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.