Answer:
[tex]F_y=96.4N[/tex]
Explanation:
Hello.
In this case, considering the force diagram shown on the attached picture, we can see that the component of his force is directed downwards is:
[tex]F_y=F\times sin (\theta)[/tex]
Because the other component is the horizontal one:
[tex]F_x=F\times cos(\theta)[/tex]
In this case, the y-component force turns out:
[tex]F_y=150N\times sin (40\°)\\\\F_y=96.4N[/tex]
Moreover, the x-component force is also computed if required:
[tex]F_x=150N\times cos(40\°)\\\\F_x=114.9N[/tex]
Best regards.
