1. Given vector $\vec{F} = (6, \frac{2\pi}{3})$, find its magnitude: $$\|\vec{F}\| = \sqrt{6^2 + \left(\frac{2\pi}{3}\right)^2} = \sqrt{36 + \frac{4\pi^2}{9}} = \sqrt{36 + \frac{4\cdot 9.8696}{9}} = \sqrt{36 + 4.3865} \approx \sqrt{40.3865} \approx 6.35$$ The closest answer is c) 6. 2. Two forces 8 and 3 N with included angle 150°; magnitude of resultant: $$R = \sqrt{8^2 + 3^2 + 2 \cdot 8 \cdot 3 \cdot \cos 150^\circ} = \sqrt{64 + 9 + 48 \cdot (-0.866)} = \sqrt{73 - 41.57} = \sqrt{31.43} \approx 5.6$$ None of the given answers match the calculated resultant; check options again. 3. Two perpendicular forces 12 N and 5 N; magnitude of resultant: $$R = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$ Answer: b) 13. 4. Two forces one is 16 N, resultant 26 N, angle 120°, find other force $F$: Using law of cosines: $$26^2 = 16^2 + F^2 + 2 \cdot 16 \cdot F \cdot \cos 120^\circ$$ $$676 = 256 + F^2 - 16F$$ $$F^2 - 16F + 256 - 676 = 0$$ $$F^2 - 16F - 420 = 0$$ Solve quadratic: $$F = \frac{16 \pm \sqrt{256 + 1680}}{2} = \frac{16 \pm \sqrt{1936}}{2} = \frac{16 \pm 44}{2}$$ $$F = 30 \text{ or } -14 \Rightarrow F=30$$ Answer: a) 30. 5. Two forces 12 N and 15 N, angle $\theta$ with $\cos \theta = -\frac{4}{5}$ find angle between resultant and first force: Using formula for resultant magnitude: $$R = \sqrt{12^2 + 15^2 + 2 \cdot 12 \cdot 15 \cdot \cos \theta} = \sqrt{144 + 225 - 288} = \sqrt{81} = 9$$ Angle $\alpha$ between resultant and first force: $$\tan \alpha = \frac{15 \sin \theta}{12 + 15 \cos \theta}$$ $$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(-\frac{4}{5}\right)^2} = \frac{3}{5}$$ $$\tan \alpha = \frac{15 \cdot \frac{3}{5}}{12 + 15 \cdot -\frac{4}{5}} = \frac{9}{12 - 12} = \frac{9}{0} \to \alpha = 90^\circ$$ Answer: d) 90°. 6. Two forces 8 and F gm.wt with angle $]0, \pi[$, resultant bisects angle; find F: When resultant bisects the angle, magnitudes are equal: $$8 = F$$ Answer: c) 8. 7. Forces $3F - 1$ and $F + 5$, resultant bisects angle; find $F$: Equal magnitude forces when bisector resultant: Set $3F-1 = F+5$: $$3F - 1 = F + 5 \Rightarrow 2F = 6 \Rightarrow F=3$$ Answer: b) 3. 8. Two equal forces 6 gm.wt each, resultant 6 gm.wt, find angle between: $$R = 2F \cos \frac{\theta}{2} = 6$$ $$2 \cdot 6 \cos \frac{\theta}{2} = 6 \Rightarrow \cos \frac{\theta}{2} = \frac{1}{2} \Rightarrow \frac{\theta}{2} = 60^\circ \Rightarrow \theta = 120^\circ$$ Answer: b) 120°. 9. Resultant of 6 and 8 N could be: Max sum: 14, min difference: 2 Options: 20 no, 15 no, 12 yes, 1 no Answer: c) 12. 10. Maximum resultant angle: Max resultant when forces aligned: Answer: c) 0°. 11. Resultant zero angle: Opposite vectors 180°. Answer: a) 180°. 12. Forces 4F and 5F, resultant 9F; find included angle: $$ (9F)^2 = (4F)^2 + (5F)^2 + 2 \cdot 4F \cdot 5F \cdot \cos \theta $$ $$81F^2 = 16F^2 + 25F^2 + 40F^2 \cos \theta$$ $$81 = 41 + 40 \cos \theta \Rightarrow \cos \theta = 1$$ Answer: a) 0°. 13. Forces $F$ and 6 N, resultant perpendicular to $F$, angle 120°, find $F$: Resultant perpendicular to $F$ means: $$R^2 = F^2 + 36 + 2 \cdot F \cdot 6 \cdot \cos 120^\circ$$ $$R^2 \perp F \implies F^2 + 72 \cos 120^\circ + 36 = R^2$$ Using vector properties, $F$ equals: $$F = 6 \sqrt{2}$$ Answer: c) $6 \sqrt{2}$. 14. Two forces F each, angle 60°, resultant: $$R = \sqrt{F^2 + F^2 + 2F^2 \cos 60^\circ} = \sqrt{2F^2 + 2F^2 \cdot \frac{1}{2}} = \sqrt{2F^2 + F^2} = \sqrt{3}F$$ Answer: c) $\sqrt{3}F$. 15. Resultant $R \in [10,22]$ with $F_1 < F_2$ and $R = (F_1, F_2)$: Answer: a) (10,22).