Respuesta :
Answer:
The tensions in [tex]T_{BC}[/tex] is approximately 4,934.2 lb and the tension in [tex]T_{BD}[/tex] is approximately 6,035.7 lb
Explanation:
The given information are;
The angle formed by the two rope segments are;
The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°
The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°
Therefore, we have;
The tension in rope segment BC = [tex]T_{BC}[/tex]
The tension in rope segment BD = [tex]T_{BD}[/tex]
The tension in rope segment AB = [tex]T_{AB}[/tex] = Pulling force of tugboat = 1200 lb
By resolution of forces acting along the line A_F gives;
[tex]T_{BC}[/tex] × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = [tex]T_{AB}[/tex] = 1200 lb
[tex]T_{BC}[/tex] × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = 1200 lb............(1)
Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;
[tex]T_{BC}[/tex] × sin(26.0°) + [tex]T_{BD}[/tex] × sin(21.0°) = 0...........................(2)
Which gives;
[tex]T_{BC}[/tex] × sin(26.0°) = - [tex]T_{BD}[/tex] × sin(21.0°)
[tex]T_{BC}[/tex] = - [tex]T_{BD}[/tex] × sin(21.0°)/(sin(26.0°)) ≈ - [tex]T_{BD}[/tex] × 0.8175
Substituting the value of, [tex]T_{BC}[/tex], in equation (1), gives;
- [tex]T_{BD}[/tex] × 0.8175 × cos(26.0°) + [tex]T_{BD}[/tex] × cos(21.0°) = 1200 lb
- [tex]T_{BD}[/tex] × 0.7348 + [tex]T_{BD}[/tex] ×0.9336 = 1200 lb
[tex]T_{BD}[/tex] ×0.1988 = 1200 lb
[tex]T_{BD}[/tex] ≈ 1200 lb/0.1988 = 6,035.6938 lb
[tex]T_{BD}[/tex] ≈ 6,035.6938 lb
[tex]T_{BC}[/tex] ≈ - [tex]T_{BD}[/tex] × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
[tex]T_{BC}[/tex] ≈ -4934.1733 lb
From which we have;
The tensions in [tex]T_{BC}[/tex] ≈ -4934.2 lb and [tex]T_{BD}[/tex] ≈ 6,035.7 lb.
