9.2 Buoyant Force and Archimedes' Principle | General Physics

Chad provides a physics lesson on the buoyant force and Archimedes' Principle which states that the buoyant force is equal to the weight of displaced fluid. The lessons begins with an explanation of the nature of the buoyant force and a derivation of the formula for the buoyant force. He then compares the formulas for calculating an object's weight and the buoyant force and explains why objects that are more dense than a fluid sink, while objects that are less dense float. Finally, Chad derives a formula for calculating the percentage of a floating object that is submerged below the surface before solving several buoyant force problems.

The first problem is a simple calculation of the buoyant force give the volume of the object and the density of the fluid. The next two problems demonstrate how an object's density affects the percentage of its volume that is submerged below the surface. The next two problems demonstrate how the buoyant force explains why objects feel lighter under water. Finally, the last problem is a calculation of the density of an object from its apparent weight under water.

00:00 Lesson Introduction
00:47 The Buoyant Force Formula Derivation
05:00 Buoyant Force vs Weight (Float or Sink)
06:51 The Volume Submerged for Floating Objects
12:22 How to Calculate Buoyant Force
14:25 How to Calculate the Percent Submerged for a Floating Object Problem #1
17:28 How to Calculate the Percent Submerged for a Floating Object Problem #2
19:23 How to Calculate the Normal Force for a Submerged Object
23:00 How to Calculate Apparent Weight for a Submerged Object
26:55 How to Calculate the Density of a Submerged Object

[pariahafezi-bakhtiari]

Hey Chat @13:20 how does an object with density less than water, would completely submerge ? was that a mistake?

[RamiRustum]

Amazing effort and explaining! Thank you

[gangagsk]

Good evening Sir
We derive formula for Archemedes principle by using the formula P = hpg.,
Where p is density and h is the height of fluid from the surface we are measuring
therefore F=hpgA
If we take a cubical container with dimension d,
The buoyant force must be the subtracted value of force acting on the top and bottom of the container which is
F=(h+d)pg-hpg
F=dpgA.
But I have a doubt. We derive p=hpg by taking the force mg acting on it by the whole fluid above it, ie in a cubical container at a depth h,
F=mg
P=mg/A
=mgh/Ah
=mgh/V
=hpg
But if we take a cube, then on the bottom side we cant take mg of fluid above it only, as the bottom surface is in a depth of h+d, till height h we can account it for the weight of the fluid, but for height d, there is no fluid infact in height d there is the solid part of the cube, then how can we write (h+d)pg

[meghanspencer]

Hey Chad I was hoping you could answer my question about floating objects. I am currently working on writing a program that will calculate the weight to off set a floating object to bring it back to neutral buoyancy. Specifically this project is for turtles who develop air masses inside of them (positive buoyancy syndrome) and they float causing them no longer to dive under the water. I was hoping you could guide me in the direction of what equation to use in order to solve this problem.

Thank you!

[viveharc]

thanks for detailed video that is very useful....

[Dreamy]

you are so good at explaining u literally save me everytime 3>

[ARTDEVGRU]

Thanks for the video, Chad. By the way, do you have videos on physics 2 concepts such as electricity, coulomb's law...etc?