What Does 860g @ 50 Degrees North of East Look Like on a Graph?

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The discussion focuses on visualizing a vector of 860 grams at an angle of 50 degrees north of east on a graph. The term "north of east" indicates a counter-clockwise angle from the positive x-axis, which is standard in polar coordinates. Participants clarified the distinction between polar and navigation conventions, emphasizing the importance of understanding vector components. The correct calculations for the north and east components of the vector were confirmed as approximately 553 grams east and 659 grams north.

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can anyone show me how 860g @ 50 degrees North of East would look like on a graph?

I don't understand what "north of east" means.

Thanks a lot
 
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On a map or graph, east would be a vector pointed to the right. 50 degrees "north of east" would be a vector that points 50 degrees counter-clockwise from the right. This is the convention used for polar coordinates, 0 degrees pointing to the right, along the "X" axis, and positive angles associated with counter clockwise angles. East is 0, north is 90, west is 180, south is 270.

Navigation uses a different convention. 0 degrees means to the north ("Y" axis), and positive angles are associated with clockwise angles. North is 0, east is 90, south is 180, west is 270 (degrees). Airport runways are normally named after the direction divided by 10. An east / west runway would be called runway 90 if approached from the west heading east, and runway 27 if approached from the east headed west.
 
Last edited:
.NoStyle,

Welcome to Physics Forums.

The phrase "north of east" is actually very nice because it is much more specific than just "northeast". For example, if the problem stated "50 degrees in the northeast direction" then it would leave you to wonder whether that means 50 degrees from the positive y-axis or 50 degrees from the positive x axis. Since you are given "north of east", you aren't left wondering. The normal direction is east. The angle (50 degrees) sends you more north, so the angle must be made with respect to the positive x axis.

See the picture enclosed. I hope this answers your question.

David
 

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thanks a lot guys, that is what I figured, but I solved the problem accordingly and I came out with an incorrect answer. I'm going to try it out again, and see what I get. Thanks for the help guys, I appreciate it.
 
Post the problem if you don't come out with the right answer again...
 
Daveyman,

the question asks:

6. I have vector A = 860 grams at 50 degrees north of east; what is A's component in the North direction?


to which I've used the SinTheta=O/H to find that "A.sub.y" or NORTH = 226

then the next question asks:

7. I have vector A = 860 grams at 50 degrees north of east; what is A's component in the East Direction?


I used the Pythagorean Theorem to find A.sub.x or EAST = 830. Thanks
 
Last edited:
seems like I did something wrong. For the east, I should get 553...

and for north, I would get, 659.
 
ok guys, got it right now. Thanks for the help.
 
I'm glad you got it.
 

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