As the board is raised by an angle θ, the block will be held in place by 3 forces:
• its weight, pointing downward (magnitude w )
• the normal force of the board pushing up on the block, pointing in the direction perpendicular to the board (mag. n)
• static friction, pointing in the direction parallel to the board, opposite the direction the block would slip (mag. f )
Decompose the vectors into components that are parallel and perpendicular to the board (taking the direction of n to be the positive direction perpendicular to the board, and the direction of f to be the negative directoin parallel to it), so that by Newton's second law, we have
• net parallel force:
∑ F = w (//) + f
∑ F = m g sin(θ) - µ n = 0
where µ = 0.45 is the coefficient of static friction, g = 9.80 m/s² is the mag. of the acceleration due to gravity, and m = 15.0 kg is the mass of the block.
• net perpendicular force:
∑ F = n + w (⟂)
∑ F = n - m g cos(θ) = 0
Solve for n in the equation for net perpendicular force:
n = m g cos(θ)
Substitute this into the equation for net parallel force:
m g sin(θ) - 0.45 m g cos(θ) = 0
Solve for θ :
sin(θ) - 0.45 cos(θ) = 0
sin(θ) = 0.45 cos(θ)
tan(θ) = 0.45
θ = tan⁻¹(0.45)
θ ≈ 24.2°
So the maximum angle the board can be lifted before the block starts to slide is about 24.2°, since the coefficient of static friction µ is such that
f = µ n
where f is the maximum magnitude of the static friction force.