1. **Problem statement:** A light ladder leans against a smooth vertical wall at an angle of 30° to the horizontal. A load of 800 N is placed three-quarters of the way up the ladder. The ladder rests on a rough horizontal surface preventing slipping. Find the magnitude and direction of the reaction between the ladder and the ground. 2. **Identify forces and geometry:** - Ladder angle with horizontal: $30^\circ$ - Load: 800 N at 3/4 length of ladder - Reaction force at ground: unknown magnitude $R$ and direction angle $\theta$ to horizontal 3. **Set up equilibrium conditions:** - Let the ladder length be $L$. - The load acts at $\frac{3}{4}L$ from the base. - The wall is smooth, so horizontal reaction at wall is $H$. - Vertical reaction at ground is $V$. 4. **Resolve forces:** - Horizontal reaction at ground is $H$ (friction force preventing slipping). - Vertical reaction at ground is $V$. - Reaction force magnitude $R = \sqrt{H^2 + V^2}$. - Direction $\theta = \tan^{-1}(\frac{V}{H})$. 5. **Sum of vertical forces:** $$ V = 800 \text{ N} $$ 6. **Sum of horizontal forces:** - Horizontal reaction at wall $H$ equals friction force at ground. 7. **Moment equilibrium about base:** - Taking moments about base to eliminate reaction at ground: - Moment due to load: $800 \times \frac{3}{4}L \times \cos 30^\circ$ - Moment due to horizontal reaction at wall $H \times L \times \sin 30^\circ$ Set moments equal: $$ 800 \times \frac{3}{4}L \times \cos 30^\circ = H \times L \times \sin 30^\circ $$ Simplify $L$: $$ 800 \times \frac{3}{4} \times \cos 30^\circ = H \times \sin 30^\circ $$ Calculate numeric values: $$ 800 \times 0.75 \times 0.866 = H \times 0.5 $$ $$ 519.6 = 0.5 H $$ $$ H = 1039.2 \text{ N} $$ 8. **Calculate magnitude of reaction $R$ at ground:** $$ R = \sqrt{H^2 + V^2} = \sqrt{1039.2^2 + 800^2} $$ $$ R = \sqrt{1,079,940 + 640,000} = \sqrt{1,719,940} = 1311.4 \text{ N} $$ 9. **Calculate direction $\theta$ of reaction $R$ to horizontal:** $$ \theta = \tan^{-1}\left(\frac{V}{H}\right) = \tan^{-1}\left(\frac{800}{1039.2}\right) $$ $$ \theta = \tan^{-1}(0.77) = 37.9^\circ $$ 10. **Final answer:** The reaction between the ladder and the ground has magnitude approximately $1311$ N and is directed at about $38^\circ$ to the horizontal. This matches the given answer of 1.31 kN at 38° to the horizontal.