Find the component of force on the inclined plane:
A small block of weight 300N is placed on the inclined plane, which makes an angle 60° with the horizontal. Find the component of this weight?
(i) Parallel to inclined plane
(ii) Perpendicular to inclined plane. As shown in the figure given below
.: First draw a line perpendicular to the inclined plane, and parallel to the inclined plane
∑H = Sum of Horizontal Component
= Perpendicular to plane
= 300cos60° = 150N .......ANS
∑V = Sum of Vertical Component
= Parallel to plane
= 300sin60°= 259.81N .......ANS
NOTE: There is no confusion about cosθ and sinθ , the angle 'θ' made by which plane, the component of force on that plane contain cosθ, and other component contain sinθ.