A cargo of 40 tons is to be lifted with a...

Question 1 / 2066636eec28f7522a1c51c082

Question 1 of 2066636eec28f7522a1c51c082

A cargo of 40 tons is to be lifted with a boom located 40 feet from the ship's centerline. The ship's displacement including the suspended cargo is 8,000 tons and the GM is 2 feet with cargo suspended. What will the list of the vessel be with the cargo suspended?

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🔍 Key Concepts

• Transverse upsetting moment created by a suspended load (weight × horizontal distance from centerline) • Use of GM (metacentric height) to find the heeling arm (GZ) for small angles: GZ ≈ GM × sin(list angle) • Relationship between heeling moment and righting moment at equilibrium: Displacement × GZ = Heeling moment

💭 Think About

• First, calculate the heeling moment produced by the 40‑ton load at 40 feet off the centerline. How do you express that as a moment? • Next, express the righting arm (GZ) in terms of the given GM and the unknown list angle. How do you use the small‑angle approximation? • Set the heeling moment equal to the righting moment (Displacement × GZ). What equation do you get for sin(list angle), and how can you turn that into an angle in degrees?

✅ Before You Answer

• Be sure you use total displacement (8,000 tons) when computing the righting moment, not the cargo weight alone. • Confirm that you keep feet with feet and tons with tons so units are consistent when forming the moment. • After solving for the angle, check that your result is a reasonable small angle (under about 10°) and matches one of the choices.

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Question 1 / 2066636eec28f7522a1c51c082

Question 1 of 2066636eec28f7522a1c51c082

Back

Question 1 of 20

66636eec28f7522a1c51c082

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🔍 Key Concepts

💭 Think About

✅ Before You Answer

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