1. **Problem statement:** A body weighing 50 N is on a rough inclined plane. A force $P$ acts up the slope along the line of greatest slope. The force $P$ that keeps the body in equilibrium lies in the interval $[20,30]$ N. We need to find the coefficient of static friction $\mu$ between the body and the plane. 2. **Set up the forces:** Let the angle of the incline be $\theta$. The weight $W=50$ N acts vertically downward. 3. **Resolve weight into components:** - Component down the slope: $W \sin \theta = 50 \sin \theta$ - Component perpendicular to slope: $W \cos \theta = 50 \cos \theta$ 4. **Equilibrium condition:** The force $P$ acts up the slope to balance the component of weight down the slope and friction. 5. **Friction force:** The friction force $F = \mu N = \mu (50 \cos \theta)$, acting up the slope to oppose motion down. 6. **Force balance along slope:** $$P + F = 50 \sin \theta$$ $$P + \mu (50 \cos \theta) = 50 \sin \theta$$ 7. **Rearranged:** $$P = 50 \sin \theta - 50 \mu \cos \theta$$ 8. **Given $P$ lies in $[20,30]$, so:** $$20 \leq 50 \sin \theta - 50 \mu \cos \theta \leq 30$$ 9. **Divide entire inequality by 50:** $$0.4 \leq \sin \theta - \mu \cos \theta \leq 0.6$$ 10. **We want to find $\mu$. To do this, we need $\theta$.** 11. **From the options, try to find $\theta$ that satisfies the interval for $P$.** 12. **Try option (a): $\mu = \frac{\sqrt{3}}{15} \approx 0.115$** 13. **Rewrite inequality:** $$0.4 \leq \sin \theta - 0.115 \cos \theta \leq 0.6$$ 14. **Try to find $\theta$ satisfying this.** 15. **Try $\theta = 30^\circ$ (common angle):** $$\sin 30^\circ = 0.5, \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$$ $$\sin \theta - \mu \cos \theta = 0.5 - 0.115 \times 0.866 = 0.5 - 0.0995 = 0.4005$$ 16. **This is within $[0.4,0.6]$, so option (a) is plausible.** 17. **Check other options quickly:** - (b) $\mu=0.5$ too large, would reduce $\sin \theta - \mu \cos \theta$ too much. - (c) $\mu=\frac{15}{\sqrt{3}} \approx 8.66$ too large. - (d) $\mu=\frac{1}{3} \approx 0.333$ larger than (a), likely outside interval. 18. **Therefore, the coefficient of static friction is $\boxed{\frac{\sqrt{3}}{15}}$.**