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Vectors in a box

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data

    3 Forces/vectors in a box going from a corner.

    F1=260N F2=75N F3=60N

    then it says calculate F0 which is F0=F1+F2+F3
    problem is that the answer is not the sum of the vectors its 347.3N

    its also wants the angles of the vectors to the corner

    problem is no cordinates are given for any of the vectors. and theres a picture of the box with measurements but does not specify the cordinates of any of the vectors. they all seem to be half way almost.


    2. Relevant equations

    is it possible to solve, dont you need the cordinates?

    picture:

    Vbox.jpg


    3. The attempt at a solution

    non
     
    Last edited: Apr 5, 2013
  2. jcsd
  3. Apr 5, 2013 #2
    The way you have it drawn, the *angle of the* vector on the right hand side would have to be ##θ=tan^{-1}{\frac{3}{4}}##. However, that is assuming that your interpertation of the vector splitting the side down the center is accurate.
     
    Last edited: Apr 5, 2013
  4. Apr 5, 2013 #3

    SammyS

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    It would help us help you, if you would give us the exact wording of the problem as it was given to you.

    Your figure does not label which of the red vectors is F1, which is F2, or which is F3 .

    One of those vectors is "anti-parallel" to the x-axis, i.e. it's in the negative x direction.

    Another is parallel to a vector going from corner A to the opposite corner, corner B.

    The remaining vector is parallel to a vector from corner A on the right face of the block, to the opposite corner of that face.

    From the dimensions of the box, you should be able to construct a unit vector in the direction each of the three force vectors.
     
    Last edited by a moderator: May 6, 2017
  5. Apr 5, 2013 #4
    ya sorry changed it. this is the exact wording. no cordinates are given. and it wants exact answers it has given me the F0 and the angles precisely they are 107.6 149.7 and 66.2
     
  6. Apr 5, 2013 #5

    SteamKing

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    Hello! If you have the dimensions of the box, you can find the coordinates by setting up your own coordinate system.
     
  7. Apr 5, 2013 #6

    SammyS

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    There are x, y, and z axes drawn in the box. As I look at this perspective drawing, it looks to me like it's drawn so that the coordinate system is drawn inside the box, with the origin being in the lower left-hand corner to the rear of the box. If that is the intent, then this is a left-handed coordinate system. -- That's very strange !

    As SteamKing and I hinted at earlier, what is the displacement vector from corner A to corner B ? F1 is parallel to this vector.


    By the way, this problem is very straight forward -- except for that left-handed system.
     
  8. Apr 5, 2013 #7
    care to explain or give tips? i mean the cordinates of the vectors F1 F2 F3 are not give only their magnitudes.
     
  9. Apr 5, 2013 #8

    SammyS

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    Start by answering this question:

    What is the displacement vector from corner A to corner B ?
     
  10. Apr 5, 2013 #9
    its not given and no cordinates are given thats the point of the assignment did you read my initial post???
     
  11. Apr 5, 2013 #10

    SammyS

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    In the figure that you posted in your OP, I see a box that's 12 meters wide, 4 meters high and 3 meters deep. I also has a coordinate system sketched in a corner.
     
    Last edited: Apr 5, 2013
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