A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A that splits

Question

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A that splits at point B and attaches to the ship at points C and D. The two rope segments BC and BD angle away from the center of the ship at angles of ϕ = 26.0 ∘ and θ = 21.0 ∘, respectively. The tugboat pulls with a force of 1200 lb . What are the tensions TBC and TBD in the rope segments BC and BD?

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Nho 5 years 2021-07-23T03:06:57+00:00 1 Answers 21 views 0

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  1. RobertKer
    0
    2021-07-23T03:08:08+00:00
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    Answer:

    The tensions in T_{BC} is approximately 4,934.2 lb and the tension in T_{BD} 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 = T_{BC}

    The tension in rope segment BD = T_{BD}

    The tension in rope segment AB = T_{AB} = Pulling force of tugboat = 1200 lb

    By resolution of forces acting along the line A_F gives;

    T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = T_{AB} = 1200 lb

    T_{BC} × cos(26.0°) + T_{BD} × 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;

    T_{BC} × sin(26.0°) + T_{BD} × sin(21.0°) = 0………………………(2)

    Which gives;

    T_{BC} × sin(26.0°) = – T_{BD} × sin(21.0°)

    T_{BC} = – T_{BD} × sin(21.0°)/(sin(26.0°))  ≈ – T_{BD} × 0.8175

    Substituting the value of, T_{BC}, in equation (1), gives;

    – T_{BD} × 0.8175 × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb

    – T_{BD} × 0.7348  + T_{BD} ×0.9336 = 1200 lb

    T_{BD} ×0.1988 = 1200 lb

    T_{BD} ≈ 1200 lb/0.1988 = 6,035.6938 lb

    T_{BD} ≈ 6,035.6938 lb

    T_{BC} ≈ – T_{BD} × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

    T_{BC} ≈ -4934.1733 lb

    From which we have;

    The tensions in T_{BC} ≈ -4934.2 lb and  T_{BD} ≈ 6,035.7 lb.

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