If \(\theta\) is equal to \(\alpha\) at the time the box is just about to slide down the plank, the coefficient of static friction between the plank and box...

Answer Details

When the box is just about to slide down the plank, the force of static friction acting on the box is equal and opposite to the component of the weight of the box parallel to the surface of the plank. This component is given by mg sin\(\alpha\), where m is the mass of the box, g is the acceleration due to gravity and \(\alpha\) is the angle of inclination of the plank. Since \(\theta\) is equal to \(\alpha\), the angle of inclination of the plank is also \(\theta\). Therefore, the coefficient of static friction \(\mu\) is given by: \(\mu = \frac{\text{Force of friction}}{\text{Normal force}} = \frac{mg \sin \theta}{mg \cos \theta} = \tan \theta = \tan \alpha\) Hence, the correct option is tan\(\alpha\).