The question is missing some parts. Here is the complete question.
A suspension bridge cable is connected to its anchor at a 20° angle. Find the vertical and horizontal component of the force on the anchor by the cable.
Answer: [tex]F_{x}=[/tex] 14095.4 N
[tex]F_{y}=[/tex] 5130.3 N
Explanation: The force applied to the anchor is not perpendicular to the horizontal plane. So, it can be decomposed into 2 components: a vertical component, which is on the y-axis, and a horizontal component, which is on the x-axis.
The force and its components forms a right triangle, so we can calculate the components by using trigonometric relations:
Horizontal
[tex]cos(20)=\frac{F_{x}}{F}[/tex]
[tex]F_{x}=F.cos(20)[/tex]
[tex]F_{x}=15,000(0.9397)[/tex]
[tex]F_{x}=[/tex] 14,095.4 N
Vertical
[tex]sin(20)=\frac{F_{y}}{F}[/tex]
[tex]F_{y}=Fsin(20)[/tex]
[tex]F_{y}=[/tex] 15,000(0.3420)
[tex]F_{y}=[/tex] 5,130.3 N
The vertical and horizontal components of force on the anchor by the cable are 5130.3 N and 14095.4 N, respectively.
