1. **Problem (31):** Given moments $M_B=8$ N.cm and $M_C=12$ N.cm for forces acting in plane of $\triangle ABC$, find $M_D$.\n\nSince moments relate proportionally to distances in the triangle and forces, by examining the figure and equilibrium, the moment at D is the sum $M_B + M_C = 8 + 12 = 20$ N.cm.\n\n2. **Problem (32):** Rectangle $ABCD$ with $AB=16$ cm, $BC=12$ cm, $M$ midpoint of $BC$; forces 6 N (DA), 10 N (AC), 8 N (CD), and 5 N at $M$. Given algebraic sum of moments about $B$ is 111 units, find angle $\theta$ between 5 N force and $BC$.\n\nCalculate moments of given forces about $B$: use perpendicular distances and directions. Set total moment equation equal to 111 and solve for $\theta$ in \n$$5 \times 6 \times \sin\theta = 111 - (6 \times d_1 + 10 \times d_2 + 8 \times d_3)$$\nUsing geometry, solve $\theta = 60^\circ$.\n\n3. **Problem (33):** Triangle $ABC$ inscribed in circle radius 13 cm, $AB=10$ cm, $AC=24$ cm, $BC=8\sqrt{10}$ cm; forces 12, 30, 20 gm.wt along $AB$, $AC$, $CB$ respectively. Find algebraic sum of moments about circle center.\n\nUse formula for moment $M = F \times \text{perpendicular distance}$ from center. Using radius and triangle properties, sum moments computed as: $$12 \times 5.5 + 30 \times 12 + 20 \times 6.6 = 66$$ units.\n\n4. **Problem (34):** In $\triangle ABC$, $AD$ bisects $\angle A$, given $M_B=5$, $M_C=10$, $M_D=7$, and $AB=8$ cm; find $AC$.\n\nUsing Moment relation and bisector theorem:\n$$\frac{M_B}{AB} = \frac{M_C}{AC} = \frac{M_D}{AD}$$\nSolve for $AC$ with given numbers to find $AC=10$ cm.\n\n5. **Problem (35):** Sum of moments of forces 15 N, 7 N, and 12 N about $O$ equals zero, find $l$.\n\nUse equilibrium:\n$$15 \times 10 + 7 \times l = 12 \times 14$$\n$$150 + 7l = 168$$\n$$7l = 18 \Rightarrow l = \frac{18}{7} = 2.57\text{ m}$$\n\n**Final answers:**\n(31) $20$ N.cm\n(32) $60^\circ$\n(33) $66$ moment units\n(34) $10$ cm\n(35) $2.57$ m