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(A) To find the maximum angle $	heta$ for which the block remains at rest, consider the forces acting on the block of mass $m$:$1$. The gravitational

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$\mu=	an 	heta$

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(A) To find the maximum angle $	heta$ for which the block remains at rest, consider the forces acting on the block of mass $m$:$1$. The gravitational force $mg$ acts vertically downwards.$2$. The normal reaction $N$ acts perpendicular to the inclined surface.$3$. The static friction force $f_s$ acts up the incline to oppose the tendency of motion.Resolving the weight $mg$ into components:- Component perpendicular to the incline: $mg \cos 	heta$- Component parallel to the incline: $mg \sin 	heta$For equilibrium perpendicular to the incline:$N = mg \cos 	heta$For the block to remain motionless, the driving force must be less than or equal to the limiting friction:$mg \sin 	heta \leq f_{s, 	ext{max}}$Since $f_{s, 	ext{max}} = \mu N = \mu mg \cos 	heta$, we have:$mg \sin 	heta \leq \mu mg \cos 	heta$Dividing both sides by $mg \cos 	heta$ (assuming $\cos 	heta 
eq 0$):$	an 	heta \leq \mu$Thus, the maximum value of $	heta$ is given by $	an 	heta = \mu$.

$A$ block rests on a fixed wedge inclined at an angle $\theta$. The coefficient of friction between the block and the plane is $\mu$. The maximum value of $\theta$ for the block to remain motionless on the wedge is

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