1. **Problem 1:** Given two like parallel forces $\vec{F_1}$ and $\vec{F_2}$ acting at points A and B respectively, with resultant $\vec{R}$ acting at point C on line $\overline{AB}$. Given $F_2 = 6$ N, $AC = 24$ cm, and $AB = 56$ cm, find $F_1$ and $R$. 2. **Formula and rules:** For two parallel forces acting in the same direction, the resultant force magnitude is the sum: $$R = F_1 + F_2$$ The point of action of the resultant $C$ divides $AB$ such that: $$R \times AC = F_2 \times AB$$ This is the moment balance about point A. 3. **Calculate $F_1$:** Given $F_2 = 6$, $AC = 24$, $AB = 56$, and $R = F_1 + 6$. Using moment balance: $$R \times 24 = 6 \times 56$$ $$24(F_1 + 6) = 336$$ $$24F_1 + 144 = 336$$ $$24F_1 = 192$$ $$F_1 = \frac{192}{24} = 8$$ 4. **Calculate $R$:** $$R = F_1 + F_2 = 8 + 6 = 14$$ 5. **Answer for Problem 1:** $F_1 = 8$ N, $R = 14$ N (Option a). --- 1. **Problem 2:** Two parallel forces $\vec{F_1}$ and $\vec{F_2}$ act in the same direction at points A and B. Given $F_1 = 8$ N, $R = 13$ N, and $AC = 10$ cm, find $AB$. 2. **Formula:** $$R = F_1 + F_2$$ Moment balance: $$R \times AC = F_2 \times AB$$ 3. **Calculate $F_2$:** $$F_2 = R - F_1 = 13 - 8 = 5$$ 4. **Calculate $AB$:** $$13 \times 10 = 5 \times AB$$ $$130 = 5 AB$$ $$AB = \frac{130}{5} = 26$$ 5. **Answer for Problem 2:** $AB = 26$ cm (Option c). --- 1. **Problem 3:** Two parallel forces $\vec{F_1} = 7$ N and $\vec{F_2} = 9$ N act in opposite directions. The distance between the resultant and the second force is 35 cm. Find the distance between the two forces. 2. **Formula:** For opposite forces, $$R = |F_2 - F_1| = 2 \text{ N}$$ Moment balance about $F_2$: $$R \times d = F_1 \times AB$$ where $d = 35$ cm is distance from resultant to $F_2$. 3. **Calculate $AB$:** $$2 \times 35 = 7 \times AB$$ $$70 = 7 AB$$ $$AB = 10$$ 4. **Answer for Problem 3:** Distance between forces is 10 cm (Option a). --- 1. **Problem 4:** Two parallel forces of magnitudes 15 N and $F$ N act in opposite directions. The resultant magnitude is 25 N. The known force and resultant act in opposite directions. Find $F$. 2. **Formula:** $$R = |F - 15| = 25$$ Since resultant and known force act opposite, $F$ must be greater than 15. 3. **Calculate $F$:** $$F - 15 = 25$$ $$F = 40$$ 4. **Answer for Problem 4:** $F = 40$ N (Option d). --- 1. **Problem 5:** Given a right triangle with forces $F_1 = 150$ N at A, resultant $R = 60$ N at the right angle vertex, and distance 30 cm from right angle vertex to B. Find $F_2$ and its distance from A. 2. **Using moment balance:** Let distance from A to B be $x$, and $F_2$ acts at B. Moment about right angle vertex: $$F_1 \times 30 = F_2 \times d$$ Resultant $R$ acts at right angle vertex. 3. **Using force balance:** $$F_1 + F_2 = R$$ But $F_1 = 150$, $R = 60$, so forces act in opposite directions: $$|F_1 - F_2| = 60$$ $$150 - F_2 = 60$$ $$F_2 = 90$$ 4. **Calculate distance $d$ from A to $F_2$:** $$150 \times 30 = 90 \times d$$ $$4500 = 90 d$$ $$d = 50$$ 5. **Answer for Problem 5:** $F_2 = 90$ N, distance from A is 50 cm (Option c).