1. **State the problem:** We have three blocks P, Q, and R with masses 25 kg, 20 kg, and m kg respectively, held in equilibrium by strings OP, OQ, and OR passing over pulleys A and B. The angles given are 30° between OA and the vertical, and $\alpha$° between BOQ. We need to find the value of $m$ and $\alpha$. 2. **Set up the forces:** Each block's weight acts vertically downward: - Weight of P: $W_P = 25g$ - Weight of Q: $W_Q = 20g$ - Weight of R: $W_R = mg$ where $g$ is acceleration due to gravity. 3. **Analyze tensions and directions:** Since the system is in equilibrium, the vector sum of tensions at point O is zero. - Tension in OP acts along OA at 30° to vertical. - Tension in OQ acts vertically downward. - Tension in OR acts along OB at angle $\alpha$ to vertical. 4. **Express tensions:** Tensions equal weights of hanging masses: - $T_{OP} = 25g$ - $T_{OQ} = 20g$ - $T_{OR} = mg$ 5. **Resolve tensions into components:** Taking vertical direction as y-axis and horizontal as x-axis: - $T_{OP}$ components: - Horizontal: $25g \sin 30^\circ = 25g \times 0.5 = 12.5g$ - Vertical: $25g \cos 30^\circ = 25g \times \frac{\sqrt{3}}{2} = 21.65g$ - $T_{OQ}$ components: - Horizontal: 0 - Vertical: $-20g$ (downward) - $T_{OR}$ components: - Horizontal: $mg \sin \alpha$ - Vertical: $-mg \cos \alpha$ (downward) 6. **Apply equilibrium conditions:** Sum of horizontal components = 0: $$12.5g - mg \sin \alpha = 0 \implies mg \sin \alpha = 12.5g$$ Sum of vertical components = 0: $$21.65g - 20g - mg \cos \alpha = 0 \implies 1.65g = mg \cos \alpha$$ 7. **Simplify equations:** Divide both equations by $g$: $$m \sin \alpha = 12.5$$ $$m \cos \alpha = 1.65$$ 8. **Find $m$ and $\alpha$:** Square and add: $$m^2 (\sin^2 \alpha + \cos^2 \alpha) = 12.5^2 + 1.65^2$$ $$m^2 = 156.25 + 2.7225 = 158.9725$$ $$m = \sqrt{158.9725} \approx 12.61$$ Find $\alpha$: $$\tan \alpha = \frac{12.5}{1.65} \approx 7.58$$ $$\alpha = \tan^{-1}(7.58) \approx 82.5^\circ$$ **Final answers:** - $m \approx 12.6$ kg - $\alpha \approx 82.5^\circ$