1. **Problem No.1**: Given the resultant force $R = 750$ N along the positive x-axis, find magnitude $F$ and direction $\theta$ of force $F$. 2. Forces given: - 325 N force at angle from right triangle with sides 5 (horizontal), 12 (vertical), 13 (hypotenuse), so angle $\alpha = \arcsin(12/13)$ above horizontal. - 600 N force at 45° below horizontal (to the right). - Force $F$ at angle $\theta$ above horizontal. 3. Resolve known forces into components: - $325$ N force components: $$F_x = 325 \times \frac{5}{13} = 125$$ N, $$F_y = 325 \times \frac{12}{13} = 300$$ N. - $600$ N force components: $$F_x = 600 \times \cos 45^\circ = 600 \times \frac{\sqrt{2}}{2} \approx 424.26$$ N, $$F_y = -600 \times \sin 45^\circ = -424.26$$ N (downward). 4. Let force $F$ components be: $$F_x = F \cos \theta$$ and $$F_y = F \sin \theta.$$ 5. The total x-component of resultant forces must be $750$ N, and total y-component must be $0$ (since resultant along x-axis): $$125 + 424.26 + F \cos \theta = 750$$ $$300 - 424.26 + F \sin \theta = 0.$$ 6. Simplify: - $F \cos \theta = 750 - 549.26 = 200.74$ - $F \sin \theta = 124.26$ 7. Find $F$ and $\theta$: $$F = \sqrt{(200.74)^2 + (124.26)^2} \approx \sqrt{40296 + 15450} = \sqrt{55746} \approx 236.03$$ N $$\theta = \arctan\left(\frac{124.26}{200.74}\right) \approx 31.34^\circ.$$ --- 8. **Problem No.2**: Given $\phi = 30^\circ$, $F_2 = 3$ kN, $F_1 = 4$ kN at $30^\circ$ above positive x-axis, and $F_3 = 5$ kN downward at angle from 3-4-5 triangle (angle $\beta = \arcsin(3/5) = 36.87^\circ$ below horizontal). 9. Resolve forces to components: - $F_1$: $$F_{1x} = 4 \times \cos 30^\circ = 4 \times 0.866 = 3.464\text{ kN},$$ $$F_{1y} = 4 \times \sin 30^\circ = 4 \times 0.5 = 2.0\text{ kN}.$$ - $F_2$ is horizontal: $$F_{2x} = 3.0\text{ kN}, F_{2y} = 0.$$ - $F_3$ at $36.87^\circ$ below horizontal: $$F_{3x} = 5 \times \cos 36.87^\circ = 5 \times 0.8 = 4.0\text{ kN},$$ $$F_{3y} = -5 \times \sin 36.87^\circ = -5 \times 0.6 = -3.0\text{ kN}.$$ 10. Sum components: $$R_x = 3.464 + 3 + 4 = 10.464\text{ kN},$$ $$R_y = 2 + 0 - 3 = -1.0\text{ kN}.$$ 11. Resultant magnitude: $$R = \sqrt{(10.464)^2 + (-1)^2} = \sqrt{109.45 + 1} = \sqrt{110.45} \approx 10.51 \text{ kN}.$$ 12. Direction $\theta$ measured clockwise from positive x-axis: Since $R_y$ is negative, angle below x-axis: $$\theta = \arctan \left( \frac{|-1|}{10.464} \right) = \arctan(0.0956) \approx 5.47^\circ.$$ **Final answers:** - Problem 1: $F \approx 236$ N, $\theta \approx 31.3^\circ$ above x-axis. - Problem 2: Resultant $\approx 10.5$ kN, direction $\theta \approx 5.5^\circ$ clockwise from x-axis.