Cross product

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Homework Statement


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

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  • #2
haruspex
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1.) How to get 0.7379? i only managed to get (6j + k ) / (3 √10 )
It's from a dot product, so I don't know how you ended up with a vector. Please post all your steps.
2.) For Fba , which is equal to F.Uba , why cant i do in (80 ) ( (-2i -2j +k)/3 ) as above (80)(rBC / rBC) ?
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?
 
  • #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 ?
--
 
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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?
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 ?
 
  • #5
haruspex
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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 ?
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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