# Resolving Forces With Vectors. Find the Resultant Force

In this system, a particle on a horizontal surface is acted on by two forces, $$F_1$$ and $$F_2$$. Find the $$\hat i$$ and $$\hat j$$ components of the resultant force, where $$\hat j$$ represents due north and $$\hat i$$ due east. $$F_1 = 3\hat i + 2\hat j$$, $$F_2$$ has a magnitude $$5$$ N and acts at a bearing of 60 degrees. I thought that I had to resolve $$F_2$$ into horizontal and vertical components $$5\cos(60^\circ)$$ and $$5\sin(60^\circ)$$ and then add these to the $$\hat i$$ and $$\hat j$$ parts of $$F_1$$, giving me $$5.5\hat i$$ and $$6.33\hat j$$ for the resultant force, but the mark scheme says that the answer is $$7.33\hat i + 4.5\hat j$$. I don't know how to get to these, and I don't think I've made any stupid mistakes, but if I have, please let me know.

• the angled $90^\circ-60^\circ =30^\circ$ should be used instead of $60^\circ$ Jan 18 '21 at 12:34
The possible way of getting the answer you have in the key is considering that $$F_{2}$$ "acts at a bearing of 60 degrees" means that this angle is not formed with the positive direction of x-axis, but with the one of the y-axis. Equivalently, we can say that the direction of the applied force is not north of east relative to the coordinate system, but the east of north. Then, The horizontal component of $$F_{2}$$ is $$5sin(60^\circ) = 5*1.73/2 = 4.33$$, and the vertical one is $$5cos(60^\circ) = 5*0.5 = 2.5$$. Then, adding these components to the corresponding ones of $$F_{1}$$: $$R = (3+4.33)\hat{i} + (2+2.5)\hat{j} = 7.33\hat{i} + 4.5\hat{j}$$