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
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!
![[FINALE] 💸ที่แท้พ่อเป็นผู้ถือหุ้นบริษัทพันล้าน เขากลับใจตอนนี้ก็สายไปแล้ว🙇🏻♂️#ละครคุณธรรม #short](http://i.ytimg.com/vi/yFwgjTFISUI/mqdefault.jpg)
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
Hey Chat @13:20 how does an object with density less than water, would completely submerge ? was that a mistake?
No, 12:30. is *held* completely submerged under water, presumably by a hand.
Amazing effort and explaining! Thank you
You're welcome and Thank You.
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!
Thanks for the video, Chad. By the way, do you have videos on physics 2 concepts such as electricity, coulomb's law...etc?
you are so good at explaining u literally save me everytime 3>
Glad the channel is helping you - Happy Studying!
thanks for detailed video that is very useful....
Excellent!
great thank you
You're welcome
brain left the chat
too complex