Respuesta :
Answer:
Tₓ = 42 N
Ty = 80 N
Explanation:
The pole, guy-wire and ground make a right-angled triangle, as shown in the attached diagram. It is clear from figure that:
tan θ = Perp/Base = 95 ft/50 ft
θ = tan⁻¹ (1.9)
θ = 62.24°
Now, for the horizontal component of tension:
Tₓ = T Cos θ
where,
T = Tension = 90 lb
Tₓ = Horizontal Component of Tension = ?
Therefore,
Tₓ = (90 lb)(Cos 62.24°)
Tₓ = 42 N
Now, for the vertical component of tension:
Ty = T Sin θ
where,
T = Tension = 90 lb
Ty = Vertical Component of Tension = ?
Therefore,
Ty = (90 lb)(Sin 62.24°)
Ty = 80 N

The horizontal and vertical components (in lb) of the force of tension in the wire is 42 and 80 lb respectively.
Given the following data:
- Height of pole = 95 ft.
- Distance = 50 ft.
- Force of tension = 90 lb.
To calculate the horizontal and vertical components (in lb) of the force of tension in the wire:
The horizontal and vertical components of force.
First of all, we would determine the angle of inclination between the pole and the guy-wire by using this formula:
[tex]Tan (\theta) = \frac{Opp}{Adj} \\\\\theta= tan^{-}(\frac{Opp}{Adj})\\\\\theta= tan^{-}(\frac{95}{50})\\\\\theta= tan^{-}(1.9)[/tex]
Angle = 62.24°
For the horizontal component of tensional force:
Mathematically, the horizontal component of tensional force is given by this formula:
[tex]T_x=Tcos\theta\\\\T_x = 90cos(62.24)\\\\T_x = 90 \times 0.4658[/tex]
Tx = 41.92 ≈ 42 lb.
For the vertical component of tensional force:
Mathematically, the vertical component of tensional force is given by this formula:
[tex]T_y=Tsin\theta\\\\T_y = 90sin(62.24)\\\\T_x = 90 \times 0.8849[/tex]
Tx = 79.64 ≈ 80 lb.
Read more on horizontal component here: https://brainly.com/question/4080400