How Does Fluid Mechanics Explain the Sinking Treasure Box Scenario?

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SUMMARY

The discussion focuses on the physics of a sinking treasure box scenario involving fluid mechanics principles. The box, measuring 40.0 cm by 25.0 cm by 30.0 cm, has a density three times that of seawater and sinks at a constant velocity of 1.15 m/s. Key calculations include a buoyant force of 294 N when submerged, a weight of 882 N, and a drag force that can be determined using the equilibrium condition due to constant velocity. The change in gravitational potential energy is calculated as -10,901.5 J, which also represents the total energy change of the box.

PREREQUISITES
  • Understanding of fluid mechanics principles, specifically buoyancy and drag forces.
  • Familiarity with gravitational potential energy calculations.
  • Knowledge of free-body diagrams and force equilibrium concepts.
  • Basic proficiency in algebra for solving physics equations.
NEXT STEPS
  • Study the concept of buoyant force in detail, particularly Archimedes' principle.
  • Learn how to derive drag force equations, including the significance of drag coefficients.
  • Explore the relationship between velocity and forces acting on submerged objects.
  • Investigate energy conservation principles in fluid dynamics scenarios.
USEFUL FOR

Students studying physics, particularly those focusing on fluid mechanics, as well as educators and professionals involved in teaching or applying these principles in real-world scenarios.

litheira
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Homework Statement


Spanish galleon is about to be boarded by bloodthirsty pirates in the
shallows of a Caribbean island. To save a box of treasure on board, the
captain orders his crew to secretly toss the box overboard, planning to come
back for it later. The rectangular box is waterproof and measures 40.0 cm by
25.0 cm by 30.0 cm. It is made of wood and has mostly gold pieces inside,
resulting in an average box density three times that of seawater.

Sinking below the surface, the box moves at constant vertical velocity of 1.15
m/s for 12.0 m before hitting the bottom. (a) Draw the free-body diagram for
the box, (b) determine the magnitudes of the forces on the box, and (c)
calculate the work done by each force and the net work done on the box. (d)
Calculate the change in the box’s gravitational potential energy. (e) What is
the change in the box’s total energy ?

Homework Equations


V= lwh
P= M/V
Buoyant force= pgV for floating
completely submerged buoyant force= (P of fluid/P of object)weight of object
W= Fd
PE= mgh
E= PE + KE
Drag force= .5pCAv^2

The Attempt at a Solution


i don't need help with part a.

part b) i got the buoyant force to be 908.46 when it is floating. for submerged, it is 294 N
Drag force= 138.88 (not sure, because i couldn't find the coefficient in my textbook)
weight= 908.46

for part c) i just multiplied y= -12 to each of the forces to find the work

for part d) i just did PE= mg(deltah) and got -10901.5 J

for part e i know how to do it (E= PE + KE) but I'm not getting the correct answer. I think I'm doing something wrong in either part b or part c.
I used the buoyant force when it submerged.

the correct answer for part e is -10901.5 J
 
Last edited:
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litheira said:

Homework Statement


Spanish galleon is about to be boarded by bloodthirsty pirates in the
shallows of a Caribbean island. To save a box of treasure on board, the
captain orders his crew to secretly toss the box overboard, planning to come
back for it later. The rectangular box is waterproof and measures 40.0 cm by
25.0 cm by 30.0 cm. It is made of wood and has mostly gold pieces inside,
resulting in an average box density three times that of seawater.

Sinking below the surface, the box moves at constant vertical velocity of 1.15
m/s for 12.0 m before hitting the bottom. (a) Draw the free-body diagram for
the box, (b) determine the magnitudes of the forces on the box, and (c)
calculate the work done by each force and the net work done on the box. (d)
Calculate the change in the box’s gravitational potential energy. (e) What is
the change in the box’s total energy ?


Homework Equations


V= lwh
P= M/V
Buoyant force= pgV for floating
completely submerged buoyant force= (P of fluid/P of object)weight of object
W= Fd
PE= mgh
E= PE + KE
Drag force= .5pCAv^2

The Attempt at a Solution


i don't need help with part a.

part b) i got the buoyant force to be 908.46 when it is floating. for submerged, it is 294 N
Drag force= 138.88 (not sure, because i couldn't find the coefficient in my textbook)
weight= 908.46

for part c) i just multiplied y= -12 to each of the forces to find the work

for part d) i just did PE= mg(deltah) and got -10901.5 J

for part e i know how to do it (E= PE + KE) but I'm not getting the correct answer. I think I'm doing something wrong in either part b or part c.
I used the buoyant force when it submerged.

the correct answer for part e is -10901.5 J

There are 3 forces acting on the box: the weight (downward), the upward buoyant force, and drag force (upward). You already calculated the buoyant force (294N) and the weight (908 N... this actually should probably have been 882N =3 x 294N ). The key to solving part b are the words "the box moves at constant vertical velocity of 1.15 m/s." This means that the box is not accelerating, and, thus, that it is in force equilibrium. You now have enough information to calculate the drag force, without needing to estimate the drag coefficient.
 

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