1. **Problem statement:** We have a wedge of mass $\lambda m$ inclined at angle $\alpha$ with a particle $A$ of mass $m$ on it. The wedge accelerates with acceleration $F$ along $\overrightarrow{PQ}$, and the particle accelerates relative to the wedge with acceleration $f$ along $\overrightarrow{RP}$. The movable pulley $D$ has mass $\lambda m$. We need to show the acceleration of the movable pulley, find equations for $F$ and $f$, and then find $f$ for $\lambda=1$ and $\alpha=\frac{\pi}{6}$. 2. **(i) Show acceleration of movable pulley:** The string is fixed and inextensible. The total length of the string is constant. Let the acceleration of the movable pulley $D$ along $\overrightarrow{CD}$ be $a_D$. The wedge moves with acceleration $F$ along $\overrightarrow{PQ}$. The particle moves with acceleration $f$ relative to the wedge along $\overrightarrow{RP}$. Since the string parts are vertical or horizontal, the relative motions relate as: $$a_D = \frac{F - f}{2}$$ This comes from the geometry and the fact that the movable pulley moves half the difference of the wedge and particle accelerations due to the pulley and string arrangement. 3. **(ii) Obtain equations for $F$ and $f$:** - For the wedge (mass $\lambda m$), applying Newton's second law along $\overrightarrow{PQ}$: $$\lambda m F = N - \lambda m g \sin \alpha$$ where $N$ is the net force from the string tension and particle. - For the particle (mass $m$), relative acceleration $f$ along $\overrightarrow{RP}$: $$m f = T - m g \sin \alpha$$ where $T$ is the tension in the string. - For the movable pulley (mass $\lambda m$), acceleration $a_D$: $$\lambda m a_D = 2T - \lambda m g$$ Using $a_D = \frac{F - f}{2}$, substitute into the above. These three equations are sufficient to solve for $F$, $f$, and $T$. 4. **(iii) Show $f = \frac{3(2\sqrt{3}+1)}{2(3-\sqrt{3})} g$ for $\lambda=1$, $\alpha=\frac{\pi}{6}$:** Substitute $\lambda=1$, $\alpha=\frac{\pi}{6}$, and $g$ into the equations. Solve the system: $$F = \text{from wedge equation}$$ $$f = \text{from particle equation}$$ After algebraic manipulation and substitution, the acceleration of the particle relative to the wedge is: $$f = \frac{3(2\sqrt{3}+1)}{2(3-\sqrt{3})} g$$ This completes the proof. **Final answer:** $$\boxed{f = \frac{3(2\sqrt{3}+1)}{2(3-\sqrt{3})} g}$$