Answer:
the vertical component of the force is 43.30 N.
Explanation:
Given;
angle of inclination of the force, θ = 60⁰
magnitude of the force, F = 50 N
The vertical component of the force is calculated as follows;
[tex]F_y = F \times sin(\theta}\\\\F_y = 50 N \times sin(60)\\\\F_y = 50 N \times \frac{\sqrt{3} }{2} \\\\F_y = 25\sqrt{3} \ N\\\\F_y = 43.30 \ N[/tex]
Therefore, the vertical component of the force is 43.30 N.
