Twenty-five hundred (2500) tons of iron ore...

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Question 1 of 2066636ee728f7522a1c518d40

Twenty-five hundred (2500) tons of iron ore with a stowage factor of 17 is stowed in a cargo hold. The dimensions of the hold are 55 feet long and 45 feet wide and 35 feet high. What is the height of the center of gravity of the ore above the bottom of the hold?

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Question 1 of 20

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🔍 Key Concepts

• How to convert weight of cargo (tons) and stowage factor (ft³/ton) into a total volume in cubic feet • How to find the depth of cargo loaded in a rectangular hold when you know the volume and the hold’s length × breadth × height • For a uniformly distributed cargo, the center of gravity in height is at half the cargo depth above the bottom

💭 Think About

• First, calculate the total volume of the ore using the given weight and stowage factor. Then compare that to the volume capacity of the hold to find how deep the ore actually fills it. • Once you know the depth of the ore, think about where the center of gravity of a uniformly filled block of material would be located vertically. • Check whether the ore fills the hold completely or only partially; this will determine whether you use half the hold height or half the cargo depth for the vertical center of gravity.

✅ Before You Answer

• Verify the multiplication for total cargo volume: tons × stowage factor (ft³/ton). • Confirm the formula for depth of cargo: depth = total cargo volume ÷ (length × breadth). • Once you have the cargo depth, ensure you take half of that depth (not half the hold height) to find the vertical center of gravity above the bottom.

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