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Distance and direction of one leg of a sailboat sail.

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A sailboat sails 2km east, then 5km at 40° west of north, then sails off in a third direction. After the third leg of the trip, the sailboat returns to its starting point by sailing 15 km at 60° west of north. How far and in what direction did the sailboat sail during the third leg of its trip?

    Leg 1: 2km to the right (east) 0°
    Leg 2: 5km up and to the left (northwest) 40°
    Leg 3: unknown
    Leg 4: 15km up and to the left (northwest) 60°

    Since I was told a direction, I assumed the kilometers would be the force in these equations. (This could be wrong, but I didn't know what else to try.)

    2. Relevant equations
    I thought I could use [itex]\Sigma[/itex]F=0
    F1 + F2 +F3 + F4 = 0
    F3 = -F1 - F2 - F4

    then I could use pythagoreans theorem and arc tangent to solve for magnitude and direction.


    3. The attempt at a solution
    F1 + F2 +F3 + F4 = 0
    F3x = -2cos0 - 5cos40 - 15cos60
    F3x = -13.33


    F1 + F2 +F3 + F4 = 0
    F3y = -2sin0 - 5sin40 - 15sin60
    F3y = 9.77

    From here you would draw two arrows on a set of coordinate axis then use F3x and F3y to find F3. But the answer I come out to isn't equalling the answer in the book. I just have no idea what I'm doing wrong or how to go about this question correctly. Any help would be very much appreciated.
     
  2. jcsd
  3. Nov 15, 2012 #2

    haruspex

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    It isn't about forces, but it is about vectors, so although it's quite wrong to say these are forces it should give the right answer.
    It would help if you were to state your choice of co-ordinate system. I.e. is the +x direction N, S, E or W? Whichever, I think your equation above takes an inconsistent view.
     
  4. Nov 15, 2012 #3
    Ok, so a vector has both a magnitude and direction. I'm given direction of each of the vectors. In the book I'm working with states that an equation I could use is:

    Vectorx = (vector magnitude)cos[itex]\Theta[/itex]
    Vectory = (vector magnitude)sin[itex]\Theta[/itex]

    This leads me to the idea (if i set up the axis like the compass, i.e. north is +y, east is +x) that I can do the distance(cos[itex]\Theta[/itex]) So I would count the angle from the +x axis to the vector:

    V1 2cos0 = 2
    V2 5cos130 = -3.21
    V4 15cos300 = 7.5

    So V1 + V2 + V4 = V3 ?

    But when I follow this, I don't get the correct distance.
     
  5. Nov 15, 2012 #4

    haruspex

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    A few problems there.
    First, I don't think the 300 is right - check that.
    Second, need to be careful with the signs. It's probably less confusing to write V1 + V2 + V3 + V4 = 0 and use a different angle in the V4 equation.
    Third, what is the equation for V3? There is an angle involved again.
    Fourth, you will need an equation for the NS direction too.
     
  6. Nov 16, 2012 #5
    For the 300, it is listed that the boat went 15km in the SE direction..if it didn't then the 3rd direction wouldn't be correct. So since it was listed at 60°, counting from +x to that point would be 300, unless I'm missing some way in figuring out that angle.

    So if I have V1 + V2 + V3 + V4 = 0, in what way am I supposed to calculate the equation for V3 if I'm not given any information about it? Unless I am assuming it is 30°; 90° (SE corner) - 60° (listed from leg 4).

    How would I figure out an equation for the NS direction, if nothing is specifically given on this? I feel like I am just missing a piece of information.
     
  7. Nov 16, 2012 #6

    haruspex

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    No, it says 60o W of N. From the x+ axis that's 150o.
    By rearranging as V3 = -(V1 + V2 + V4)
    The above is a vector equation, so you can use it to produce two scalar equations, one NS and one EW, with sines and cosines.
     
  8. Nov 17, 2012 #7
     
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