A board leaning against a wall

A uniform board is leaning against a smooth vertical wall. The board is at an angle above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is 0.680. Find the smallest value for the angle , such that the lower end of the board does not slide along the ground.

I answered a similar question a few months ago (see ref. 1). I'll adapt the answer I gave there. We'll use the same formulas, except that there we needed to calculate the μs between the ground and the board, and here we need to solve for θ.

Approach:

Analyze the forces and moments on the system just before the board begins to slip. At this point in time the forces and torques must be in balance, since nothing is moving. Then solve for θ in terms of μs, the coefficient of static friction. There is an illustration with some limited animation at reference 2; I'll use the same notation, even if it seems non-standard in places.

At the base of the board, there are two forces operating:
An upward force normal to the ground Fn supporting the board; and
a horizontal frictional force H that keeps the board from slipping.

At the top of the ladder, there is a force F normal to the wall to keep that part of the board from moving.

Let:
L = the length of the board
W = the weight of the board
θ = the angle the board makes with the ground

Both horizontal and vertical forces must balance; therefore
(1) H = F
(2) Fn = W

The torques about the axis where the board meets the ground must also balance:
(3) F * L * sin(θ) = W * 0.5L * cos(θ)

The static friction coefficient is defined as:
(4) μs = H / Fn

Solving (3) for θ:

(5) W * 0.5L * cos(θ) = F * L * sin(θ) ⇒

(6) cos(θ) / sin(θ) = F * L / (W * 0.5L )

= F / (W * 0.5 )

= 2 * F / W

Substituting (1) and (2) into (6):