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Trouble finding the resultant using rectangular components

  1. Sep 26, 2015 #1
    Hello, I am having trouble finding the direction of the resultant of this statics homework problem. The picture uploaded is exactly how it is shown in my book. The problem states:

    1. The problem statement, all variables and given/known data
    For the forces shown, S = 3.81 kip, T = 4.73 kip, and U = 3.65 kip. Determine the resultant.
    statics_zpsunwfvgge.png

    The answers to this problem are 8.05 kip @ -169 degrees.

    2. Relevant equations
    Pythagorean Theorem (to find resultant force): C2 = A2 + B2
    x-component = cos(angle)*force
    y-component = sin(angle)*force
    magnitude of resultant = arctan(y-component / x-component)


    3. The attempt at a solution

    I have gotten the magnitude via trigonometry as such:

    x-component of S = -cos(33.6901 deg) * 3.81 = -3.17011
    x-component of T = -4.73
    x-component of U = 0
    ______________________________
    = -7.90011

    y-component of S = +sin(33.6901 deg) * 3.81 = 2.11341
    y-component of T = 0
    y-component of U = -3.65
    ______________________________
    = -1.53659

    I then used the Pythagorean Theorem to find the magnitude of the resultant:

    R2 = √(-7.90011)2 + (-1.53659)2 = 8.05 kip

    I then took the arctan of the y-component over the x-component, which equals +11.0067 deg which is the wrong direction...

    I am very lost as to why I am getting the wrong direction if I have gotten the correct magnitude. Any help?
     
  2. jcsd
  3. Sep 26, 2015 #2

    gneill

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    Hi dlacombe13, Welcome to Physics Forums!

    Note that the signs of both of your components are negative. In what quadrant does that place the resulting angle?

    When you provide an argument to the arctangent function (or atan) on your calculator it cannot distinguish a positive argument from one obtained by dividing a negative by a negative (which yields a positive). It's up to you to place the result in the right quadrant by adjusting the result returned.

    Some calculators offer an atan2(x,y) function which takes two arguments and can sort out the quadrant automatically. Alternatively, some calculators offer rectangular to polar vector conversion features which will accomplish the same thing.
     
  4. Sep 26, 2015 #3
    Wow! thank you so much, I now get that it is in the third quadrant, and I must subtract 11.0067 from 180 which will give me about 169 degrees. However, why is it negative 169? How do I determine whether my direction is negative when solving a problem like this?
     
  5. Sep 26, 2015 #4

    SteamKing

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    That's not how you get an angle in the third quadrant. If the angle is less than 180°, which quadrant is it in?
    You can take your angle reference from 0° or 360°.
     
  6. Sep 26, 2015 #5
    That would put it in the second quadrant...now I am confused. So my x-component and y-components are both negative, and (-,-) is in the third quadrant. All I know is that the direction as stated in the answers section of my book is -169 degrees. When I subtracted 11.0067 from 180 I got 168.993, but it is supposed to be negative. Could someone maybe walk me through exactly how I would go about getting the direction, as well as the sign of it.

    EDIT: Well I understand that if I measure an angle clockwise it would be negative, I'm just confused as how to get to -169 from the 11.0067 that I got when I took the arctan. I assumed I would subtract it from 180, but now I'm confused about where the negative comes from.
     
  7. Sep 26, 2015 #6

    SteamKing

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    First, since the x and y components of the resultant are both negative and indicate that it is located in the third quadrant, why are you subtracting the angle given by the arctan from 180° ? You're going in the wrong direction.

    If an angle is located in the third quadrant, what would the angle be if you measured the angle CCW from 0° ? What about if you measured to the same angle CW from 360° ?
     
  8. Sep 26, 2015 #7
    I think I'm getting it now. So I would add 11.0067 to 180 to get 191.007, then if I subtract 191.007 I would get 168.993 ≈ 169. Since I'm measuring it CW, it would be -169 right? If so, how would I even know that the angle should be negative not knowing the answer already? The reasoning for coming up with the number is making sense now, but not so much where the sign of the angle is being determined.
     
  9. Sep 26, 2015 #8

    SteamKing

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    It's often helpful if you draw the resultant vector on a piece of paper. Use the components to help you locate where
    It's often helpful if you draw the resultant vector on a piece of paper. Use the components to help you locate where the vector should be in the four quadrants. After some practice, you may not need to draw things out, but it is a valuable tool to check your work.
     
  10. Sep 26, 2015 #9
    Well I am clear it is in the third quadrant. My whole confusion is arising because just concerning the resultant as a vector on a plane, it's angle could be either ≈191 measuring from CCW, or ≈-169 measuring from CW, right? I am confused as to what is telling me that the angle should be the negative measurement vs. its positive measurement. I think I am having a hard time perhaps treating this as a force, since my mind is set to treating these as simply sides of triangles and such...
     
  11. Sep 26, 2015 #10

    SteamKing

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    Working with vectors aside, figuring out the angle is something which you'll also encounter in trigonometry.

    The inverse trig functions, like arcsine, arccosine, and arctangent, give what are called principal angles in the range -90° ≤ θ ≤ 90°. You still must figure out in which quadrant the principal angle lies, given the circumstances of a particular problem.

    The choice of whether to measure angles outside the first quadrant starting at 0° and going CCW or 360° going CW is really immaterial. The trig functions are normally shown derived for angles in the first quadrant, but with a little care, these functions can be extended to operate in all four quadrants.

    Vectors will be found in all four quadrants. You should not let this fact confuse their nature as forces or whatever.
     
  12. Sep 26, 2015 #11
    Right, but the problem is I don't understand why the answer is negative. I seriously don't get it. I drew a picture, and I can see that the resultant is in quadrant III. But I can't see how the answer is -169. I am reading online that I am supposed to add 180 to the answer I get from the arctan calculation, but that gives me 191. However, the book states the answer as -169 (which is the negative corresponding angle of 191). Could you please explain to me why they chose the negative angle VS. the positive angle?
     
  13. Sep 26, 2015 #12

    SteamKing

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    It could be as simple as the personal preference of who wrote the problem or the solution. Perhaps they wanted to keep |θ| ≤ 180°

    There is no quantitative difference, as far as the vector is concerned, if you express the angle as 191° or -169°. If you calculate sin (191°) or sin (-169°), you'll get the same result. Likewise with calculating the cosine of these angles.
     
  14. Sep 26, 2015 #13
    I thought so...I am sorry for all of the confusion. I mean it makes sense. A force may have a positive or negative magnitude, but whether the angle of the force measured is positive or its negative corresponding angle, it still is acting in the same direction. I wish the book made that clearer and showed that it doesn't matter. Thanks for all of your help, I really do appreciate it.
     
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