What's the volume of water required inside the lock if

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SUMMARY

The volume of water required inside the lock is calculated based on the dimensions of the lock (40m long, 3.5m height, 6m width) and the total weight of the ship and its load, which is 700T. Using Archimedes' principle, the mass of water displaced equals the weight of the ship, resulting in a displaced volume of 700m³. The total volume of the lock, allowing for a height of 0.5m, is determined to be (3m x 40m x 6m) - 700m³. If the ship empties its load, the new displaced volume will need to be recalculated based on the density of the load, which is 755kg/m³.

PREREQUISITES
  • Understanding of Archimedes' principle
  • Basic knowledge of volume calculations
  • Familiarity with weight and density concepts
  • Ability to perform unit conversions (e.g., from tons to kilograms)
NEXT STEPS
  • Calculate the new displaced volume when the ship empties its load using the density of 755kg/m³.
  • Explore the implications of varying the dimensions of the lock on water displacement.
  • Investigate the effects of different ship weights on lock water volume requirements.
  • Learn about the design considerations for locks in maritime engineering.
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Students studying fluid mechanics, civil engineers involved in lock design, and maritime professionals managing ship operations in locks.

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What's the volume of water required inside the lock if...

Homework Statement



The dimension of the lock is:40 long x 3.5 height x 6 width
allowing 0.5 height.

The ship weights 400T plus 300T(LOAD)
total= 700T

whats the amount of water used per lock allowing allowing 0.5 m from the height?

What if the ship empties its load on the lock ? the density of the load is 755kg/m^3.

Homework Equations



Archimedes principle
Mass of water displaced

The Attempt at a Solution



Mass of water displaced by the boat mass = 700T
VOLUME = 700M^3
Total volume inside the lock allowing 0.5

(3x40x6)m^3 - 700M^3

Is that correct ?

How about if the ship empties its load on the lock ?
 
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davekardle said:

Homework Statement



The Attempt at a Solution



Mass of water displaced by the boat mass = 700T
VOLUME = 700M^3
Total volume inside the lock allowing 0.5

(3x40x6)m^3 - 700M^3

Is that correct ?
Yes.

How about if the ship empties its load on the lock ?

The new displaced volume of water will be how much? So the level of the lock rises by how much?
 


Duplicate posting in several other forums.
 

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