Complete Question
The complete question(reference (chegg)) is shown on the first uploaded image
Answer:
The magnitude of the resultant force is [tex]F = 199.64 \ N[/tex]
The direction of the resultant force is [tex]\theta = 4.1075^o[/tex] from the horizontal plane
Explanation:
Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be [tex]Fcos(\theta )[/tex] while if the force is moving away from the angle then the resolved force is [tex]Fsin (\theta )[/tex]
Now from the diagram let resolve the forces to their horizontal component
So
[tex]\sum F_x = 150 cos(30) + 100cos(15) -80sin (20)[/tex]
[tex]\sum F_x = 199.128 \ N[/tex]
Now resolving these force into their vertical component can be mathematically evaluated as
[tex]\sum F_{y} = 150 sin(30) - 100sin(15) -110 +80 cos(20)[/tex]
[tex]\sum F_{y} = 14.30[/tex]
Now the resultant force is mathematically evaluated as
[tex]F = \sqrt{F_x^2 + F_y^2}[/tex]
substituting values
[tex]F = \sqrt{199.128^2 + 14.3^2}[/tex]
[tex]F = 199.64 \ N[/tex]
The direction of the resultant force is evaluated as
[tex]\theta = tan^{-1}[\frac{F_y}{F_x} ][/tex]
substituting values
[tex]\theta = tan^{-1}[\frac{ 14.3}{199.128} ][/tex]
[tex]\theta = 4.1075^o[/tex] from the horizontal plane
