1. **Problem statement:** A uniform ladder of length $L$ leans against a building. A person of mass $m=75$ kg stands on it. The ladder's mass is $M=10$ kg. The coefficients of static friction are $\mu_{s1}=0.5$ (ground) and $\mu_{s2}=0.3$ (wall). Find the minimum angle $\theta$ so the ladder does not slip. 2. **Key concepts and formulas:** - Forces acting: weight of ladder $Mg$, weight of person $mg$, friction forces at ground and wall, normal forces at ground and wall. - Static friction force $f_s \leq \mu_s N$ where $N$ is the normal force. - Equilibrium conditions: sum of forces and torques must be zero. 3. **Set up forces:** - Let the ladder make angle $\theta$ with the ground. - Normal force at ground: $N_1$ vertical reaction. - Friction force at ground: $f_1$ horizontal, preventing slipping. - Normal force at wall: $N_2$ horizontal reaction. - Friction force at wall: $f_2$ vertical. 4. **Force equilibrium:** - Horizontal: $N_2 = f_1$ - Vertical: $N_1 + f_2 = Mg + mg$ 5. **Friction limits:** - $f_1 \leq \mu_{s1} N_1$ - $f_2 \leq \mu_{s2} N_2$ 6. **Torque equilibrium about base (point where ladder touches ground):** - Taking counterclockwise positive: - Torque by ladder weight: $\frac{L}{2} Mg \cos\theta$ (clockwise, negative) - Torque by person weight: $d mg \cos\theta$ where $d$ is distance along ladder from base - Torque by wall normal force: $L N_2 \sin\theta$ (counterclockwise, positive) Set sum to zero: $$L N_2 \sin\theta = \frac{L}{2} Mg \cos\theta + d mg \cos\theta$$ 7. **Assuming person stands at top of ladder, $d=L$:** $$N_2 = \frac{Mg/2 + mg}{\tan\theta}$$ 8. **From horizontal equilibrium:** $$f_1 = N_2$$ 9. **From friction limit at ground:** $$f_1 \leq \mu_{s1} N_1 \Rightarrow N_2 \leq \mu_{s1} N_1$$ 10. **From vertical equilibrium:** $$N_1 + f_2 = Mg + mg$$ 11. **From friction limit at wall:** $$f_2 \leq \mu_{s2} N_2$$ 12. **At slipping threshold, friction forces reach maximum:** $$f_1 = \mu_{s1} N_1, \quad f_2 = \mu_{s2} N_2$$ 13. **Substitute into vertical equilibrium:** $$N_1 + \mu_{s2} N_2 = Mg + mg$$ 14. **From horizontal friction limit:** $$N_2 = \mu_{s1} N_1 \Rightarrow N_1 = \frac{N_2}{\mu_{s1}}$$ 15. **Substitute $N_1$ into vertical equilibrium:** $$\frac{N_2}{\mu_{s1}} + \mu_{s2} N_2 = Mg + mg$$ $$N_2 \left( \frac{1}{\mu_{s1}} + \mu_{s2} \right) = Mg + mg$$ 16. **Solve for $N_2$:** $$N_2 = \frac{Mg + mg}{\frac{1}{\mu_{s1}} + \mu_{s2}}$$ 17. **Recall from torque equilibrium:** $$N_2 = \frac{Mg/2 + mg}{\tan\theta}$$ 18. **Equate expressions for $N_2$ and solve for $\tan\theta$:** $$\frac{Mg/2 + mg}{\tan\theta} = \frac{Mg + mg}{\frac{1}{\mu_{s1}} + \mu_{s2}}$$ 19. **Rearranged:** $$\tan\theta = \frac{(Mg/2 + mg) \left( \frac{1}{\mu_{s1}} + \mu_{s2} \right)}{Mg + mg}$$ 20. **Plug in values:** - $g=9.8$ m/s$^2$ - $M=10$, $m=75$ - $\mu_{s1}=0.5$, $\mu_{s2}=0.3$ Calculate numerator: $$Mg/2 + mg = 10 \times 9.8 / 2 + 75 \times 9.8 = 49 + 735 = 784$$ Calculate denominator: $$Mg + mg = 98 + 735 = 833$$ Calculate friction term: $$\frac{1}{0.5} + 0.3 = 2 + 0.3 = 2.3$$ 21. **Calculate $\tan\theta$:** $$\tan\theta = \frac{784 \times 2.3}{833} = \frac{1803.2}{833} \approx 2.164$$ 22. **Find $\theta$:** $$\theta = \arctan(2.164) \approx 65.3^\circ$$ **Final answer:** The minimum angle $\theta$ so the ladder does not slip is approximately **65.3 degrees**.