1. **Problem statement:** Given two blocks A and B with masses $m_A = 30$ kg and $m_B = 70$ kg, and coefficient of friction $\mu_s = 0.1$, find the largest force $F$ that can be applied without causing the blocks to slip. 2. **Equilibrium equations:** For block A: $$\sum F_x: R - F \sin 30^\circ - \mu_s P \cos 20^\circ - P \sin 20^\circ = 0$$ $$\sum F_y: -\mu_s R - F \cos 30^\circ - m_A g + P \cos 20^\circ - \mu_s P \cos 20^\circ = 0$$ For block B: $$\sum F_x: P \sin 20^\circ - \mu_s P \cos 20^\circ - \mu_s N = 0$$ $$\sum F_y: -P \cos 20^\circ - \mu_s P \sin 20^\circ - m_B g + N = 0$$ 3. **Known values:** - $m_A = 30$ - $m_B = 70$ - $g = 9.8$ - $\mu_s = 0.1$ - Angles: $30^\circ$ and $20^\circ$ 4. **Substitute and solve:** Using the given values and solving the system yields: $$R = 212\ \mathrm{N},\quad P = 456\ \mathrm{N},\quad N = 1130\ \mathrm{N},\quad F = 197\ \mathrm{N}$$ 5. **Answer:** The largest force $F$ that can be applied without causing the blocks to slip is $$F = 197$$ This force ensures equilibrium without slipping, considering friction and angles involved.