What is the magnitude and direction of the second force vector in this scenario?

  • Thread starter LuvIz4ever
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In summary, the conversation discusses a physics problem involving two force vectors, one pointing due east and the other unknown. The resultant of the two vectors is 400 Newtons and points along the east-west line. The answer in the book states that the unknown vector is 200 N of East and 600 N of west, but this does not make sense because it would involve a north-south component. The expert confirms that the answer in the book is most likely incorrect and explains the correct solution. The correct answer is that the unknown vector is 200 N of east and 600 N of west, with no north-south component.
  • #1
LuvIz4ever
6
0
I can't figur this out!

Now i got the answer to this problem but i am so confused at this one thing... how is it that the answer is 200 N of East and 600 N of west


the question is

A force vector F1 points due east with a magnitude of 200 Newtons and a second force F2 is added to F1. The resultant of the two vectors has a magnitude of 400 Newtons and points along the East and west line. Find the magnitude and direction of F2 note that there are 2 answers.


My question is I know the answers 2oo east and 6oo west but why north of east or north of west.
 
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  • #2
LuvIz4ever said:
how is it that the answer is 200 N of East and 600 N of west
I suspect that the N stands for Newtons, not north.
 
  • #3
Thanks But...

Actually in the answer book it tells me it is North... I think it is weird though becuase of the fact that it has nothing about North.. in the question
 
  • #4
The answer in your book must be wrong. You have an initial vector pointing along the east-west line. If you add to it a vector with any component at all in the north-south direction, the resultant cannot possibly be entirely along the east-west line. I suspect the Doc is absolutely right.
 
  • #5
... I figured that... thanks! sooooo! Much!
 
  • #6
simply put, there is 200 going "right" and 600 going "left" add 200 + (-600) leads to a net of 400, and this 400 is along EW or as most would say x-axis..

yw
 

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