Two forces 80 N and 100 N acting at an angle of 60° to each other pull on an object. What single force would have the same effect? (Answer: 156.2 N at an angle of 26.3° to the 100 N force (angle of 33.7° to the 80 N force) Really not sure how they got this answer...
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 [itex]\theta = \arctan{(\frac{y}{x})}[/itex] to get [itex]\theta[/itex] for the resultant vector and the magnitude ('M') is just [itex]x^2 + y^2 = M^2[/itex].
Firstly, you get to make an arbitrary choice of where the 'x' and 'y' axes point. This is because vectors are physical objects that are independent of the choice of coordinate systems used to represent them (indeed this is the entire reason why we use vectors in physics instead of just coordinate systems points to represent everything.) I would just say that the 80N force is along the positive 'x' axis. If we measure all angles from the positive 'x' axis going counterclockwise (as is per usual), then the 100N force is 30° from the 'y' axis in the first quadrant. Thus in this situation your 80N force has coordinates <80, 0> and your 100N force has coordinates <100cos60°, 100sin60°>. You can then add them to get the net force which is a new force that is perfectly equivalent to the original 2 forces.
You can also use a formula to add two vectors together (which comes from the law of cosines) which is [itex]V_R^2 = V_1^2 + V_2^2 + 2V_1V_2\cos{\theta}[/itex], where θ is the angle between the two vectors. Just plug in your data in [itex]V_1, V_2[/itex] and [itex] \theta[/itex] 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.
Thanks for the replies - I managed to get the magnitude but I tried plugging in the x and y values of the 100N force into the formula for getting the direction and I didn't get the correct answer so i suppose I was putting in the wrong values?
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 [itex] \theta = \arctan{\frac{86.6}{130}}=33.7° [/itex]. Is that clear? :)