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Homework Help: How to find magnitude of two forces

  1. Feb 15, 2015 #1
    Two forces are given and are applied to a car in an effort to accelerate it: 414N at 9 degrees and 340N at 25 degrees. What is magnitude of the resultant of the two forces? Answer in units of N.

    So I've added together X= (414cos9) + (340cos25), and got -40.
    For the Y, Y= (414sin9) + (340sin25)=125.
    I took square root of (-40)^2 + (125)^2, and got 131.24.

    However, that is not the correct answer. What am I doing wrong? Help please?!
     
  2. jcsd
  3. Feb 15, 2015 #2

    TSny

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    Note the angles are in degrees, not radians.
     
  4. Feb 15, 2015 #3

    SteamKing

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    You've added two positive quantities together and obtained a negative result. How? o0)

     
  5. Feb 15, 2015 #4

    phinds

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    Well you are computing the right values to add together for X and also for Y but the answers you get for those computation are just weird.
     
  6. Feb 15, 2015 #5
    Steamking, 414 x cos 9 degrees = - 377.2.
     
  7. Feb 15, 2015 #6

    TSny

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    Check your calculator to make sure it is in degree mode.
     
  8. Feb 15, 2015 #7
    So what you are saying Tsny is that I should convert the answers to radians?
     
  9. Feb 15, 2015 #8

    TSny

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    Keep the angles in degrees and just switch the angle mode of your calculator to degrees instead of radians. Your calculator is assuming that your angles are in radians.
     
  10. Feb 15, 2015 #9
    I've switched the mode to degrees, but still am getting the wrong answer. :(
     
  11. Feb 15, 2015 #10

    TSny

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    what does your calculator now give you for cosine of 9 degrees?
     
  12. Feb 15, 2015 #11
    .9876883
     
  13. Feb 15, 2015 #12

    TSny

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    OK good. Was a diagram included in the problem? Are the two angles measured from the x axis? Are both angle above the x axis, or is one angle above the axis and the other angle below the axis?
     
  14. Feb 15, 2015 #13
    340 N @ 25 degrees is below x-axis. 414 N @ 9 degrees I above x-axis.
     
  15. Feb 15, 2015 #14

    TSny

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    Are the Y components of both forces positive? If not, which force has a negative Y component?
     
  16. Feb 15, 2015 #15
    Yes
     
  17. Feb 15, 2015 #16

    TSny

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    The 340 N points below the x axis. Draw this force and draw its Y component. Does the Y component point up or down?
     
  18. Feb 15, 2015 #17
    I
    It is pointing up
     
  19. Feb 15, 2015 #18

    TSny

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  20. Feb 15, 2015 #19
    Ye
    I think once I get a good look at this tomorrow, it'll make better sense. As far as I can see, the vector does move in a negative direction towards F2.
     
  21. Feb 15, 2015 #20
    Thanks for all your help TSny!
     
  22. Feb 15, 2015 #21

    TSny

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    OK, good luck with it. I'll just leave you with this.

    In the picture below, the red vector has both positive X and Y components. (If you walked from the tail to the head of the vector you would be going in the positive X and positive Y direction.) But the blue vector has a positive X component and a negative Y component. (If you walked from the tail to the head of this vector you would be going in the positive X but in the negative Y direction.)
     

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  23. Feb 15, 2015 #22
    Yes, thank you for this! I had actually gone off of a sketch I had, and it was not accurate. After you had mentioned so, I looked back at the hw and it's sketched the way you have it. Thanks for all your explanations!
     
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