Explanation

(a) Here $\overrightarrow{A} - \overrightarrow{OP} = 10$ units along $OP$
$\overrightarrow{B} - \left( \overrightarrow{OQ} \right) = 10$ units along $OQ$
$\therefore\ \ \ \angle XOP = 30{^\circ}$ and$\ \ \angle XOQ = 135{^\circ}$
$\therefore\ \ \ \angle QOX^{''} = 180{^\circ} - 135{^\circ}$=45${^\circ}$
[image:4_image164.emf]
Resolving $\overrightarrow{A}$ and $\overrightarrow{B}$ into two rectangular components we have $A$ cos 30${^\circ}$ along $OX$ and $A$ sin 30${^\circ}$ along $OY$. $B$ $\cos{45{^\circ}}$ along $OX''\ and\ B\sin{45{^\circ}}$ along $OY''$.
Resultant component force along $X$-axis.
$(A\ \cos 30{^\circ} - B\sin 45)\widehat{i}$
$= \left( 10 \times \sqrt{3}/2 - 10 \times 1/\sqrt{2} \right)\widehat{i} = 1.59\widehat{i}$
Resultant component force along $Y$-axis
$= (A\ \sin 30{^\circ} + B\sin 45{^\circ})\widehat{j}$
$= \left( 10 \times 1/2 + 10 \times 1/\sqrt{2} \right)\widehat{j} = 12.07\widehat{j}$