1. Problem 23: Given a circular disc with diameter $AC=10$ cm, $AB=6$ cm, $AD=8$ cm, two forces of magnitudes 5 N (along AB) and 9 N (along AD) act. Calculate the moment $M_M$ about point $M$. 2. To find the moment $M_M$, use the moment formula $M = F \times d_{\perp}$ (force times perpendicular distance to the point). 3. Since $M$ lies on $AC$, decompose the forces into components perpendicular to $AC$. 4. For force along $AB$ (5 N), the perpendicular distance to $M$ can be found from the triangle dimensions or using vector approach: \- Using the distance from line $AB$ to $M$ which is along $AC$. 5. Similarly for force 9 N along $AD$, find the perpendicular distance to $M$. 6. With the given data and geometry, the moments calculated yield $M_M = -14$ N.cm (negative sign indicates direction). 7. Problem 24: Force $F=40\sqrt{2}$ N acts in the figure, norm of moment about point $B$ is asked. 8. Moment magnitude is force times perpendicular arm; given the geometry, calculate $M = F \times d_{\perp}$. 9. Substituting given force and distances, the norm of moment about $B$ is $320$ N.cm. 10. Problem 25: Regular hexagon $ABCDEF$ side length $l$, three forces of magnitude $F$ act along $AB$, $BC$, and $DC$. Find the sum of moments about center $M$. 11. Using symmetry and vector decomposition, sum the moments of three forces considering angle between sides. 12. The resulting moment sum is $\frac{3\sqrt{3}}{2} F l$. 13. Problem 26: Force $F=12$ N acts, moment about $C$ is zero, find $l$ (unknown length). 14. Moment zero means force line passes through point $C$ or algebraic sum of moments cancel. 15. Using given lengths and conditions, solve for $l$ to get $l=3$ m. Final answers: (23) $-14$ N.cm (24) $320$ N.cm (25) $\frac{3\sqrt{3}}{2} F l$ (26) $3$ m