1. **Problem 1: Find vector C such that A + B + C = 0 for each figure.** **Top-left figure:** - Vector A has magnitude 4 m and angle 40° above positive x-axis. - Vector B has magnitude 2 m and angle 20° below positive x-axis. **Step 1:** Write components of A and B. $$A_x = 4 \cos 40^\circ, \quad A_y = 4 \sin 40^\circ$$ $$B_x = 2 \cos (-20^\circ), \quad B_y = 2 \sin (-20^\circ)$$ **Step 2:** Calculate numeric values. $$A_x = 4 \times 0.7660 = 3.064$$ $$A_y = 4 \times 0.6428 = 2.571$$ $$B_x = 2 \times 0.9397 = 1.879$$ $$B_y = 2 \times (-0.3420) = -0.684$$ **Step 3:** Since $A + B + C = 0$, then $C = - (A + B)$. $$C_x = -(A_x + B_x) = -(3.064 + 1.879) = -4.943$$ $$C_y = -(A_y + B_y) = -(2.571 - 0.684) = -1.887$$ **Answer for top-left:** $$\boxed{C = (-4.943, -1.887)}$$ --- **Top-right figure:** - Vector A: length 2 m, 15° with negative x-axis vertically upward means angle $180^\circ - 15^\circ = 165^\circ$ from positive x-axis. - Vector B: length 4 m, 15° above positive x-axis. **Step 1:** Components: $$A_x = 2 \cos 165^\circ, \quad A_y = 2 \sin 165^\circ$$ $$B_x = 4 \cos 15^\circ, \quad B_y = 4 \sin 15^\circ$$ **Step 2:** Calculate numeric values. $$A_x = 2 \times (-0.9659) = -1.932$$ $$A_y = 2 \times 0.2588 = 0.518$$ $$B_x = 4 \times 0.9659 = 3.864$$ $$B_y = 4 \times 0.2588 = 1.035$$ **Step 3:** Find $C$: $$C_x = -(A_x + B_x) = -(-1.932 + 3.864) = -1.932$$ $$C_y = -(A_y + B_y) = -(0.518 + 1.035) = -1.553$$ **Answer for top-right:** $$\boxed{C = (-1.932, -1.553)}$$ --- 2. **Problem 2: Forces on tilted floor** - $F_1 = 3.0$ N at $135^\circ$ from horizontal (left upward) - $F_2 = 6.0$ N vertically upward ($90^\circ$) - $F_3 = 5.0$ N at $30^\circ$ below horizontal to the right ($-30^\circ$) Assuming floor is horizontal, components parallel to floor are horizontal components, perpendicular are vertical components. **Step 1:** Find components of each force. $$F_{1x} = 3.0 \cos 135^\circ = 3.0 \times (-0.7071) = -2.121$$ $$F_{1y} = 3.0 \sin 135^\circ = 3.0 \times 0.7071 = 2.121$$ $$F_{2x} = 6.0 \cos 90^\circ = 0$$ $$F_{2y} = 6.0 \sin 90^\circ = 6.0$$ $$F_{3x} = 5.0 \cos (-30^\circ) = 5.0 \times 0.8660 = 4.330$$ $$F_{3y} = 5.0 \sin (-30^\circ) = 5.0 \times (-0.5) = -2.5$$ **Step 2:** Sum components for net force. $$F_{net,x} = F_{1x} + F_{2x} + F_{3x} = -2.121 + 0 + 4.330 = 2.209$$ $$F_{net,y} = F_{1y} + F_{2y} + F_{3y} = 2.121 + 6.0 - 2.5 = 5.621$$ **Step 3:** (a) Component parallel to floor is horizontal component: $$\boxed{F_{net,\parallel} = 2.209\ \text{N}}$$ (b) Component perpendicular to floor is vertical component: $$\boxed{F_{net,\perp} = 5.621\ \text{N}}$$ **Step 4:** (c) Magnitude and direction of net force: $$|F_{net}| = \sqrt{(2.209)^2 + (5.621)^2} = \sqrt{4.878 + 31.601} = \sqrt{36.479} = 6.04$$ Direction angle $\theta$ from horizontal: $$\theta = \tan^{-1} \left( \frac{5.621}{2.209} \right) = \tan^{-1}(2.545) = 68.5^\circ$$ **Answer:** $$\boxed{|F_{net}| = 6.04\ \text{N}, \quad \theta = 68.5^\circ \text{ above horizontal}}$$