Forces and Equilibrium Questions and Answers

Measure each direction anticlockwise from the positive $x$-axis. The 2N force along the $y$-axis is at $90^{\circ}$, the 3N force at $90^{\circ}+30^{\circ}=120^{\circ}$, the 5N force at $90^{\circ}-30^{\circ}=60^{\circ}$, and the fourth 2N force, drawn at right angles to the 5N force, at $60^{\circ}-90^{\circ}=-30^{\circ}$.

Resolving along $\mathbf{j}$:

$3\sin 120^{\circ}+2\sin 90^{\circ}+5\sin 60^{\circ}+2\sin(-30^{\circ})=\dfrac{3\sqrt{3}}{2}+2+\dfrac{5\sqrt{3}}{2}-1$

$=\dfrac{8\sqrt{3}}{2}+1=1+4\sqrt{3}=7,93$.

Substituting $P=5$, $Q=8$ and $\theta=60^{\circ}$, with $\cos60^{\circ}=0.5$:
$R^{2}=5^{2}+8^{2}+2(5)(8)(0.5)=25+64+40=129$.

So $R=\sqrt{129}=11.358$, that is $11.4$ N to 3 significant figures.

Place the two forces tail to tail with 40 degrees between them. In the triangle of forces the included angle is $180^{\circ}-40^{\circ}=140^{\circ}$, so by the cosine rule

$R^{2}=4^{2}+9^{2}-2(4)(9)\cos 140^{\circ}=16+81+72\cos 40^{\circ}$

$=97+72(0,766044)=97+55,155=152,155$

$R=\sqrt{152,155}=12,3$ N.

The same answer comes from components: $R_{x}=9+4\cos 40^{\circ}=12,064$ and $R_{y}=4\sin 40^{\circ}=2,571$, giving $R=\sqrt{12,064^{2}+2,571^{2}}=12,3$ N.