1. **Problem 9:** Find the minimum number of non-equal coplanar forces in equilibrium. - In physics, for forces to be in equilibrium, their vector sum must be zero. - The minimum number of non-equal coplanar forces that can be in equilibrium is 3 because two non-equal forces cannot balance each other unless they are equal and opposite. **Answer:** 3 (ج) 2. **Problem 10:** Three coplanar forces not on the same line and concurrent at a point, two forces have magnitudes 7 N and 3 N. Find the possible magnitude of the third force. - For three concurrent forces in equilibrium, the third force must satisfy the triangle inequality with the other two. - The third force magnitude $F_3$ must satisfy: $$|7 - 3| < F_3 < 7 + 3$$ $$4 < F_3 < 10$$ - From the options 10, 4, 5, 3, only 5 satisfies this inequality. **Answer:** 5 (ج) 3. **Example:** Three forces of magnitudes 60 N, 40 N, and $K$ N are concurrent and coplanar. The angle between the first and second forces is 120°, and between the second and third is 90°. Find the magnitudes $Q$ and $K$. - Given forces: $F_1 = 60$, $F_2 = 40$, $F_3 = K$. - Angles: $ heta_{12} = 120^\\circ$, $ heta_{23} = 90^\\circ$. - For equilibrium, the vector sum is zero: $$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$$ - Using the law of cosines for the triangle formed by the forces: $$F_3^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos(\theta_{12})$$ $$K^2 = 60^2 + 40^2 + 2 \times 60 \times 40 \times \cos(120^\circ)$$ - Calculate: $$\cos(120^\circ) = -0.5$$ $$K^2 = 3600 + 1600 + 2 \times 60 \times 40 \times (-0.5)$$ $$K^2 = 5200 - 2400 = 2800$$ $$K = \sqrt{2800} = 52.92$$ - To find $Q$, use the angle between $F_2$ and $F_3$ (90°) and the law of cosines: $$Q^2 = F_2^2 + F_3^2 + 2 F_2 F_3 \cos(90^\circ)$$ $$Q^2 = 40^2 + 52.92^2 + 0$$ $$Q = \sqrt{1600 + 2800} = \sqrt{4400} = 66.33$$ **Final answers:** - $K \approx 52.92$ - $Q \approx 66.33$