Let's break down the solution step by step, **with explanation and final answer**. --- ### **1. Vector Definitions** Let's first convert all vectors to the same unit and component form (x and y components). We'll use the **scale 1 cm = 2 units** (where "units" can be meters, newtons, etc., depending on the vector). #### **Given Vectors** - \(\vec{a} = 10\, m\), up - \(\vec{b} = 8\, m\) @ 20° below the left horizon - \(\vec{c} = 2\, m\), left - \(\vec{d} = 12\, m\) @ 30° left of vertical - \(\vec{e} = 8\, N\), right - \(\vec{f} = 600\, cm = 6\, m\), down - \(\vec{g} = 40\, N\) @ 15° above the left horizon - \(\vec{h} = 65\, ft = 19.81\, m\), left --- ### **2. Write All Vectors in Component Form** Let’s define right as +x, up as +y. #### **a.** \(\vec{a} = 10\, m\) up - \(x_a = 0\) - \(y_a = +10\) #### **b.** \(\vec{b} = 8\, m\) @ 20° below the left horizon - "Left horizon" = negative x-axis, "below" = negative y - \(x_b = -8 \cos(20^\circ) = -7.52\) - \(y_b = -8 \sin(20^\circ) = -2.74\) #### **c.** \(\vec{c} = 2\, m\) left - \(x_c = -2\) - \(y_c = 0\) #### **d.** \(\vec{d} = 12\, m\) @ 30° left of vertical - "Left of vertical" = 30° counterclockwise from up (positive y), into the second quadrant. - \(x_d = -12 \sin(30^\circ) = -6\) - \(y_d = 12 \cos(30^\circ) = 10.39\) #### **e.** \(\vec{e} = 8\, N\), right - \(x_e = +8\) - \(y_e = 0\) #### **f.** \(\vec{f} = 6\, m\), down - \(x_f = 0\) - \(y_f = -6\) #### **g.** \(\vec{g} = 40\, N\) @ 15° above the left horizon - "Left horizon" = negative x-axis, "above" = positive y - \(x_g = -40 \cos(15^\circ) = -38.64\) - \(y_g = +40 \sin(15^\circ) = +10.35\) #### **h.** \(\vec{h} = 19.81\, m\), left - \(x_h = -19.81\) - \(y_h = 0\) --- ### **3. Write the Required Expression** \[ \vec{R} = 2\vec{a} - 2\vec{b} + \vec{h} - \vec{c} + \vec{f} - \vec{d} \] Let’s expand each term: - \(2\vec{a} = (0, 20)\) - \(-2\vec{b} = -2 \times (-7.52, -2.74) = (15.04, 5.48)\) - \(+\vec{h} = (-19.81, 0)\) - \(-\vec{c} = (2, 0)\) - \(+\vec{f} = (0, -6)\) - \(-\vec{d} = -(-6, 10.39) = (6, -10.39)\) --- ### **4. Sum the Components** #### **X-Components:** \[ 0 + 15.04 + (-19.81) + 2 + 0 + 6 = 3.23 \] #### **Y-Components:** \[ 20 + 5.48 + 0 + 0 + (-6) + (-10.39) = 9.09 \] --- ### **5. Find the Resultant Vector** \[ \vec{R} = (3.23, 9.09) \] #### **Magnitude:** \[ |\vec{R}| = \sqrt{(3.23)^2 + (9.09)^2} = \sqrt{10.43 + 82.65} = \sqrt{93.08} \approx 9.65 \] #### **Direction (angle from x-axis):** \[ \theta = \tan^{-1} \left( \frac{9.09}{3.23} \right ) \approx \tan^{-1}(2.82) \approx 70.6^\circ \text{ above the x-axis} \] --- ### **6. Graphic Method** To **draw this on graph paper** with scale \(1\,\text{cm} = 2\,\text{units}\): - Magnitude: \(9.65\,\text{units} / 2 = 4.83\,\text{cm}\) - Starting from the origin, draw a vector \(4.83\,\text{cm}\) long at \(70.6^\circ\) above the x-axis (to the right and up). --- ## **Final Answer** \[ \boxed{ \vec{R} \approx 9.65\,\text{units} \text{ at } 70.6^\circ \text{ above the x-axis (right and up direction)} } \] **Explanation**: Each vector was broken into x and y components, the required operations were performed, and the resultant vector was calculated in both magnitude and direction. Use the given scale to draw this on your graph paper.