Coplanar forces, of magnitudes $$\(F \mathrm{~N}, 3 \mathrm{~N}, 6 \mathrm{~N}\)$$ and $$\(4 \mathrm{~N}\)$$, act at a point $$\(P\)$$, as shown in the diagram. Given instead that the four forces are in equilibrium, find the values of $$\(F\)$$ and $$\(\alpha\)$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_m20_qp_42 Year：2020 Question No：5(b)
Answer:
Resolve forces either vertically or horizontally
$F \sin \alpha+4 \sin 30-6=0$
and
$F \cos \alpha+3-4 \cos 30=0$
$\left[F^{2}=4^{2}+0.464^{2}\right]$
or
$\left[F=\frac{4}{\sin 83.4}=\frac{0.464}{\cos 83.4}\right]$
$\left[\alpha=\tan ^{-1}\left(\frac{4}{0.464}\right)\right]$
or
$\left[\alpha=\sin ^{-1}\left(\frac{4}{4.03}\right)=\cos ^{-1}\left(\frac{0.464}{4.03}\right)\right]$
$F=4.03$ and $\alpha=83.4$
$F \sin \alpha+4 \sin 30-6=0$
and
$F \cos \alpha+3-4 \cos 30=0$
$\left[F^{2}=4^{2}+0.464^{2}\right]$
or
$\left[F=\frac{4}{\sin 83.4}=\frac{0.464}{\cos 83.4}\right]$
$\left[\alpha=\tan ^{-1}\left(\frac{4}{0.464}\right)\right]$
or
$\left[\alpha=\sin ^{-1}\left(\frac{4}{4.03}\right)=\cos ^{-1}\left(\frac{0.464}{4.03}\right)\right]$
$F=4.03$ and $\alpha=83.4$
Knowledge points:
4.1.3 use the principle that, when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently, that the sum of the components in any direction is zero （Solutions by resolving are usually expected, but equivalent methods (e.g. triangle of forces, Lami’s Theorem, where suitable) are also acceptable; these other methods are not required knowledge, and will not be referred to in questions.）
Solution:
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