Question

Resolve a weight of 10N in two directions which are parallel and perpendiular to a slope inclined at 30^@ to the horizontal.

Answer 1

Given;

Weight W = 10 N

Angle to resolve θ  = 30°

To find;

Resolved weights

Solution;

If a vector W is to be resolved into two elements about an angle θ ,

It is resolved as

i)   Wsin θ = Vertical (perpendicular) resolution

ii)  Wcos θ= Horizontal (parallel) resolution                  (Equation 1)

                                                                               (As given in figure)

So, for W = 10 N and  θ = 30°

(putting in equation 1 )

Vertical (perpendicular) resolution = 10sin(30°) = $10\times \frac{1}{2}$ = 5 N

Horizontal (parallel) resolution = 10cos(30°) = $10\times \frac{\sqrt 3}{2}= 5\sqrt{3}$ N