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Solving vectors using component method 
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#1
Jul511, 07:22 AM

P: 2

1. The problem statement, all variables and given/known data
Add the following vectors using component method: Vector A=175km; 30 degrees north of east Vector B=153km; 20 degrees west of north Vector C=195km; West a)Find x and y component of each vector b)Solve for the magnitude of resultant vector c) Solve for direction(angle) 2. The attempt at a solution So i these are what i got Vector A: x=175cos 30....x=151.56 y=175sin 30......y=87.5 Vector B: x=153cos 20....x=143.77 y=153sin 20.....y=52.33 (are these correct?) so after getting this i added all the x's and y's but then i realized that the question had said that vector C goes 195km west, so i now got confused what to do next. please help thank you for your time. 


#2
Jul511, 07:49 AM

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P: 5,365

There is nothing special about "in a westerly direction". A westerly vector contributes a component D.cos 180^{o} in the xdirection, and D.sin 0^{o} in the ydirection.



#3
Jul511, 07:54 AM

P: 2

well that was stupid of me, thanks for the reply finally solved it :D



#4
Jul511, 07:58 AM

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P: 39,568

Solving vectors using component method
If [itex]\theta[/itex] is measured counterclockwise from the positive xaxis then a vector of length r and angle [itex]\theta[/itex] has components [itex]rcos(\theta)[/itex] and [itex]r sin(\theta)[/itex]. Strictly speaking, you are free to choose the "positive xaxis" any way you want as long as you are consistent but the usual convention is that the positive xaxis points East.
For the first vector you are given that r= 175km and the directon is 30 degrees north of east. "north of east" is counterclockwise from east so the angle is [itex]\theta= 30[/itex] degrees. For the second vector you are given that r= 153km and the direction is 20 degrees west of north. West is clockwise of north but north itself is 90 degrees clockwise of east. The angle is [itex]\theta= 90+20= 110[/itex] degrees. For the third vector you are given that r= 195km and the direction is west. West is exactly opposite east so the angle is [itex]\theta= 180[/itex] degrees. 


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