If the co-efficient of static friction is 0.75, the angle through an i

1.	If the co-efficient of static friction is 0.75, the angle through an inclined plane must be raised sothat the block placed on it may begin to slide is sin 375
Answer» $37^{\circ}$ Answer: Explanation: Given Coefficient of static friction is $\mu =0.75$ Block will REMAIN in EQUILIBRIUM until $\tan \theta \leq \mu$ At CRITICAL point $\tan \theta =0.75$ $\theta =\tan ^{-1}(0.75)$ $\theta =36.86\approx 37^{\circ}$ after $\theta$ is greater than $37^{\circ}$ block BEGIN to SLIDE