Answer:
The x-component of [tex]F_{3}[/tex] is 56.148 newtons.
Explanation:
From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:
[tex]\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O[/tex] (1)
Where:
[tex]\vec F_{1}[/tex], [tex]\vec F_{2}[/tex], [tex]\vec F_{3}[/tex] - External forces exerted on the ring, measured in newtons.
[tex]\vec O[/tex] - Vector zero, measured in newtons.
If we know that [tex]\vec F_{1} = (70.711,70.711)\,[N][/tex], [tex]\vec F_{2} = (-126.859, 46.173)\,[N][/tex], [tex]F_{3} = (F_{3,x},F_{3,y})[/tex] and [tex]\vec O = (0,0)\,[N][/tex], then we construct the following system of linear equations:
[tex]\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N[/tex] (2)
[tex]\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N[/tex] (3)
The solution of this system is:
[tex]F_{3,x} = 56.148\,N[/tex], [tex]F_{3,y} = -116.884\,N[/tex]
The x-component of [tex]F_{3}[/tex] is 56.148 newtons.
