A particle of mass 10 kg is placed on an inclined plane at an angle of 30 degrees. The coefficient of static friction is 0.4. Determine the forces involved and whether the particle remains at rest.

To solve this problem, we first need to determine the components of the gravitational force acting on the particle parallel and perpendicular to the inclined plane, and then calculate the maximum static friction force.

Given:
Mass $m = 10 \text{ kg}$
Angle of inclination $\theta = 30^\circ$
Coefficient of static friction $\mu = 0.4$
Acceleration due to gravity $g = 10 \text{ m/s}^2$

Step 1: Calculate the components of the gravitational force and the normal force.
The weight of the particle is $W = mg$.

$$W = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100 \text{ N}$$

The component of the weight parallel to the plane (acting down the plane) is $W_{\parallel}$.

$$W_{\parallel} = W \sin \theta = 100 \text{ N} \times \sin 30^\circ = 100 \text{ N} \times 0.5 = 50 \text{ N}$$

The component of the weight perpendicular to the plane is $W_{\perp}$.

$$W_{\perp} = W \cos \theta = 100 \text{ N} \times \cos 30^\circ = 100 \text{ N} \times \frac{\sqrt{3}}{2} \approx 100 \text{ N} \times 0.866 = 86.6 \text{ N}$$

The normal force $N$ is equal to the perpendicular component of the weight.

$$N = W_{\perp} = 86.6 \text{ N}$$

The maximum static friction force $f_s$ is given by $\mu N$.

$$f_s = \mu N = 0.4 \times 86.6 \text{ N} = 34.64 \text{ N}$$