Question

A force vector has a magnitude of 720 N and a direction of 38 degree north of east. Determine the magnitude and direction of the components of the force that point along the north-south line and along east-west line.

Answer 1

Answer:

443.3 N

567.4 N

Explanation:

Consider the triangle ABC

AC = hypotenuse = magnitude of force vector = F = 720 N

AB = adjacent = Component of force along east-west line = $F_{x}$

BC = Opposite = Component of force along north-south line = $F_{y}$

θ = Angle = 38 deg

In triangle ABC

$Sin38 = \frac{BC}{AC}$

$Sin38 = \frac{F_{y}}{F}$

$0.616 = \frac{F_{y}}{720}$

$F_{y}$ = 443.3 N

Also, In triangle ABC

$Cos38 = \frac{AB}{AC}$

$Cos38 = \frac{F_{x}}{F}$

$0.788 = \frac{F_{x}}{720}$

$F_{x}$ = 567.4 N