli
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    can u plz help me with this

    • one year ago
  9. imron07
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    The chair doesn't touch any surface right? Then there's no normal force. To decide choosing cos or sin, you need to know what angle is given first.

    • one year ago
  10. JakeV8
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    I don't think you need a normal force for this one. (yeah, what @imron07 said!)

    • one year ago
  11. ksaimouli
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    what do u mean by what angle

    • one year ago
  12. JakeV8
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    For the cos and sin relations, don't think of the y-coordinate... just look at the diagram and remember that cos = adj/hyp and sin = opp/hyp . You should see it as @imron07 drew it...

    • one year ago
  13. ksaimouli
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    yes i know that cos =... and sin=.. but ..

    • one year ago
  14. ksaimouli
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    ?

    • one year ago
  15. ksaimouli
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    can u draw it @JakeV8

    • one year ago
  16. JakeV8
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    I can't tell what you are asking... @imron07 has the right diagram. The vertical component of the force exerted by the 12m line is T cos(65) because T is the hyp and the vertical component is the adjacent side.

    • one year ago
  17. ksaimouli
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    |dw:1350337034801:dw|

    • one year ago
  18. JakeV8
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    X = T cos(65), right?

    • one year ago
  19. ksaimouli
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    yuppppp

    • one year ago
  20. ksaimouli
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    thx

    • one year ago
  21. ksaimouli
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    i was doing other way but same

    • one year ago
  22. JakeV8
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    glad to help. I'm pretty out of practice on these, but the diagram helped me remember a little.

    • one year ago
  23. ksaimouli
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    |dw:1350337344014:dw|

    • one year ago
  24. JakeV8
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    yes, that's what I get. I have to take off, just signed on for a minute... good luck!

    • one year ago
  25. ksaimouli
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    so how to find the tension of the rope @imron07

    • one year ago
  26. ksaimouli
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    Tcos 65=mg

    • one year ago
  27. imron07
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    Yes, you can find T then.

    • one year ago
  28. ksaimouli
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    then how to find the velocity

    • one year ago
  29. ksaimouli
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    -T sin 65= mv^2/r

    • one year ago
  30. ksaimouli
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    ?

    • one year ago
  31. imron07
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    Correct :)

    • one year ago
  32. ksaimouli
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    got it thx u

    • one year ago
  33. ksaimouli
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    hold on what is the answer for tension

    • one year ago
  34. ksaimouli
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    @imron07

    • one year ago
  35. ksaimouli
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    5102

    • one year ago
  36. ksaimouli
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    but book gave 5.1*10^3N

    • one year ago
  37. imron07
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    \[T=\frac{mg}{\cos65}=5101N=5.1\times10^3N\]

    • one year ago
  38. ksaimouli
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    and the last one velocity |dw:1350338528028:dw|

    • one year ago
  39. ksaimouli
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    so what did u get @imron07 this is last one

    • one year ago
  40. imron07
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    Actually, you can omit the minus sign:\[v=\frac{Tr\sin65}{m}\]where r equals:\[r=l\sin65\]|dw:1350338783066:dw|

    • one year ago
  41. imron07
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    sorry,\[v=\sqrt{\frac{Tr\sin65}{m}}\]

    • one year ago
  42. ksaimouli
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    how can i omit the -

    • one year ago
  43. imron07
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    the centripetal force is radially inward. This force itsel is minus, so: \[-\frac{mv^2}{r}=-T\sin65 \]

    • one year ago
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