1. **State the problem:** Find the magnitude, direction, and location of the resultant force with respect to point A for the given system of forces. 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 $45^\circ$ below horizontal (down-right), $F_2$ at $30^\circ$ below horizontal (down-left), $F_3$ at $30^\circ$ from vertical (down-right), $F_4$ vertical downward, $F_5$ horizontal right - Distances from point A: see problem description 3. **Step 1: Resolve each force into horizontal ($F_x$) and vertical ($F_y$) components.** - $F_1$: angle $45^\circ$ below horizontal to the right $$F_{1x} = 70 \cos 45^\circ = 70 \times 0.7071 = 49.50$$ $$F_{1y} = -70 \sin 45^\circ = -70 \times 0.7071 = -49.50$$ - $F_2$: angle $30^\circ$ below horizontal to the left $$F_{2x} = -45 \cos 30^\circ = -45 \times 0.8660 = -38.97$$ $$F_{2y} = -45 \sin 30^\circ = -45 \times 0.5 = -22.50$$ - $F_3$: angle $30^\circ$ from vertical down-right means $60^\circ$ from horizontal to right $$F_{3x} = 45 \cos 60^\circ = 45 \times 0.5 = 22.50$$ $$F_{3y} = -45 \sin 60^\circ = -45 \times 0.8660 = -38.97$$ - $F_4$: vertical downward $$F_{4x} = 0$$ $$F_{4y} = -40$$ - $F_5$: horizontal right $$F_{5x} = 45$$ $$F_{5y} = 0$$ 4. **Step 2: Sum horizontal and vertical components** $$\Sigma F_x = 49.50 - 38.97 + 22.50 + 0 + 45 = 78.03$$ $$\Sigma F_y = -49.50 - 22.50 - 38.97 - 40 + 0 = -150.97$$ 5. **Step 3: Calculate magnitude and direction of resultant force** Magnitude: $$R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} = \sqrt{78.03^2 + (-150.97)^2} = \sqrt{6088.68 + 22792.44} = \sqrt{28881.12} = 169.94$$ Direction (angle $\theta$ from horizontal, positive counterclockwise): $$\theta = \tan^{-1} \left( \frac{\Sigma F_y}{\Sigma F_x} \right) = \tan^{-1} \left( \frac{-150.97}{78.03} \right) = -62.56^\circ$$ Since $\Sigma F_x > 0$ and $\Sigma F_y < 0$, angle is $62.56^\circ$ below horizontal to the right. 6. **Step 4: Calculate moments about point A (counterclockwise positive)** Coordinates of forces from A (horizontal right positive, vertical up positive): - $F_1$ at top-left corner: horizontal $0$, vertical $8$ m - $F_2$ at top edge 3 m left of top-right corner (top-right corner is 8 m right), so $8 - 3 = 5$ m right, vertical $8$ m - $F_3$ on right vertical side 2 m above bottom: horizontal $8$ m, vertical $2$ m - $F_4$ bottom-right corner: horizontal $8$ m, vertical $0$ m - $F_5$ left vertical side 1 m above A: horizontal $0$ m, vertical $1$ m Moments: - $M_{A1} = x_1 F_{1y} - y_1 F_{1x} = 0 \times (-49.50) - 8 \times 49.50 = -396.00$ kNm - $M_{A2} = 5 \times (-22.50) - 8 \times (-38.97) = -112.50 + 311.76 = 199.26$ kNm - $M_{A3} = 8 \times (-38.97) - 2 \times 22.50 = -311.76 - 45.00 = -356.76$ kNm - $M_{A4} = 8 \times (-40) - 0 \times 0 = -320.00$ kNm - $M_{A5} = 0 \times 0 - 1 \times 45 = -45.00$ kNm Sum moments: $$\Sigma M_A = -396.00 + 199.26 - 356.76 - 320.00 - 45.00 = -918.50$$ 7. **Step 5: Calculate location $d$ of resultant force from point A along horizontal axis** Using moment equilibrium: $$\Sigma M_A = -918.50 = -d \times R_y + 0 \times R_x$$ Since $R_x$ and $R_y$ components of resultant are: $$R_x = 78.03, \quad R_y = -150.97$$ Moment arm $d$ is vertical distance from A to line of action of resultant: $$d = \frac{\Sigma M_A}{\Sigma F_x} = \frac{-918.50}{78.03} = -11.77 \text{ m}$$ Negative sign means location is 11.77 m below point A horizontally. 8. **Final answers:** - Magnitude of resultant force: $169.94$ kN - Direction: $62.56^\circ$ below horizontal to the right - Location: $11.77$ m below point A along horizontal axis