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

Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 04

Years : MAY 2015

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 x 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 x 9.8) x \sin 18.4 = 31.55 N ↖ θ = 18.4 $