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NotTim

Determine resultant: 25N at 10degrees to the horizontal and 30N horizontally.

  • 2 years ago
  • 2 years ago

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  1. NotTim
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    relatively simple. again, i ask for the angle only.

    • 2 years ago
  2. NotTim
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    |dw:1333533030424:dw|

    • 2 years ago
  3. experimentX
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    |dw:1333522316234:dw|

    • 2 years ago
  4. experimentX
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    try using cosine law

    • 2 years ago
  5. NotTim
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    but there are no angles hen (10 isnt' actual an angle within the triangle)

    • 2 years ago
  6. experimentX
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    it's 170

    • 2 years ago
  7. NotTim
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    wut. how?

    • 2 years ago
  8. experimentX
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    because 25 is applied 10 degree to horizontal ... and obviously, it makes 10 degree or 180-10 degree with the horizontal

    • 2 years ago
  9. NotTim
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    ok...but how do i apply 170 in this triangle|dw:1333533509958:dw|

    • 2 years ago
  10. NotTim
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    (im more familiar with that format)

    • 2 years ago
  11. experimentX
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    |dw:1333522738607:dw|

    • 2 years ago
  12. NotTim
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    Like

    • 2 years ago
  13. NotTim
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    |dw:1333533595849:dw|Like I understand how it applies outside, but how do i calculate the angle within?

    • 2 years ago
  14. experimentX
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    30N is parallel to horizontal

    • 2 years ago
  15. experimentX
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    |dw:1333522901720:dw|

    • 2 years ago
  16. NotTim
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    oh. the z,c, f pattern. got it.

    • 2 years ago
  17. NotTim
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    what's our next step then, cos law?

    • 2 years ago
  18. experimentX
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    yep ... we have all parameters required to calculate

    • 2 years ago
  19. dumbcow
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    |dw:1333525849053:dw| another way of looking at it

    • 2 years ago
  20. dumbcow
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    \[x^{2} = 25^{2}+30^{2}-2(25)(30)\cos 170\]

    • 2 years ago
  21. NotTim
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    thank you! i hadnt thought of this

    • 2 years ago
  22. NotTim
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    hey guys. I'm having problems figuring out the angle (again!)

    • 2 years ago
  23. NotTim
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    I manged to get 5.4 degrees (the closest yet!), while the actual answer is 4.5 degrees.

    • 2 years ago
  24. NotTim
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    i did so by using sin-1(30sin170/54.8)

    • 2 years ago
  25. Carniel
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    Yea he is correct

    • 2 years ago
  26. NotTim
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    what?

    • 2 years ago
  27. NotTim
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    @Carniel Wat?

    • 2 years ago
  28. Carniel
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    What they all said to you is correct

    • 2 years ago
  29. NotTim
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    yeah, but i still got the wrong angle...

    • 2 years ago
  30. Carniel
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    How so? Have you tried Tan,Cos, and Sin?

    • 2 years ago
  31. NotTim
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    the trig ratios? it only applies to right angle triangles...doesnt it?

    • 2 years ago
  32. Carniel
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    Not if I re-call

    • 2 years ago
  33. NotTim
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    Sorry if I sound incredulous, but I really do think it applies only to right angle triangles.

    • 2 years ago
  34. Carniel
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    Do you have notes? If soo check to make sure.

    • 2 years ago
  35. NotTim
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    i dont have the notes specifically on that (its buried in a box somewhere); ill check the internet.

    • 2 years ago
  36. Carniel
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    There you go then =D

    • 2 years ago
  37. NotTim
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    but...i can't use it here...(unless i do component method, which im not allowed to)

    • 2 years ago
  38. Carniel
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    Which methods do you recall :)?

    • 2 years ago
  39. dumbcow
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    you need to use Law of Sines \[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\]

    • 2 years ago
  40. NotTim
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    yeah. i used that and got 5.4 degrees.

    • 2 years ago
  41. NotTim
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    @dumbcow yeah. i used that and got 5.4 degrees.

    • 2 years ago
  42. NotTim
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    hello?

    • 2 years ago
  43. dumbcow
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    ok the 5.4 degrees is the angle inside the triangle, they want the angle of resultant vector --> 10 - 5.45 = 4.55

    • 2 years ago
  44. NotTim
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    thanks again

    • 2 years ago
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