Determine the magnitude & direction of the friction force exerted by the roof surface on wooden block& normal force exerted by the roof on the wooden block.

Angle of plane, $θ = \tan^{-1}(4/12) = 18.4$

Solution:

FBD can be drawn as follows

Taking Y and X directions as shown,

$∑Fy = 0 \\ N – W\cos18.4 = 0 \\ N – (10.2 \times 9.8) \cos 18.4 = 0 $

(Mass is taken as 10.2 kg because both the box and block will act as one body, assuming the box doesn’t slide)

$N = 94.85 N ↗ 71.60 $ (Normal reaction is perpendicular to plane. Hence angle $= 90 – θ)\\ ∑Fx = 0 \\ W\sin 18.4 – F = 0\\ F = (10.2 \times 9.8) \times \sin18.4 = 31.55 N ↖ θ = 18.4$