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Math Help - Statics/Physics question.

  1. #1
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    Statics/Physics question.

    I know this is technically not a trig question, but it involves trig and I'm sure there are a lot of engineering majors/graduates here, so here goes. I'm in a statics class (this is a question from my physics class) but we haven't actually talked about this specific type of problem yet.

    The problem says:

    "A sailor in a small sailboat encounters shifting winds. She sails 2.00km east, then 3.50km southeast, and then an additional distance in an unknown direction. Her final position is 5.80km directly east of her starting point. Find the magnitude and direction of the third leg of her journey."

    So it looks like I've given 3 of the 4 parts of a vector addition equation that involves three vectors and a resultant. The resultant and two vectors are given, and you have to find the third. I know how to add vectors (graphically, with trig, and using x and y components) but I'm not sure about how to do this one.

    EDIT: Should I break each known vector down into it's components and substitute a variable in for the unknown components of the third vector and from there solve for the unknown vectors components? Basically working backward to find the vector?
    Last edited by JonDick13926; January 28th 2014 at 10:37 AM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Statics/Physics question.

    You're on the right track. If we call the first leg A, the second B, the third C, and the final result D, you have:

     \vec A + \vec B + \vec C = \vec D

    So to find vector C just rearange:

     \vec C = \vec D - \vec A - \vec B

    can you take it from here?
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  3. #3
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    Re: Statics/Physics question.

    Thanks. I think I got it. I broke each known vector into components. I subtracted the x-component of the two known vectors from the x-component of the resultant and did the same for the y-components and ended up with an x- and y-component of the third vector, which led to a vector of approx 2.81km at 62 degrees in standard position. I don't have a way of checking this since it's an even number problem from the book, but I double checked and got the same thing again.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Statics/Physics question.

    Looks good, except be careful with how you specify direction. Your angle of 62 degrees is measured from east, but bearings are typically measured from north (or south). In this case the angle is to the east of due north, and is specified as "north 28 degrees east."
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