The net force is 2.2 N at 80.9 degrees south of east
Explanation:
In order to find the net force, we have to resolve each force along the x-y direction, and then add the components in each direction.
Taking east as positive x-direction and north as positive y-direction, we have:
- First force:
[tex]F_{1x} = -3.0 N\\F_{1y} = 0[/tex]
- Second force:
[tex]F_{2x} = (4.0)(cos (-33^{\circ})=3.35 N\\F_{2y} = (4.0)(sin (-33^{\circ})=-2.18 N[/tex]
Therefore, the components of the net force are
[tex]F_x = F_{1x}+F_{2x}=-3.0+3.35=+0.35 N\\F_y = F_{1y}+F_{2y}=0+(-2.18)=-2.18 N[/tex]
And so, the magnitude of the net force is
[tex]F=\sqrt{F_x^2+F_y^2}=\sqrt{(0.35)^2+(-2.18)^2}=2.2 N[/tex]
And the direction is
[tex]\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-2.18}{0.35})=-80.9^{\circ}[/tex]
which means 80.9 degrees south of east.
Learn more about vector addition:
brainly.com/question/4945130
brainly.com/question/5892298
#LearnwithBrainly
