1. **Problem Statement:** Two identical cylindrical rollers each of mass 10 kg and diameter 100 mm (radius 50 mm) are placed in a box. The rollers touch the box at points A, B, and C, and each other at point D. All contacts are smooth (frictionless). We need to find the reaction forces at points A, B, C, and D. 2. **Known Data:** - Mass of each roller, $m = 10$ kg - Diameter = 100 mm = 0.1 m, so radius $r = 0.05$ m - Gravity acceleration, $g = 9.81$ m/s$^2$ - Angle at point D between line A-D and horizontal is $30^\circ$ 3. **Forces and Reactions:** - Weight of each roller: $W = mg = 10 \times 9.81 = 98.1$ N acting vertically downward at the center of each roller. - Reaction forces at points A, B, C, and D are unknown. 4. **Assumptions:** - All contacts are smooth, so reaction forces act normal to the surfaces. - The rollers are in static equilibrium. 5. **Set up equilibrium equations for each roller:** **Left roller:** - Forces: Reaction at A ($R_A$) horizontal to the left wall, reaction at B ($R_B$) vertical at bottom, reaction at D ($R_D$) at the contact with right roller. - Weight $W$ acts downward at center. **Right roller:** - Forces: Reaction at C ($R_C$) horizontal to the right wall, reaction at D ($R_D$) from left roller (equal and opposite to left roller's $R_D$), weight $W$ downward. 6. **Geometry and directions:** - Since the rollers touch at D, the line of action of $R_D$ is along the line connecting centers. - The angle between A-D and horizontal is $30^\circ$, so the line D-C is also at $30^\circ$ but mirrored. 7. **Equilibrium equations for left roller:** - Horizontal forces: $R_A - R_D \cos 30^\circ = 0$ - Vertical forces: $R_B + R_D \sin 30^\circ - W = 0$ - Moment about B (taking counterclockwise positive): $$R_A \times 0.1 - W \times 0.05 = 0$$ (since radius is 0.05 m, and distance from B to center is radius vertically) 8. **Solve moment equation for $R_A$:** $$R_A \times 0.1 = 98.1 \times 0.05$$ $$R_A = \frac{98.1 \times 0.05}{0.1} = 49.05 \text{ N}$$ 9. **Solve horizontal force equation for $R_D$:** $$49.05 - R_D \cos 30^\circ = 0 \Rightarrow R_D = \frac{49.05}{\cos 30^\circ} = \frac{49.05}{0.866} = 56.64 \text{ N}$$ 10. **Solve vertical force equation for $R_B$:** $$R_B + 56.64 \times \sin 30^\circ - 98.1 = 0$$ $$R_B + 56.64 \times 0.5 = 98.1$$ $$R_B + 28.32 = 98.1 \Rightarrow R_B = 69.78 \text{ N}$$ 11. **Right roller equilibrium:** - Horizontal: $R_C - R_D \cos 30^\circ = 0 \Rightarrow R_C = R_D \cos 30^\circ = 56.64 \times 0.866 = 49.05$ N - Vertical: $R_D \sin 30^\circ - W = 0 \Rightarrow 56.64 \times 0.5 - 98.1 = 0$ which is not balanced, so vertical reaction at bottom is zero for right roller (no bottom contact), so vertical equilibrium is just weight balanced by $R_D$ vertical component. 12. **Final reaction forces:** - $R_A = 49.05$ N (horizontal left wall) - $R_B = 69.78$ N (vertical bottom) - $R_C = 49.05$ N (horizontal right wall) - $R_D = 56.64$ N (contact between rollers at 30° angle) These forces satisfy static equilibrium for both rollers with smooth contacts.