A body of mass 12 kg is placed on an inclined plane with angle of inclination 30 degrees. The coefficient of friction is 2/3. Calculate the force P required to move the body down the plane.

The weight of the body is $W = mg$.

$$W = 12 \text{ kg} \times 10 \text{ ms}^{-2} = 120 \text{ N}$$

The component of weight perpendicular to the plane is $W_{\perp} = W \cos \theta$. This is balanced by the normal reaction force $R$.

$$R = W \cos \theta = 120 \cos 30^\circ = 120 \times \frac{\sqrt{3}}{2} = 60\sqrt{3} \text{ N}$$

The maximum static frictional force $F_f$ is given by $F_f = \mu R$.

$$F_f = \frac{2}{3} \times 60\sqrt{3} = 40\sqrt{3} \text{ N}$$

Using $\sqrt{3} \approx 1.732$:

$$F_f \approx 40 \times 1.732 = 69.28 \text{ N}$$

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The component of weight parallel to the plane, acting downwards, is $W_{\parallel} = W \sin \theta$.

$$W_{\parallel} = 120 \sin 30^\circ = 120 \times \frac{1}{2} = 60 \text{ N}$$

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When the body is at the point of moving down the plane, the frictional force $F_f$ acts up the plane, opposing the impending downward motion.
Since $W_{\parallel} (60 \text{ N})$ is less than $F_f (69.28 \text{ N})$, the body would not slide down on its own. Therefore, the external force $P$ must be acting down the plane to assist the motion.
For equilibrium (point of impending motion):

$$W_{\parallel} + P = F_f$$ $$P = F_f - W_{\parallel}$$ $$P = 40\sqrt{3} - 60$$ $$P \approx 69.28 - 60$$ $$P = \boxed{9.28 \text{ N}}$$

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When the body is at the point of moving up the plane, the frictional force $F_f$ acts down the plane, opposing the impending upward motion. The external force $P$ must be acting up the plane to cause this motion.
For equilibrium (point of impending motion):

$$P = W_{\parallel} + F_f$$ $$P = 60 + 40\sqrt{3}$$ $$P \approx 60 + 69.28$$ $$P = \boxed{129.28 \text{ N}}$$