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I am sure this is easy but i am stuck!

  1. Jan 22, 2009 #1
    The question: A grasshopper makes four jumps. The displacement vectors are (1) 27.0 cm due west (2) 23.0 cm, 35.0[tex]^{o}[/tex] due south of west (3) 28.0 cm, 55.0[tex]^{o}[/tex] south of east and (4) 35.0 cm, 63.0 [tex]^{o}[/tex] north of east. Find the magnitude and direction of the resultant displacement. Express the direction with respect to due west.

    Vector magnitude direction x component y component
    (1) 27.0 cm 0 [tex]^{o}[/tex] 27 cos 0[tex]^{o}[/tex] 27 sin 0[tex]^{o}[/tex]
    (2) 23.0 cm 35.0 [tex]^{o}[/tex] 23 cos 35[tex]^{o}[/tex] 23 sin 35[tex]^{o}[/tex]
    (3) 28.0 cm 55.0 [tex]^{o}[/tex] 28 cos 55[tex]^{o}[/tex] 28 sin 55[tex]^{o}[/tex]
    (4) 35.0 cm 63.0 [tex]^{o}[/tex] 35 cos 63[tex]^{o}[/tex] 35 sin 63[tex]^{o}[/tex]

    r= resultant variable

    I add all of the x components and I get 62 cm.
    I add all of the y components and I get 49.5 cm.

    r=[tex]\sqrt{x^{2}+ y^{2}}[/tex]
    r=111.5 cm

    the direction i get 38.6[tex]^{o}[/tex].

    Thats not the right answer! Please help!!!
     
  2. jcsd
  3. Jan 22, 2009 #2

    LowlyPion

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    Welcome to PF.

    I'm not sure you have properly accounted for the signs of your magnitudes. For instance with positive x East then your west vector would be negative.
     
  4. Jan 22, 2009 #3

    I am still not getting the right answer with this! I am really confused now!
     
  5. Jan 22, 2009 #4
    I haven't done vector addition in a while but here it goes!

    I use Dx to label the vectors corresponding to the jumps in sequence. I set up my unit so that west is -x and south is -y which is standard I suppose.

    D1 = -27cm (negative because it's going west)

    D2x = -13.19cm (you get this by sin(35)*23cm, keep in mind that this is a negative because it's moving westward in the x direction.)
    D2y = -18.84cm (Same process of above except cos(35)*23cm)

    Now that you got the hang of things I'm gonna run down to my solution.

    D3x = 16.06cm
    D3y = -22.93cm

    D4x = 15.89cm
    D4y = 31.18cm

    Sum of X's = 45.76cm
    Sum of Y's = -10.59cm

    Make a triangle using those two components and your hypotenuse is the composite vector. 46.97cm 13.03o S of E

    Is that correct?
     
  6. Jan 22, 2009 #5
    I dont think so because it asks me to express the direction with respect to due west.
     
    Last edited: Jan 22, 2009
  7. Jan 22, 2009 #6
    Well that is actually really simple. 13.03o S of E can be converted to W by simple subtraction. 180o - 13.03o = 166.97o
    That is to say the new direction is 166.97o S of W.

    I hope this helps.
     
  8. Jan 22, 2009 #7
    Thank You very much but I am still confused
     
  9. Jan 22, 2009 #8
    Solving vectors for me is basically 3 steps.
    1. Setup two equations. One that adds together all the x components and one that adds together all the y components. For you problem my equations looked like this:

    Xtotal = D1 + D2x + D3x + D4x
    Ytotal = D2y + D3y + D4y

    I use the subscripts to number each vector, and the x/y's to label the x or y component. Notice that since D1 is completely due west it is only present in the x equation because it has no y component.

    2. Solve for the components. Use trigonometry to draw out each vector and break it into components. Be mindful of the direction of the vector. If it's moving downwards it's y component will have a negative. If it's moving to the right it's x component will have a negative as well.

    3. Plug and chug. Plug all your x and y values for D1...D4 into your two equations from step one. This gives you the total displacement in the x and y directions. Using these values (and watching the signs) you can draw a traingle which becomes your composite vector. All you need now are the Pythagorean theorem and an inverse sine or cosine to find the direction.

    Hope this clears up my process.
     
  10. Jan 22, 2009 #9

    LowlyPion

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    Sorry. I think you got your x,y and sin,cos reversed.

    Otherwise it looks ok ... except for the result.
     
  11. Jan 22, 2009 #10
    You're right I did flip those.
    For my new sums I get 23.79 for x and -4.94 for y.
    24.29cm @ 11.64o S of E or 168.36o S of W

    I think that's better.
     
  12. Jan 22, 2009 #11

    LowlyPion

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    I didn't run the numbers.

    The OP is encouraged to verify to be certain of the proper magnitude.
     
  13. Jan 23, 2009 #12

    nvn

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    mmacholl: That still doesn't look quite right yet. I currently got R = 14.744 cm at 19.589 deg S of W.
     
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