1. **Problem 4: Calculate the magnitude and direction of the resultant force of three forces A, B, C acting on a ring.** Given: - Force A = 3 N at 40° above the negative X-axis (up-left) - Force B = 7 N upward along positive Y-axis - Force C = 10 N along positive X-axis 2. **Resolve each force into components along X and Y axes:** - Force A components: - $A_x = -3 \cos 40^\circ$ (negative X direction) - $A_y = 3 \sin 40^\circ$ (positive Y direction) - Force B components: - $B_x = 0$ - $B_y = 7$ - Force C components: - $C_x = 10$ - $C_y = 0$ 3. **Calculate each component numerically:** - $A_x = -3 \times 0.7660 = -2.298$ N - $A_y = 3 \times 0.6428 = 1.928$ N 4. **Sum the components along each axis:** - $R_x = A_x + B_x + C_x = -2.298 + 0 + 10 = 7.702$ N - $R_y = A_y + B_y + C_y = 1.928 + 7 + 0 = 8.928$ N 5. **Calculate the magnitude of the resultant force:** $$ R = \sqrt{R_x^2 + R_y^2} = \sqrt{7.702^2 + 8.928^2} = \sqrt{59.36 + 79.72} = \sqrt{139.08} = 11.8 \text{ N} $$ 6. **Calculate the direction (angle) of the resultant force relative to the positive X-axis:** $$\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right) = \tan^{-1} \left( \frac{8.928}{7.702} \right) = \tan^{-1} (1.159) = 49.3^\circ \text{ above X-axis} $$ --- 7. **Problem 5: Find the resultant displacement from city A to city C, given two legs of the trip.** Given: - From A to B: 800 km due east (angle 0°) - From B to C: 600 km at 40° north of east 8. **Resolve each displacement into components:** - $D_{ABx} = 800 \cos 0^\circ = 800$ km - $D_{ABy} = 800 \sin 0^\circ = 0$ km - $D_{BCx} = 600 \cos 40^\circ = 600 \times 0.7660 = 459.6$ km - $D_{BCy} = 600 \sin 40^\circ = 600 \times 0.6428 = 385.7$ km 9. **Sum components to get resultant displacement $D_{AC}$:** - $D_x = 800 + 459.6 = 1259.6$ km - $D_y = 0 + 385.7 = 385.7$ km 10. **Calculate magnitude of resultant displacement:** $$ D = \sqrt{1259.6^2 + 385.7^2} = \sqrt{1587100 + 148784} = \sqrt{1735884} = 1317 \text{ km} $$ 11. **Calculate direction angle north of east:** $$ \theta = \tan^{-1} \left( \frac{385.7}{1259.6} \right) = \tan^{-1} (0.306) = 17^\circ $$ **Final answers:** - Resultant force magnitude = 11.8 N, direction = 49.3° above positive X-axis - Resultant displacement = 1317 km at 17° north of east