1. **Problem statement:** Two blocks A and B with masses 10 kg and 6 kg respectively are placed on a 30° inclined plane connected by a link. The coefficients of kinetic friction are $\mu_A=0.1$ and $\mu_B=0.3$. We need to find the force developed in the link when the blocks are released. 2. **Forces acting on each block:** - Weight components along the incline: $W_A = m_A g \sin 30^\circ$, $W_B = m_B g \sin 30^\circ$ - Friction forces opposing motion: $f_A = \mu_A m_A g \cos 30^\circ$, $f_B = \mu_B m_B g \cos 30^\circ$ - Tension in the link: $T$ 3. **Assuming block A moves down and block B moves up the incline:** - For block A (down the slope): $m_A g \sin 30^\circ - f_A - T = m_A a$ - For block B (up the slope): $T - m_B g \sin 30^\circ - f_B = m_B a$ 4. **Calculate weight components and friction:** - $g = 9.8$ m/s$^2$ - $W_A = 10 \times 9.8 \times \sin 30^\circ = 10 \times 9.8 \times 0.5 = 49$ N - $W_B = 6 \times 9.8 \times 0.5 = 29.4$ N - $f_A = 0.1 \times 10 \times 9.8 \times \cos 30^\circ = 0.1 \times 10 \times 9.8 \times 0.866 = 8.48$ N - $f_B = 0.3 \times 6 \times 9.8 \times 0.866 = 15.22$ N 5. **Write equations of motion:** $$ 10a = 49 - 8.48 - T = 40.52 - T $$ $$ 6a = T - 29.4 - 15.22 = T - 44.62 $$ 6. **Add the two equations to eliminate $T$:** $$ 10a + 6a = 40.52 - T + T - 44.62 $$ $$ 16a = -4.1 $$ $$ a = \frac{-4.1}{16} = -0.25625 \text{ m/s}^2 $$ 7. **Find tension $T$ using block A's equation:** $$ 10(-0.25625) = 40.52 - T $$ $$ -2.5625 = 40.52 - T $$ $$ T = 40.52 + 2.5625 = 43.08 \text{ N} $$ **Final answer:** The force developed in the link is approximately **43.1 N**.