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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.

Any advice would be appreciated.

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  • $\begingroup$ the angled $90^\circ-60^\circ =30^\circ$ should be used instead of $60^\circ$ $\endgroup$
    – 5201314
    Jan 18 '21 at 12:34
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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 $F_{1}$: $$R = (3+4.33)\hat{i} + (2+2.5)\hat{j} = 7.33\hat{i} + 4.5\hat{j}$$

Because of this ambiguity, in this kind of problems, besides the angle, it is usually indicated the direction too (in our example, north of east vs. east of north), so the condition is not really accurate.

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