SOLUTION: Solve A block weighing 90 lb rests on a 15 degree incline. Find the magnitude of the components of the block's weight perpendicular and parallel to the incline. Please help, I'm s

Algebra ->  Vectors -> SOLUTION: Solve A block weighing 90 lb rests on a 15 degree incline. Find the magnitude of the components of the block's weight perpendicular and parallel to the incline. Please help, I'm s      Log On


   

Question 928226: Solve
A block weighing 90 lb rests on a 15 degree incline. Find the magnitude of the components of the block's weight perpendicular and parallel to the incline. Please help, I'm so confuse about this problem
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The weight (90 lb force), and its components are represented by arrows.
The angles between the weight and the component perpendicular to the inclined plane is the same 15%5Eo angle as between the horizontal and the inclined plane (just rotated 90%5Eo and shifted).
The components of the weight are represented by the arrows along the legs of those green right triangles that have the weight arrow as the hypotenuse,
and a 15%5Eo angle marked by a red arc.
The perpendicular component's magnitude is
%22%2890%22%22lb%29%22cos%2815%5Eo%29=86.9lb
The parallel component's magnitude is
%22%2890%22%22lb%29%22sin%2815%5Eo%29=23.3lb