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Net force using parallelograms

  1. Jun 20, 2014 #1
    1. The problem statement, all variables and given/known data
    There are only 3 forces acting on an object. The forces are 20.0 dynes E15°, 30.0 dynes W15°, and 40.0 dynes S15°W


    2. Relevant equation
    I've tried using the parallelogram method


    3. The attempt at a solution
    When i did the parallelogram method i got 30^2+20^2-2*30*20*cos(115) because 15° for the drawn parallelogram 15+15=30 for 2 of the 4 total angles in every corner, then 360-30=330/2=165 for the two remaining angles that will let the Resultant be 50 dynes. After that i tried doing the same thing for the parallelogram figure but i just couldn't figure it out. I've tried looking for different equations but it wouldn't budge or maybe i'm just not seeing it simple?
     
  2. jcsd
  3. Jun 20, 2014 #2

    SteamKing

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    Your manner of representing vectors is not clear.

    Take the first vector: 20.0 dynes E15° Is this supposed to mean 15° E of north? Or what?

    Also, the parallelogram method is best used when trying to find the resultant of two vectors graphically. Using the P-method to find a resultant by calculation is very complex and prone to error, and you probably have to make a sketch to keep things straight.

    If you are trying to find the resultant of three or more vectors by calculation, it's easier to find the components of each vector, add them together, and then convert the components of the resultant back into the vector form of your choice.
     
  4. Jun 21, 2014 #3

    haruspex

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    How did you get 115?
     
  5. Jun 21, 2014 #4
    Maybe you can upload a diagram or restate your the problem more clearly. Because like SteamKing said, its not possible to understand how your vectors are oriented from the way you described them.
     
  6. Jun 21, 2014 #5
    Yes i'm sorry, it's 30 dynes North west, 20 dynes North east, and 40 dynes South west, in the problem in my book, they used the parallelogram method in order to solve it, connecting the 30 and 20 dynes together as a parallelogram and then connected it as a parallelogram (the 40 dynes) back to the point of 0 which it's where 30 dynes north west and 20 dynes north east. I've only been doing physics for a week so my understanding of equations is very dim, i'm sorry.
     
  7. Jun 24, 2014 #6
    Net force problem

    1. The problem statement, all variables and given/known data
    The following were given, 30 Dynes 15° North west, 20 Dynes 15° North East, and 40 Dynes 15° South west. Fine the Net force


    2. Relevant equations
    Parallelogram Method, Law of cosines


    3. The attempt at a solution
    You have to find the force between 30 dynes and 20 dynes. So i used parallelogram method to add 15+15=30 degrees. I inserted that in The law of cosines formula c^2=a^2+b^2=2*a*b*cos(30°) and i got 16.4, you do the same for the other side and get 32.8 in total. I followed to draw vectors connecting them and i used, again, The law of cosines. Plugging in 40^2+32.8^2-2*40*32.8*cos(48°) so i added the product 30.33+40 dynes South west and got 70.33 dynes and subtracted 32.8 dynes that is the force that's applying north. This is the part where i got lost. I guesstimated it was 50 Degrees because the way the angle applied and was drawn as it could be divided by 3 (using parallelogram method 15+15=30*2=60, 360-60=300/2=150 divided by 3 because of how the vectors are aligned) End point is that i just played with the degrees so that it could make me obtain 37 dynes South west. This problem was off a book and it didn't offer any equations to solve it, Just a picture of it and the result. Which was 37 dynes South west. I took 2 hours to solve this problem and i really wished for an answer or help. Please
     
  8. Jun 25, 2014 #7
    I don't think your description of the problem is same as the book's.
    From your problem statement

    the diagram will be as below:

    attachment.php?attachmentid=70856&stc=1&d=1403680964.png

    Is this the diagram given in your question?

    If so then the correct answer should be about 48 dyne. And the direction is 87 degree West of North.
    But as these never agree with your book's answer, I guess your question doesn't describe the books diagram properly.

    You should upload a diagram describing your problem.
     

    Attached Files:

    Last edited: Jun 25, 2014
  9. Jun 25, 2014 #8

    NascentOxygen

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    Hi Sifuejos. Mathematics as a narrative is not a gripping read. I think you should scan your neat maths solution and attach it here.

    I get your 16.148 magnitude when 2 are summed, but you haven't indicated the resultant angle. Then adding the 40 I seem to get a different result.
     
  10. Jun 25, 2014 #9
    Problem in the book

    It's in the attachments
     

    Attached Files:

  11. Jun 25, 2014 #10
    F1+F2 is equal to the resultant. I use the resultant, 32.8 and 40 into the law of cosines equation to get F1+F2+F3 which then i follow to add with 40 and then subtract the original North force 32.8 from the south force that would be 70.33 and is equal to 37.53 (The answer that is given in the book)
     
  12. Jun 25, 2014 #11
    Is it the book that stated that problem incorrectly? I still don't know how or why they got that answer
     
  13. Jun 25, 2014 #12
    Am I missing something here? How did the angle between [itex]\overline{F3}[/itex] and [itex]\overline{F1}[/itex]+[itex]\overline{F2}[/itex]+[itex]\overline{F3}[/itex] became 15 degree?
     
  14. Jun 25, 2014 #13
    I'm guessing that's the angle between F1+F2+F3 and F3, That's exactly how the problem was stated and drawn as. I really don't know what's going on
     
  15. Jun 25, 2014 #14
    :confused:
    You know what... If I were you I would leave the Parallelogram Method and the Law of cosines.
    Just find what angles F1, F2 and F3 make with X and Y axises and find their x components and y components. Then finding net force and direction is easy.

    And if this problem is making you too mad, why not just leave it for now. Ha? :devil:
     
  16. Jun 25, 2014 #15
    I guess..I'll leave it for my physics teacher to solve next year! Thank you for your help though ^-^
     
  17. Jun 25, 2014 #16
    Hey I said so because I think the problem statement, diagram, textbook answer all were creating a confusing situation.
    Never mind, did you? I was just trying to be funny. :tongue2:
     
  18. Jun 25, 2014 #17
    If there was a formula for this situation then it'll be great. I tried searching up problems like these but they're college problems and really confusing stuff. I'm not even in a physics class yet
     
  19. Jun 25, 2014 #18
    I guess you know how to do this ? Or not yet?
     
  20. Jun 25, 2014 #19
    Actually no...
     
  21. Jun 25, 2014 #20
    Okey, you are given three forces and their directions (by angles) right?
    Now look at the way I tried to draw your problem in a diagram. Are my angles correct?

    attachment.php?attachmentid=70856&stc=1&d=1403680964.png

    For F1 and F2 it seems that your diagram agrees with mine.

    But for F3, can you tell me, what's the degree of this red angle OR what's the degree of this blue angle?

    attachment.php?attachmentid=70888&stc=1&d=1403736448.png
     

    Attached Files:

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