For a block on a frictional incline with μ_s = 0.4, what is the maximum incline angle θ_max for which the block remains at rest without any other force along the plane?

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Multiple Choice

For a block on a frictional incline with μ_s = 0.4, what is the maximum incline angle θ_max for which the block remains at rest without any other force along the plane?

Explanation:
When a block sits on a frictional incline, static friction can oppose motion up to a maximum value equal to μ_s times the normal force. The normal force on a slope is N = mg cos θ, and the downslope component of gravity is mg sin θ. For the block to remain at rest without any other force along the plane, the downslope force must be balanced by static friction: mg sin θ ≤ μ_s mg cos θ. Cancel mg and rearrange to tan θ ≤ μ_s. The largest angle that still holds the block at rest is when this is an equality: tan θ = μ_s, so θ = arctan(μ_s). With μ_s = 0.4, θ_max = arctan(0.4) ≈ 21.8°. The other forms, like arcsin(μ_s) or arccos(μ_s), aren’t correct here because they don’t reflect how friction scales with the changing normal force as the angle changes. And tan^-1 is the same as arctan, just a different notation.

When a block sits on a frictional incline, static friction can oppose motion up to a maximum value equal to μ_s times the normal force. The normal force on a slope is N = mg cos θ, and the downslope component of gravity is mg sin θ. For the block to remain at rest without any other force along the plane, the downslope force must be balanced by static friction: mg sin θ ≤ μ_s mg cos θ. Cancel mg and rearrange to tan θ ≤ μ_s. The largest angle that still holds the block at rest is when this is an equality: tan θ = μ_s, so θ = arctan(μ_s). With μ_s = 0.4, θ_max = arctan(0.4) ≈ 21.8°. The other forms, like arcsin(μ_s) or arccos(μ_s), aren’t correct here because they don’t reflect how friction scales with the changing normal force as the angle changes. And tan^-1 is the same as arctan, just a different notation.

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