1. **Problem statement:** A ladder of length 10 m and weight 500 N leans against a vertical wall, resting on a horizontal surface. The coefficient of friction at both contact points (A on the wall and B on the ground) is $\mu=0.2$. A man weighing 1200 N climbs the ladder. We need to find the maximum height he can reach before the ladder slips. 2. **Known data:** - Ladder length $L=10$ m - Ladder weight $W_l=500$ N - Man's weight $W_m=1200$ N - Coefficient of friction $\mu=0.2$ - Height of wall $h=6$ m - Ladder is inclined at angle $\theta$ with the horizontal 3. **Find the angle $\theta$:** Since the ladder forms a right triangle with the wall and ground, and the vertical height is 6 m, horizontal distance $x$ can be found by Pythagoras: $$x=\sqrt{L^2 - h^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \text{ m}$$ Then, $$\theta = \arcsin\left(\frac{h}{L}\right) = \arcsin\left(\frac{6}{10}\right) = 36.87^\circ$$ 4. **For equilibrium and no slipping:** The friction force at the base must balance the tendency of the ladder to slip. The maximum friction force is $$F_f = \mu N_B$$ where $N_B$ is the normal force at the base. 5. **Forces acting:** - Ladder weight $W_l$ acts at the midpoint (5 m from base) - Man's weight $W_m$ acts at a distance $d$ along the ladder from the base - Normal forces at wall ($N_A$) and ground ($N_B$) - Friction force $F_f$ at base opposes slipping 6. **Sum of vertical forces:** $$N_B = W_l + W_m$$ 7. **Sum of horizontal forces:** $$N_A = F_f = \mu N_B$$ 8. **Taking moments about point B (base):** Clockwise moments = Counterclockwise moments - Moment due to ladder weight: $$M_l = W_l \times \left(\frac{L}{2} \cos \theta\right) = 500 \times 5 \times \cos 36.87^\circ = 500 \times 5 \times 0.8 = 2000 \text{ Nm}$$ - Moment due to man weight at distance $d$: $$M_m = W_m \times d \times \cos \theta = 1200 \times d \times 0.8 = 960 d \text{ Nm}$$ - Moment due to wall normal force $N_A$ acting at height $L \sin \theta = 10 \times 0.6 = 6$ m: $$M_A = N_A \times L \sin \theta = \mu N_B \times 6 = 0.2 (W_l + W_m) \times 6 = 0.2 (500 + 1200) \times 6 = 0.2 \times 1700 \times 6 = 2040 \text{ Nm}$$ 9. **Equating moments:** $$M_A = M_l + M_m$$ $$2040 = 2000 + 960 d$$ 10. **Solve for $d$:** $$960 d = 2040 - 2000 = 40$$ $$d = \frac{40}{960} = 0.0417 \text{ m}$$ 11. **Interpretation:** $d$ is the distance along the ladder from the base to the man. The maximum height $h_m$ the man can reach is: $$h_m = d \sin \theta = 0.0417 \times 0.6 = 0.025 \text{ m}$$ This is very small, indicating the ladder will slip almost immediately as the man starts climbing. 12. **Check if friction at wall is limiting:** If friction at wall is negligible, the limiting friction is at base only. The problem states friction at both surfaces, so this is consistent. **Final answer:** The maximum height the man can reach before the ladder slips is approximately **0.025 m** (2.5 cm) from the base.