Possible title: How Does Water Affect the Force on a Metal Cube in a Vessel?

Click For Summary
SUMMARY

The discussion centers on the forces acting on a metal cube placed in a vessel, specifically comparing scenarios with and without water. When the cube is fully immersed in water, the force exerted on the bottom of the vessel remains unchanged due to the constant downward force of gravity (F=mg) and the upward buoyant force being countered by the pressure difference. The presence of water does not increase the pressure exerted by the cube on the vessel's bottom, as the forces acting on the cube remain constant regardless of the fluid density surrounding it, provided the cube is not detached from the vessel's floor.

PREREQUISITES
  • Understanding of buoyancy and Archimedes' principle
  • Knowledge of pressure calculations (pressure = force/area)
  • Familiarity with basic physics concepts such as force, mass, and gravity
  • Concept of fluid density and its impact on pressure
NEXT STEPS
  • Research the principles of buoyancy and how they apply to submerged objects
  • Study pressure calculations in fluids, including variations with different fluid densities
  • Explore the effects of micro layers in fluid dynamics and their impact on buoyancy
  • Learn about Archimedes' principle in detail and its applications in real-world scenarios
USEFUL FOR

Students of physics, engineers, and anyone interested in fluid mechanics and the forces acting on submerged objects.

Cromptu
Messages
33
Reaction score
0
A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube :

Ans: Will remain the same.

But how? Won't the force exerted in the second case,i.e when the vessel is filled with water be less because of the buoyant force?


Please help!
 
Physics news on Phys.org
Presumably: There is no water layer beneath the cube, so it remains in direct contact with the surface of the vessel. And the water level is just up to the top of the cube, but does not cover it.
 
I have another doubt.
The force which the cube will exert on the bottom will be equal to : pressure*area , i.e., (height*density*gravitational acceleration) * area of the bottom.
Won't this force vary in the two cases? where in case 1, air is the fluid(density almost negligible) and in case 2 water is the fluid (much more dense)
 
Cromptu said:
I have another doubt.
The force which the cube will exert on the bottom will be equal to : pressure*area , i.e., (height*density*gravitational acceleration) * area of the bottom.
Won't this force vary in the two cases? where in case 1, air is the fluid(density almost negligible) and in case 2 water is the fluid (much more dense)
You're going to have to describe the situation in more detail. (See the presumptions of my last post.) Assuming that the water does not cover the cube, why would the presence of the water increase the pressure that the cube exerts on the bottom?
 
(height*density*gravitational acceleration) * area

This is the downward force due to gravity: F=mg. It stays constant.
The upward force due to buoyancy is obtained from the pressure difference between the top and the bottom of your object. The density on the bottom of your solid is the density of the 'floor'. The density on the top is the density of air. These stay the same, regardless of any density changes at the sides of the cube.

In practice however, it will be difficult to prevent the occurrence of a micro layer of water between the solid and the floor. As soon as the solid is detached from the floor, you will get buoyancy.
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K