1. **Problem Statement:** Determine the reaction forces at points A, B, C, and D for two identical cylindrical rollers each of mass 10 kg and diameter 100 mm placed in a box with smooth contacts. 2. **Given Data:** - Mass of each roller, $m = 10$ kg - Diameter of each roller, $d = 100$ mm = 0.1 m - Radius, $r = \frac{d}{2} = 0.05$ m - Distance between centers of rollers = 50 mm = 0.05 m - Angle between horizontal and line connecting centers = 30° - Gravity, $g = 9.81$ m/s$^2$ 3. **Assumptions and Formulas:** - All contacts are smooth, so friction forces are zero. - Weight of each roller, $W = mg = 10 \times 9.81 = 98.1$ N acting vertically downward at the center. - Reaction forces at points A, B, C, D are normal forces. - Use equilibrium equations for forces and moments: $$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M = 0$$ 4. **Free Body Diagram and Forces:** - At point A: reaction force $R_A$ (horizontal wall contact on left roller) - At point B: reaction force $R_B$ (vertical wall contact on left roller) - At point C: reaction force $R_C$ (vertical wall contact on right roller) - At point D: reaction force $R_D$ (contact between rollers, acts along line connecting centers at 30° angle) 5. **Equilibrium for Left Roller:** - Horizontal forces: $$R_A - R_D \cos 30^\circ = 0 \implies R_A = R_D \cos 30^\circ$$ - Vertical forces: $$R_B + R_D \sin 30^\circ - W = 0 \implies R_B = W - R_D \sin 30^\circ$$ - Moment about center of left roller (taking counterclockwise positive): Since $R_D$ acts at radius $r$ at 30°, and $R_A$, $R_B$ act at the surface, moments balance implies: $$R_B \times r - R_A \times r - R_D \times 0 = 0 \implies R_B = R_A$$ 6. **From moment equation:** $$R_B = R_A$$ 7. **Substitute $R_A$ and $R_B$ from step 5:** $$W - R_D \sin 30^\circ = R_D \cos 30^\circ$$ $$W = R_D (\sin 30^\circ + \cos 30^\circ)$$ 8. **Calculate $R_D$:** $$\sin 30^\circ = 0.5, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$$ $$R_D = \frac{W}{0.5 + 0.866} = \frac{98.1}{1.366} \approx 71.8 \text{ N}$$ 9. **Calculate $R_A$ and $R_B$:** $$R_A = R_D \cos 30^\circ = 71.8 \times 0.866 = 62.2 \text{ N}$$ $$R_B = R_A = 62.2 \text{ N}$$ 10. **Right Roller Reactions:** - Weight $W = 98.1$ N downward - Reaction at C ($R_C$) vertical wall contact - Reaction at D ($R_D$) same magnitude but opposite direction to left roller contact - Vertical equilibrium: $$R_C + R_D \sin 30^\circ - W = 0 \implies R_C = W - R_D \sin 30^\circ = 98.1 - 71.8 \times 0.5 = 98.1 - 35.9 = 62.2 \text{ N}$$ - Horizontal equilibrium: The right roller has no horizontal wall contact, so horizontal force is balanced by $R_D \cos 30^\circ$ acting opposite to left roller. **Final Reaction Forces:** - $R_A = 62.2$ N (horizontal on left roller) - $R_B = 62.2$ N (vertical on left roller) - $R_D = 71.8$ N (contact between rollers along 30° line) - $R_C = 62.2$ N (vertical on right roller) These satisfy equilibrium for both rollers with smooth contacts.