a. Consult the attached free body diagram.
As the board is lifted by some angle θ, the net forces acting on the box *just before* it begins to slide are
• parallel to the board
∑ F = mg sin(θ) - f = 0
• perpendicular to the board
∑ F = n - mg cos(θ) = 0
From the "perpendicular" equation,
n = mg cos(θ)
At this moment, static friction is maximized, and so
f = 0.46 n = 0.46 mg cos(θ)
Substitute this into the "parallel" equation and solve for θ :
mg sin(θ) - 0.46 mg cos(θ) = 0
sin(θ) - 0.46 cos(θ) = 0
sin(θ) = 0.46 cos(θ)
tan(θ) = 0.46
θ = arctan(0.46) ≈ 24.7°
b. Once the box starts sliding, the free body diagram stays the same, but now the net force acting on the box parallel to the board is no longer zero, and kinetic friction replaces static friction. So we have
• parallel to the board
∑ F = mg sin(θ) - f = ma
where a is the acceleration of the box, and
f = 0.32 n = 0.32 mg cos(θ)
Plug in the minimal angle for sliding to occur from part (a) and solve for a :
mg sin(θ) - 0.32 mg cos(θ) = ma
a = (sin(θ) - 0.32 cos(θ)) g
a = (sin(24.7°) - 0.32 cos(24.7°)) (9.80 m/s²)
a ≈ 1.25 m/s²
