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Cross product

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data
    1.) How to get 0.7379? i only managed to get (6j + k ) / (3 √10 )
    2.) For Fba , which is equal to F.Uba , why cant i do in (80 ) ( (-2i -2j +k)/3 ) as above (80)(rBC / rBC) ?

    when i do in (80 ) ( (-2i -2j +k)/3 ) , i get back F=80N

    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 15, 2015 #2

    haruspex

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    It's from a dot product, so I don't know how you ended up with a vector. Please post all your steps.
    I don't understand. F is 80(rBC/rBC), not 80(rBA/rBA).
    But either way, |F|=80N. Perhaps you can explain your question better?
     
  4. Oct 15, 2015 #3

    BvU

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    You need a little help understanding vector products !

    if you know about an inner product of two vectors, you know that it is a number

    Pity you posted a picture, now I have to spend more precious time typing it out:
    vector inner product: ##\vec a \cdot \vec b \equiv |\vec a| |\vec b| \cos \theta##
    therefore $$ \cos \theta = {\vec a \cdot \vec b \over |\vec a| |\vec b|} $$
    It's all linear (check!) so if ##\vec a = a_x \hat\imath + a_y\hat \jmath ## and ##\vec b = b_x \hat\imath + b_y\hat\jmath ## then $$
    \vec a \cdot \vec b = ( a_x \hat\imath + a_y\hat \jmath) ( b_x \hat\imath + b_y\hat \jmath) =\\ \mathstrut \\
    \quad a_x b_x \hat\imath \cdot \hat\imath + a_x b_y \hat\imath \cdot\hat \jmath + a_y b_x \hat \jmath \cdot \hat\imath + a_y b_y\hat \jmath \cdot \hat \jmath = \\ \mathstrut \\
    \quad a_x b_x (1)(1)\cos 0 + a_x b_y (1)(1)\cos \textstyle {\pi\over 2} + a_y b_x (1)(1)\cos\textstyle {\pi\over 2} + a_y b_y (1)(1)\cos 0 = \\ \mathstrut \\ \quad a_x b_x + a_y b_y $$(I did it for 2D to save time)

    So your numerator is not a vector (6j + k ) but a number -2 * 0 + -2 * 3 + 1 * 1

    Does this make sense ?
    --
     
  5. Oct 15, 2015 #4
    the author used 80(rBC/rBC) to get the cartesian form of 80N .
    Now , i wanna find the magnitude of FBA , the author used the cartesian form of 80N to find the FBA ,
    If i use the (rBA/rBA) , i will get back 80 N , but not 59.0N . why cant I use this way ?
     
  6. Oct 15, 2015 #5

    haruspex

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    F is in the direction BC, F=FBC=80(rBC/rBC). But F is not the same as FBA, so FBA is not 80(rBA/rBA).
     
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