Let w denote the weight of the box, n the magnitude of the normal force (the sheet pushing up on the box), k the magnitude of the kinetic friction force, and µ the coefficient of kinetic friction.
The kinetic friction force doesn't kick in until the sheet is tilted at an angle θ₂. Since the box is sliding down at constant speed, it's in equilibrium and the forces acting on it sum to 0.
In the horizontal direction (literally left-and-right, parallel to the ground and not the sheet), we have only the horizontal components of the friction and normal forces to worry about, so that [tex]\mathbf n_x+\mathbf f_x=\mathbf0[/tex], which reduces to
f cos(θ₂) + n cos(90º + θ₂) = 0
(where we take the right to be positive and left to be negative)
Recall that cos(90º + x) = -sin(x) for all x, and that the friction force is proportional to the normal force by a factor of µ. This gives
µ n cos(θ₂) - n sin(θ₂) = 0
Solve for µ :
µ n cos(θ₂) = n sin(θ₂)
µ = sin(θ₂) / cos(θ₂)
µ = tan(θ₂)
