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Calculating Net force

  1. Mar 15, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-3-15_2-2-5.png

    2. Relevant equations
    upload_2016-3-15_2-2-22.png
    upload_2016-3-15_2-2-36.png

    3. The attempt at a solution

    32° + 24° = 56°
    upload_2016-3-15_2-8-48.png
    I used the tip-to-tail method to add vectors
    c = ([12 N]^2 + [15 N]^2 - 2 [12 N] x [15 N] cos124°)1/2
    c = 23.9 N

    SinA/12 N = Sin124° / 23.9 N (i dont know how to solve this part)

    can someone please explain what im supposed to do for this question, and verify if what i've done so far is correct.
    thanks :)



     
  2. jcsd
  3. Mar 15, 2016 #2
    some force is pulling the cart - i would have simply resolved the forces in the direction horizontal and vertical and took the sum of horizontal and vertical components and found out the resultant force as square root of sum of the square of horizontal and vertical net forces.
    will i be wrong ?
     
  4. Mar 15, 2016 #3

    haruspex

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    Looks good so far.
    Doesn't look hard. Can't you just multiply both sides so that you only have sin A on the left?
    Alexandria has been taught to use the parallelogram approach. At this stage, I think it is better to stick to that.
     
  5. Mar 15, 2016 #4
    SinA = Sin124° / 23.9 N x (12 N)
    Sin124° = 0.83
    so 0.83/23.9 N x (12 N) =0.42 ???
     
  6. Mar 16, 2016 #5

    haruspex

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    Yes, that looks right. What is your next step?
     
  7. Mar 16, 2016 #6
    so im having a problem with figuring out the degrees. How does 0.42 convert to degrees?
     
  8. Mar 16, 2016 #7

    haruspex

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    You have sin A = 0.42, yes? Do you not have an inverse sine function of your calculator, spreadsheet, or whatever you use?
    (But I feel you should keep more significant digits than 2 or your final answer will be rather inaccurate.)
     
  9. Mar 16, 2016 #8
    So i used the inverse sine function on my calculator , and i got 24.8 degrees
    so would the direction of the force be West 24.8 derees North?
     
  10. Mar 16, 2016 #9

    haruspex

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    Look at your diagram. Which angle did you label A?
     
  11. Mar 17, 2016 #10
    wait so did i calculate the right degrees (24.8)
    im confused on how to figure out the direction, just by looking at angle a in the diagram i still cannot seem to figure out the direction,
    is it East 24.8 degrees North?
     
  12. Mar 17, 2016 #11

    haruspex

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    Take your triangle diagram in post #1 and add a line to the East from the left-hand corner. There are three angles you know in that diagram. What other angles can you deduce?
     
  13. Mar 31, 2016 #12
    i also re-did this question, can someone tell me if i am correct?
    40.


    c2 = a2 + b2 – 2abcosC

    c = (a2 + b2 – 2abcosC) ½
    upload_2016-3-31_16-40-28.png

    32 degrees + 24 degrees = 56 degrees

    c = ([12 N]2 + [15 N]2 - 2 [12 N] x [15 N] cos124°) ½

    c = 23.9 N

    SinA/a = SinB/b = SinC/c

    SinA/15 N = Sin124°/23.9 N

    SinA = (15 N) Sin124° / 23.9 N

    A = sin^-1 (0.52)

    A = 31.3 degrees

    Fnet = 23.9 N [North 31.3°East]
     
  14. Mar 31, 2016 #13

    haruspex

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    Ok until the last line. You calculated the angle A (theta in the diagram). That is not the angle between the resultant and North.
     
  15. Mar 31, 2016 #14
    so what angle in the diagram is 31.3 supposed to represent? which angle did i calculate, i need to figure out the angle between the resultant and the 12 N vector (theta) right?
     
  16. Mar 31, 2016 #15

    haruspex

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    You did calculate theta, but when you say the resultant is N x degrees E you mean x is the angle between the resultant and N. In your diagram, you have a dashed line going N. What is the angle between that and the resultant?
     
  17. Mar 31, 2016 #16
    is it 89 degrees?
     
  18. Mar 31, 2016 #17

    haruspex

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    Yes. (Well, 89.3, using the 31.3 you calculated.)
     
  19. Mar 31, 2016 #18
    i got that by doing the following
    32 - 31.3 = 0.7
    90 - 0.7 = 89.3

    but 89.3 is the angle between the resultant and the dashed line going north, what does that have to do with the angle between the resultant and the 12 N vector (theta)?
     
  20. Mar 31, 2016 #19

    haruspex

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    I don't understand your question. You just wrote out the relationship between the two in your own post: angle between resultant and N is angle between E and N (90) - angle between 12N force and E (32) + angle between resultant and 12N force (theta).
     
  21. Mar 31, 2016 #20
    I just dont understand what angle im supposed to use in my final answer, if 89.3 is the angle between the north line and the resultant, then 31.3 is the angle between the 12 N vector and the resultant, meaning Fnet is still:
    23.9 N [North 31.3°East]
    am i wrong?
     
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