# Force in 2D

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• February 27th 2009, 12:42 AM
Koops
Force in 2D
http://img407.imageshack.us/img407/5993/maths.png
http://img407.imageshack.us/img407/m...ng/1/w1127.png
thats the question, but i dont know how to find the direction,
(–7.08i – 1.47j) is the compenants, and7.23N is the magnitude,
help me please :)
• February 27th 2009, 01:01 AM
Jameson
I didn't check your work so I'm assuming it's right. The resultant vector is in the third quadrant. Draw the two components so that you can make a right triangle. Now use the fact that tan(x) = (opposite)/(adjacent). Now compute the tan-inverse of this fraction and you'll get the angle. This is not the true angle, but the angle in respect to the quadrant so make sure to add 180 degrees onto it.
• February 27th 2009, 01:04 AM
ADARSH
Quote:

Originally Posted by Koops
http://img407.imageshack.us/img407/m...ng/1/w1127.png
thats the question, but i dont know how to find the direction,
(–7.08i – 1.47j) is the compenants, and7.23N is the magnitude,
help me please :)

I hope you know what is a component i for component along x direction and J in the direction of the y axis , After you have resolved the forces in to its components you just need to find the direction of two forces along x and y axis ( ie; its components)

$tan \theta = \frac{b sin\alpha}{a+bcos\alpha}$

Where alpha is the angle(90 here) between two vectors(components in your case ,ie: a and b) and theta the angle of resultant