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?

The tensions in is approximately 4,934.2 lb and the tension in 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 =

The tension in rope segment BD =

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

By resolution of forces acting along the line A_F gives;

× cos(26.0°) + × cos(21.0°) = = 1200 lb

× cos(26.0°) + × 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;

× sin(26.0°) + × sin(21.0°) = 0...........................(2)

Which gives;

× sin(26.0°) = - × sin(21.0°)

= - × sin(21.0°)/(sin(26.0°))  ≈ - × 0.8175

Substituting the value of, , in equation (1), gives;

- × 0.8175 × cos(26.0°) + × cos(21.0°) = 1200 lb

- × 0.7348  + ×0.9336 = 1200 lb

×0.1988 = 1200 lb

≈ 1200 lb/0.1988 = 6,035.6938 lb

≈ 6,035.6938 lb

≈ - × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

≈ -4934.1733 lb

From which we have;

The tensions in ≈ -4934.2 lb and   ≈ 6,035.7 lb.

the answer is b ) square root of 5