Friction On An Inclined Plane

Drawing on NCERT Class 11, Chapter 5 (Laws of Motion), the angle at which a block on an incline is just about to slide is called the angle of repose, and at this angle the limiting friction exactly balances the gravity component along the slope. Setting mg sinθ equal to μ_s mg cosθ, the mass and g cancel, leaving μ_s = tanθ. This elegant result means the coefficient of static friction equals the tangent of the angle of repose. Substituting the given angle, μ_s = tan45° = 1.0. Option 0.5 has no basis and would correspond to about a 27° angle. Option 0.707 mistakenly uses sin45° or cos45° instead of the tangent. Option 1.414 uses √2, confusing 1/cos45° with the friction coefficient. A valuable observation is that the angle of repose is completely independent of the block's mass, since both the driving and resisting forces scale with weight and cancel; a heavier block tips at exactly the same angle as a lighter one of the same material. This is why piles of sand or grain naturally settle to a characteristic slope determined only by the material's friction. Plausibility check: the coefficient is dimensionless, and tan45° equalling exactly 1 corresponds to the special case where the sliding and balancing components are equal, which is the textbook angle-of-repose result, confirming the answer.