1. Problem No.1: Given forces on a bracket, find magnitude $F$ and direction $\theta$ such that the resultant force $\vec{R}$ is 750 N along positive x-axis. 2. Identify given forces: - $325\ \text{N}$ vertical upward at location described by sides 5, 12, 13, so direction is vertical (along y) - $600\ \text{N}$ at $45^\circ$ downward to the right (from x-axis) - Unknown force $F$ at angle $\theta$ 3. Write components of known forces: - $325\ \text{N}$ upward: $R_{x1} = 0$, $R_{y1} = 325$ - $600\ \text{N}$ at $45^\circ$: $R_{x2} = 600\cos 45^\circ = 600 \times \frac{\sqrt{2}}{2} = 424.26$, $R_{y2} = -600\sin 45^\circ = -424.26$ 4. Unknown force $F$ components: - $F_x = F \cos \theta$ - $F_y = F \sin \theta$ 5. Resultant force must be $\vec{R} = 750 \hat{i} + 0 \hat{j}$ because it is along positive x axis, 750 N magnitude. 6. Summing x-components: $$0 + 424.26 + F \cos \theta = 750$$ $$F \cos \theta = 750 - 424.26 = 325.74$$ 7. Summing y-components: $$325 - 424.26 + F \sin \theta = 0$$ $$F \sin \theta = 424.26 - 325 = 99.26$$ 8. From these two equations: $$F \cos \theta = 325.74$$ $$F \sin \theta = 99.26$$ 9. Square and add both: $$F^2 = (325.74)^2 + (99.26)^2 = 106100 + 9853 = 115953$$ $$F = \sqrt{115953} = 340.55\ \text{N}$$ 10. Find $\theta$: $$\tan \theta = \frac{F \sin \theta}{F \cos \theta} = \frac{99.26}{325.74} = 0.3047$$ $$\theta = \arctan(0.3047) = 16.95^\circ$$ --- 11. Problem No.2: Given $\phi = 30^\circ$, $F_1 = 3\ \text{kN}$, find magnitude and direction $\theta$ of resultant force on plate. 12. Forces given: - $F_1 = 3\ \text{kN}$ at $30^\circ$ (assumed from vertical y-axis) - $F_3 = 5\ \text{kN}$ at angle from a 3-4-5 triangle - $F_2$ unknown magnitude and angle 13. Resolve vectors to components with respect to positive x-axis: - $F_1$ is at $30^\circ$ from vertical, vertical is $90^\circ$ from x-axis so $F_1$ angle from x-axis is $90^\circ - 30^\circ = 60^\circ$ - $F_3$ angle corresponds to 3-4-5 triangle with sides (3,4,5): angle with x is $\theta_3 = \arctan(3/4) = 36.87^\circ$ 14. Components: - $F_1$: $$F_{1x} = 3 \times \cos 60^\circ = 3 \times 0.5 = 1.5\ \text{kN}$$ $$F_{1y} = 3 \times \sin 60^\circ = 3 \times 0.866 = 2.598\ \text{kN}$$ - $F_3$: $$F_{3x} = 5 \times \cos 36.87^\circ = 5 \times 0.8 = 4\ \text{kN}$$ $$F_{3y} = 5 \times \sin 36.87^\circ = 5 \times 0.6 = 3\ \text{kN}$$ 15. Sum components: - $R_x = F_{1x} + F_{2x} + F_{3x} = 1.5 + F_{2x} + 4 = 5.5 + F_{2x}$ - $R_y = F_{1y} + F_{2y} + F_{3y} = 2.598 + F_{2y} + 3 = 5.598 + F_{2y}$ 16. Without $F_2$ magnitude and direction, cannot find exact resultant. Problem may ask only for resultant of given $F_1$ and $F_3$ or more data needed. 17. Assuming $F_2 = 0$, approximate: 18. Resultant magnitude: $$R = \sqrt{(5.5)^2 + (5.598)^2} = \sqrt{30.25 + 31.34} = \sqrt{61.59} = 7.85\ \text{kN}$$ 19. Direction $\theta$ measured clockwise from positive x-axis: $$\tan \theta = \frac{R_y}{R_x} = \frac{5.598}{5.5} = 1.018$$ $$\theta = \arctan(1.018) = 45.6^\circ$$ Final Answers: - Problem 1: $F = 340.55$ N, $\theta = 16.95^\circ$ - Problem 2: Without $F_2$, $R = 7.85$ kN, $\theta = 45.6^\circ$ clockwise from positive x-axis.