1. **Problem statement:** We have two blocks connected by a string over a pulley. Block P (mass $2m$) is on a rough inclined plane at angle $\alpha$ where $\tan \alpha = \frac{5}{12}$. Block Q (mass $5m$) hangs vertically. The system is released from rest, and block P moves up the plane. 2. **Equation of motion for Q:** Block Q moves vertically under gravity and tension $T$. Using Newton's second law: $$5mg - T = 5ma$$ where $a$ is the acceleration of the blocks. 3. **Find $T$ in terms of $m$ and $g$:** - Calculate $\sin \alpha$ and $\cos \alpha$ from $\tan \alpha = \frac{5}{12}$: $$\sin \alpha = \frac{5}{13}, \quad \cos \alpha = \frac{12}{13}$$ - Forces on block P along the plane: - Weight component down the plane: $2mg \sin \alpha = 2mg \times \frac{5}{13} = \frac{10mg}{13}$ - Friction force opposes motion (up the plane), friction coefficient $\mu = \frac{1}{8}$: - Normal reaction $N = 2mg \cos \alpha = 2mg \times \frac{12}{13} = \frac{24mg}{13}$ - Friction force $F = \mu N = \frac{1}{8} \times \frac{24mg}{13} = \frac{3mg}{13}$ - Equation of motion for P moving up the plane: $$T - \frac{10mg}{13} - \frac{3mg}{13} = 2ma$$ Simplify: $$T - \frac{13mg}{13} = 2ma \implies T - mg = 2ma$$ - From Q's equation: $$5mg - T = 5ma$$ - Add the two equations: $$(T - mg) + (5mg - T) = 2ma + 5ma$$ $$4mg = 7ma$$ - Solve for $a$: $$a = \frac{4g}{7}$$ - Substitute $a$ back to find $T$: $$T - mg = 2m \times \frac{4g}{7} = \frac{8mg}{7}$$ $$T = mg + \frac{8mg}{7} = \frac{15mg}{7}$$ 4. **Use of inextensible string:** The acceleration of both blocks is the same magnitude because the string length does not change. 5. **Force on the pulley:** - The string exerts tension $T$ in two directions: along the incline (on P) and vertically downward (on Q). - Components of tension on pulley: - Along incline: $T$ acts up the plane. - Vertically downward: $T$ acts downward. - Resolve tension along incline into horizontal and vertical components: - Horizontal: $T \cos \alpha = \frac{15mg}{7} \times \frac{12}{13} = \frac{180mg}{91}$ - Vertical (upwards): $T \sin \alpha = \frac{15mg}{7} \times \frac{5}{13} = \frac{75mg}{91}$ - Total vertical force on pulley is sum of vertical components: - From string on P side: upward $\frac{75mg}{91}$ - From string on Q side: downward $T = \frac{15mg}{7} = \frac{195mg}{91}$ - Net vertical force: $$\frac{75mg}{91} - \frac{195mg}{91} = -\frac{120mg}{91}$$ (downward) - Horizontal force: $$\frac{180mg}{91}$$ - Magnitude of total force on pulley: $$F = mg \sqrt{\left(\frac{180}{91}\right)^2 + \left(-\frac{120}{91}\right)^2} = mg \frac{\sqrt{180^2 + 120^2}}{91} = mg \frac{\sqrt{32400 + 14400}}{91} = mg \frac{\sqrt{46800}}{91}$$ - Simplify $\sqrt{46800}$: $$\sqrt{46800} = \sqrt{468 \times 100} = 10 \sqrt{468} \approx 10 \times 21.6333 = 216.333$$ - So: $$F \approx mg \times \frac{216.333}{91} = mg \times 2.379$$ - Therefore: $$k = 2.38 \text{ (3 significant figures)}$$