1. **Problem Statement:** Find the Magnitude, Direction, and Location (MDL) of the Resultant force with respect to point A for the given forces F1 to F5 acting on a rectangular frame. 2. **Given Data:** - Forces: $F_1=70$ kN, $F_2=45$ kN, $F_3=45$ kN, $F_4=40$ kN, $F_5=45$ kN - Angles: $F_1$ at 45° downward-right, $F_2$ at 30° downward-left, $F_3$ at 30° from vertical downward-right, $F_4$ vertical downward, $F_5$ horizontal rightward - Dimensions: Rectangle base = 8 m, height = 6 m, top-right horizontal segment = 3 m, $F_3$ located 2 m above bottom, $F_5$ located 1 m above A 3. **Step 1: Calculate Horizontal and Vertical Components of Each Force** - $F_1$: angle 45° downward-right means components: $$F_{1x} = 70 \cos 45^\circ = 70 \times \frac{\sqrt{2}}{2} = 49.5\text{ kN}$$ $$F_{1y} = -70 \sin 45^\circ = -49.5\text{ kN}$$ (negative because downward) - $F_2$: angle 30° downward-left means components: $$F_{2x} = -45 \cos 30^\circ = -45 \times 0.866 = -38.97\text{ kN}$$ (left is negative) $$F_{2y} = -45 \sin 30^\circ = -45 \times 0.5 = -22.5\text{ kN}$$ - $F_3$: angle 30° from vertical downward-right means: Vertical component: $$F_{3y} = -45 \cos 30^\circ = -45 \times 0.866 = -38.97\text{ kN}$$ Horizontal component: $$F_{3x} = 45 \sin 30^\circ = 45 \times 0.5 = 22.5\text{ kN}$$ - $F_4$: vertical downward force: $$F_{4x} = 0$$ $$F_{4y} = -40\text{ kN}$$ - $F_5$: horizontal rightward force: $$F_{5x} = 45\text{ kN}$$ $$F_{5y} = 0$$ 4. **Step 2: Sum Horizontal and Vertical Components** $$\Sigma F_x = 49.5 - 38.97 + 22.5 + 0 + 45 = 78.03\text{ kN}$$ $$\Sigma F_y = -49.5 - 22.5 - 38.97 - 40 + 0 = -150.97\text{ kN}$$ 5. **Step 3: Calculate Magnitude and Direction of Resultant Force** Magnitude: $$R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} = \sqrt{78.03^2 + (-150.97)^2} = \sqrt{6088.7 + 22792.5} = \sqrt{28881.2} = 170.0\text{ kN}$$ Direction (angle $\theta$ from positive x-axis, clockwise is negative): $$\theta = \tan^{-1} \left( \frac{\Sigma F_y}{\Sigma F_x} \right) = \tan^{-1} \left( \frac{-150.97}{78.03} \right) = -62.3^\circ$$ Since $\Sigma F_x > 0$ and $\Sigma F_y < 0$, the angle is $62.3^\circ$ below the positive x-axis. 6. **Step 4: Calculate Moments About Point A** Coordinates of forces relative to A (x horizontal right, y vertical up): - $F_1$ at top-left corner: x=0, y=6 m - $F_2$ at top edge near right side: x=8 m, y=6 m - $F_3$ at right edge 2 m above bottom: x=8 m, y=2 m - $F_4$ at bottom-right corner: x=8 m, y=0 - $F_5$ at left edge 1 m above A: x=0, y=1 m Moment $M_A = x F_y - y F_x$ (counterclockwise positive) - $M_{A1} = 0 \times (-49.5) - 6 \times 49.5 = -297\text{ kNm}$ - $M_{A2} = 8 \times (-22.5) - 6 \times (-38.97) = -180 + 233.82 = 53.82\text{ kNm}$ - $M_{A3} = 8 \times (-38.97) - 2 \times 22.5 = -311.76 - 45 = -356.76\text{ kNm}$ - $M_{A4} = 8 \times (-40) - 0 \times 0 = -320\text{ kNm}$ - $M_{A5} = 0 \times 0 - 1 \times 45 = -45\text{ kNm}$ Sum moments: $$\Sigma M_A = -297 + 53.82 - 356.76 - 320 - 45 = -965.94\text{ kNm}$$ 7. **Step 5: Calculate Location (Perpendicular Distance) of Resultant from A** $$d = \frac{|\Sigma M_A|}{R} = \frac{965.94}{170.0} = 5.68\text{ m}$$ The moment is negative, indicating the resultant moment is clockwise, so the resultant force line of action is located 5.68 m from A on the side that produces this moment. --- **Final Answers:** - Magnitude of Resultant Force: $\boxed{170.0\text{ kN}}$ - Direction of Resultant Force: $\boxed{62.3^\circ}$ below the positive x-axis - Location of Resultant Force from A: $\boxed{5.68\text{ m}}$