# Basic mechanics question

by DanB1993
Tags: basic, mechanics
 P: 111 Use vector addition. Break the two vectors you are dealing with into their 'x' and 'y' components and then add them together. Use the relation $\theta = \arctan{(\frac{y}{x})}$ to get $\theta$ for the resultant vector and the magnitude ('M') is just $x^2 + y^2 = M^2$.
 P: 22 You can also use a formula to add two vectors together (which comes from the law of cosines) which is $V_R^2 = V_1^2 + V_2^2 + 2V_1V_2\cos{\theta}$, where θ is the angle between the two vectors. Just plug in your data in $V_1, V_2$ and $\theta$ However, I strongly believe that decomposing the vectors in order to add them (as described in previous posts) is a much preferable way than memorizing a formula for nearly a thousand reasons.
 P: 22 Well, you had 80+100*cos(60) in one direction and 100*sin(60) in the other direction, right? That results in 130N for one component and 86.60N for the perpendicular one - using pithagoras you get the magnitude of the resultant vector (156.2N), which is probably what you did already. For the direction, notice that you should use the values of x and y components of the resultant vector (since that's the vector whose direction you are looking for). That should give you $\theta = \arctan{\frac{86.6}{130}}=33.7°$. Is that clear? :)