1. **Problem Statement:** We have an isosceles triangle ABC with $m(\angle B) = 120^\circ$ and side $AC = 12\sqrt{3}$ cm. Forces of magnitudes 6, 7, and $8\sqrt{3}$ newtons act along sides $AC$, $CB$, and $AB$ respectively. We need to find the sum of the moments of these forces about the midpoint of $BC$. 2. **Understanding the problem:** - Since $ABC$ is isosceles with vertex angle $B = 120^\circ$, sides $AB = BC$. - The forces act along the sides, so their lines of action are along $AC$, $CB$, and $AB$. - Moment of a force about a point is $\text{Force} \times \text{perpendicular distance from the point to the line of action of the force}$. 3. **Find the lengths of sides:** - Given $AC = 12\sqrt{3}$ cm. - Since $ABC$ is isosceles with $AB = BC$, use Law of Cosines on triangle $ABC$: $$AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(120^\circ)$$ - But $AB = BC = x$ (say), and $AC = 12\sqrt{3}$. - So, $$x^2 = (12\sqrt{3})^2 + x^2 - 2 \cdot 12\sqrt{3} \cdot x \cdot \cos(120^\circ)$$ - Simplify: $$x^2 = 432 + x^2 - 2 \cdot 12\sqrt{3} \cdot x \cdot (-\frac{1}{2})$$ - Since $\cos 120^\circ = -\frac{1}{2}$, $$x^2 = 432 + x^2 + 12\sqrt{3} x$$ - Cancel $x^2$ on both sides: $$0 = 432 + 12\sqrt{3} x$$ - Rearranged: $$12\sqrt{3} x = -432$$ - This is impossible for positive length $x$, so re-examine the Law of Cosines setup. 4. **Correct Law of Cosines application:** - The angle $B$ is between sides $AB$ and $BC$. - So, $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(120^\circ)$$ - Since $AB = BC = x$, $$ (12\sqrt{3})^2 = x^2 + x^2 - 2 x^2 \cos(120^\circ)$$ - Simplify: $$432 = 2x^2 - 2x^2 \cdot (-\frac{1}{2}) = 2x^2 + x^2 = 3x^2$$ - So, $$x^2 = \frac{432}{3} = 144$$ - Therefore, $$x = 12$$ 5. **Coordinates for points:** - Place $B$ at origin $(0,0)$. - Let $BC$ lie along positive x-axis, so $C = (12,0)$. - Since $AB = 12$ and $\angle B = 120^\circ$, point $A$ is at: $$A = (12 \cos 120^\circ, 12 \sin 120^\circ) = (12 \times -\frac{1}{2}, 12 \times \frac{\sqrt{3}}{2}) = (-6, 6\sqrt{3})$$ 6. **Midpoint of BC:** $$M = \left( \frac{0+12}{2}, \frac{0+0}{2} \right) = (6,0)$$ 7. **Forces and their lines of action:** - Force along $AC$ of magnitude 6 N. - Force along $CB$ of magnitude 7 N. - Force along $AB$ of magnitude $8\sqrt{3}$ N. 8. **Calculate moments about $M$:** - Moment = Force $\times$ perpendicular distance from $M$ to line of action. 9. **Line equations:** - Line $AC$: passes through $A(-6,6\sqrt{3})$ and $C(12,0)$. - Vector $\overrightarrow{AC} = (18, -6\sqrt{3})$. - Equation of line $AC$ in form $Ax + By + C = 0$: - Slope $m = \frac{0 - 6\sqrt{3}}{12 - (-6)} = \frac{-6\sqrt{3}}{18} = -\frac{\sqrt{3}}{3}$. - Equation: $y - 6\sqrt{3} = -\frac{\sqrt{3}}{3}(x + 6)$. - Rearranged: $$y = -\frac{\sqrt{3}}{3} x - 2\sqrt{3} + 6\sqrt{3} = -\frac{\sqrt{3}}{3} x + 4\sqrt{3}$$ - Standard form: $$\frac{\sqrt{3}}{3} x + y - 4\sqrt{3} = 0$$ - Multiply by 3: $$\sqrt{3} x + 3 y - 12 \sqrt{3} = 0$$ - Line $CB$: points $C(12,0)$ and $B(0,0)$. - Line along x-axis: $y=0$. - Line $AB$: points $A(-6,6\sqrt{3})$ and $B(0,0)$. - Vector $\overrightarrow{AB} = (6, -6\sqrt{3})$. - Slope $m = \frac{0 - 6\sqrt{3}}{0 - (-6)} = \frac{-6\sqrt{3}}{6} = -\sqrt{3}$. - Equation: $$y - 0 = -\sqrt{3} (x - 0) \Rightarrow y = -\sqrt{3} x$$ - Standard form: $$\sqrt{3} x + y = 0$$ 10. **Calculate perpendicular distances from $M(6,0)$ to each line:** - Distance formula: $$d = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}$$ - For $AC$ line: $\sqrt{3} x + 3 y - 12 \sqrt{3} = 0$ - $A = \sqrt{3}$, $B=3$, $C = -12 \sqrt{3}$ - Substitute $M(6,0)$: $$|\sqrt{3} \times 6 + 3 \times 0 - 12 \sqrt{3}| = |6\sqrt{3} - 12 \sqrt{3}| = 6 \sqrt{3}$$ - Denominator: $$\sqrt{(\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2 \sqrt{3}$$ - Distance: $$d_{AC} = \frac{6 \sqrt{3}}{2 \sqrt{3}} = 3$$ - For $CB$ line: $y=0$ - Distance from $M(6,0)$ to $y=0$ is 0 (point lies on the line). - For $AB$ line: $\sqrt{3} x + y = 0$ - $A=\sqrt{3}$, $B=1$, $C=0$ - Substitute $M(6,0)$: $$|\sqrt{3} \times 6 + 1 \times 0 + 0| = 6 \sqrt{3}$$ - Denominator: $$\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$$ - Distance: $$d_{AB} = \frac{6 \sqrt{3}}{2} = 3 \sqrt{3}$$ 11. **Calculate moments:** - Moment is positive if force tends to rotate counterclockwise about $M$, negative if clockwise. - Determine direction of each force's moment: - Force along $AC$ (6 N): line passes through $A$ and $C$, force acts along $AC$. - Force along $CB$ (7 N): along $CB$, which is x-axis from $C$ to $B$. - Force along $AB$ ($8\sqrt{3}$ N): along $AB$. - Since $M$ is midpoint of $BC$, forces along $CB$ pass through $M$ (distance zero), so moment is zero. - For $AC$ and $AB$, find moment direction by vector cross product of position vector from $M$ to line and force direction. 12. **Force directions as unit vectors:** - $\overrightarrow{AC} = (18, -6\sqrt{3})$ - Magnitude: $$\sqrt{18^2 + (-6\sqrt{3})^2} = \sqrt{324 + 108} = \sqrt{432} = 12 \sqrt{3}$$ - Unit vector: $$\hat{u}_{AC} = \left( \frac{18}{12\sqrt{3}}, \frac{-6\sqrt{3}}{12\sqrt{3}} \right) = \left( \frac{3}{2\sqrt{3}}, -\frac{1}{2} \right)$$ - $\overrightarrow{AB} = (6, -6\sqrt{3})$ - Magnitude: $$\sqrt{6^2 + (-6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12$$ - Unit vector: $$\hat{u}_{AB} = \left( \frac{6}{12}, \frac{-6\sqrt{3}}{12} \right) = \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right)$$ - $\overrightarrow{CB} = (0 - 12, 0 - 0) = (-12, 0)$ - Unit vector: $$\hat{u}_{CB} = (-1, 0)$$ 13. **Position vectors from $M$ to points:** - $M = (6,0)$ - Vector $\overrightarrow{MA} = A - M = (-6 - 6, 6\sqrt{3} - 0) = (-12, 6\sqrt{3})$ - Vector $\overrightarrow{MC} = C - M = (12 - 6, 0 - 0) = (6, 0)$ - Vector $\overrightarrow{MB} = B - M = (0 - 6, 0 - 0) = (-6, 0)$ 14. **Calculate moments (torques) about $M$:** - Moment $\vec{\tau} = \vec{r} \times \vec{F}$ - Force vector $\vec{F} = F \hat{u}$ - For force along $AC$ (6 N): $$\vec{F}_{AC} = 6 \times \hat{u}_{AC} = 6 \times \left( \frac{3}{2\sqrt{3}}, -\frac{1}{2} \right) = \left( \frac{18}{2\sqrt{3}}, -3 \right) = \left( \frac{9}{\sqrt{3}}, -3 \right)$$ - Moment: $$\vec{\tau}_{AC} = \overrightarrow{MA} \times \vec{F}_{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -12 & 6\sqrt{3} & 0 \\ \frac{9}{\sqrt{3}} & -3 & 0 \end{vmatrix}$$ - Only $k$ component matters: $$\tau_{AC} = (-12)(-3) - (6\sqrt{3}) \left( \frac{9}{\sqrt{3}} \right) = 36 - 54 = -18$$ - For force along $CB$ (7 N): $$\vec{F}_{CB} = 7 \times (-1, 0) = (-7, 0)$$ - Moment: $$\vec{\tau}_{CB} = \overrightarrow{MC} \times \vec{F}_{CB} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 0 & 0 \\ -7 & 0 & 0 \end{vmatrix} = 0$$ - For force along $AB$ ($8\sqrt{3}$ N): $$\vec{F}_{AB} = 8\sqrt{3} \times \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) = \left( 4\sqrt{3}, -12 \right)$$ - Moment: $$\vec{\tau}_{AB} = \overrightarrow{MB} \times \vec{F}_{AB} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -6 & 0 & 0 \\ 4\sqrt{3} & -12 & 0 \end{vmatrix}$$ - $k$ component: $$(-6)(-12) - 0 \times 4\sqrt{3} = 72$$ 15. **Sum of moments about $M$:** $$\tau_{total} = \tau_{AC} + \tau_{CB} + \tau_{AB} = -18 + 0 + 72 = 54$$ 16. **Interpretation:** - The positive value indicates the net moment tends to rotate counterclockwise about $M$. **Final answer:** $$\boxed{54 \text{ newton-centimeters}}$$