1. **Problem statement:** Three forces of magnitudes 60 N, 80 N, and 130 N act at the vertices A, B, and C respectively of an equilateral triangle ABC with sides 40 cm each. We need to find the resultant force's magnitude, direction, and position. 2. **Setup and assumptions:** Since ABC is equilateral, all sides are 40 cm and all angles are 60°. 3. **Coordinate assignment:** Place point A at origin (0,0). Let AB lie along the positive x-axis, so B is at (40,0). Point C is at (20, 20\sqrt{3}) because height of equilateral triangle is $40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3}$ cm. 4. **Force vectors:** - At A: 80 N upwards $\Rightarrow \vec{F_A} = (0,80)$ N - At B: 60 N to the left $\Rightarrow \vec{F_B} = (-60,0)$ N - At C: 130 N diagonally downwards making 60° with horizontal. Since it forms 60° angles at B and C, the force at C points downward and to the right at 60° below horizontal. Calculate components of $\vec{F_C}$: $$F_{Cx} = 130 \cos 60^\circ = 130 \times 0.5 = 65$$ $$F_{Cy} = -130 \sin 60^\circ = -130 \times \frac{\sqrt{3}}{2} \approx -112.58$$ So $\vec{F_C} = (65, -112.58)$ N 5. **Resultant force vector:** $$\vec{R} = \vec{F_A} + \vec{F_B} + \vec{F_C} = (0 - 60 + 65, 80 + 0 - 112.58) = (5, -32.58)$$ N 6. **Magnitude of resultant:** $$|\vec{R}| = \sqrt{5^2 + (-32.58)^2} = \sqrt{25 + 1061.7} = \sqrt{1086.7} \approx 32.97 \text{ N}$$ 7. **Direction of resultant:** Angle $\theta$ with positive x-axis: $$\theta = \tan^{-1} \left( \frac{-32.58}{5} \right) = \tan^{-1}(-6.516) \approx -81.3^\circ$$ This means 81.3° below the positive x-axis. 8. **Position of resultant force (line of action):** Calculate moments about A to find the point along x-axis where resultant acts. Moments of forces about A (taking counterclockwise as positive): - $M_A(F_A) = 0$ (force at A) - $M_A(F_B) = F_{By} \times x_B - F_{Bx} \times y_B = 0 \times 40 - (-60) \times 0 = 0$ - $M_A(F_C) = F_{Cx} \times y_C - F_{Cy} \times x_C = 65 \times 20\sqrt{3} - (-112.58) \times 20$ Calculate: $$65 \times 20\sqrt{3} = 1300 \sqrt{3} \approx 2251.66$$ $$-(-112.58) \times 20 = 2251.6$$ Sum: $$M_A(F_C) = 2251.66 + 2251.6 = 4503.26 \text{ N cm}$$ Moment of resultant about A: $$M_A(R) = R_y \times x - R_x \times y$$ Since resultant acts at unknown point $(x,0)$ on x-axis (y=0), $$M_A(R) = R_y \times x = -32.58 \times x$$ Set moments equal: $$-32.58 x = 4503.26 \Rightarrow x = \frac{-4503.26}{-32.58} = 138.2 \text{ cm}$$ 9. **Interpretation:** The resultant force of magnitude approximately 32.97 N acts at 81.3° below the positive x-axis, and its line of action crosses the x-axis at 138.2 cm from point A. **Final answers:** - Magnitude: $32.97$ N - Direction: $-81.3^\circ$ (below positive x-axis) - Position: $138.2$ cm from A along x-axis