Archimedes’ principle | Description & Facts | dayline.info
Define buoyant force. State Archimedes' principle. Understand why objects float or sink. Understand the relationship between density and Archimedes' principle. Pascal's principle explains the relationship between density and buoyancy. An object will sink if its density is greater than the density of water. Multiple Choice. Sal explains the difference between liquids and gasses (both fluids). He then starts a.
Archimedes principle helps to explain the relationship between what
Let's take what we know about force-- actually about work-- and see if we can come up with any rules about force and pressure with liquids. So what do we know about work?
Work is force times distance, or you can also view it as the energy put into the system-- I'll write it down here. Work is equal to force times distance. We learned in mechanical advantage that the work in-- I'll do it with that I-- is equal to work out. The force times the distance that you've put into a system is equal to the force times the distance you put out of it.
And you might want to review the work chapters on that. That's just the little law of conservation of energy, because work in is just the energy that you're putting into a system-- it's measured in joules-- and the work out is the energy that comes out of the system.
And that's just saying that no energy is destroyed or created, it just turns into different forms. Let's just use this definition: Let's say that I pressed with some force on this entire surface. Let's say I had a piston-- let me see if I can draw a piston, and what's a good color for a piston-- so let's add a magenta piston right here.
I push down on this magenta piston, and so I pushed down on this with a force of F1. Let's say I push it a distance of D that's its initial position. Its final position-- let's see what color, and the hardest part of these videos is picking the color-- after I pushed, the piston goes this far. This is the distance that I pushed it-- this is D1. The water is here and I push the water down D1 meters. In this situation, my work in is F1 times D1.
Let me ask you a question: How much total water did I displace? Well, it's this volume? I took this entire volume and pushed it down, so what's the volume right there that I displaced?
Pascal's Law & Buoyancy - SchoolWorkHelper
The volume there is going to be-- the initial volume that I'm displacing, or the volume displaced, has to equal this distance. This is a cylinder of liquid, so this distance times the area of the container at that point.
I'm assuming that it's constant at that point, and then it changes after that, so it equals area 1 times distance 1. We also know that that liquid has to go someplace, because what do we know about a liquid? We can't compress it, you can't change its total volume, so all of that volume is going to have to go someplace else. This is where the liquid was, and the liquid is going to rise some level-- let's say that it gets to this level, and this is its new level.
It's going to change some distance here, it's going to change some distance there, and how do we know what distance that's going to be? The volume that it changes here has to go someplace.
You can say, that's going to push on that, that's all going to push, and that liquid has to go someplace. Essentially it's going to end up-- it might not be the exact same molecules, but that might displace some liquid here, that's going to displace some liquid here and here and here and here and all the way until the liquid up here gets displaced and gets pushed upward.
The volume that you're pushing down here is the same volume that goes up right here. So what's the volume-- what's the change in volume, or how much volume did you push up here?
This volume here is going to be the distance 2 times this larger area, so we could say volume 2 is going to be equal to the distance 2 times this larger area. We know that this liquid is incompressible, so this volume has to be the same as this volume. We know that these two quantities are equal to each other, so area 1 times distance 1 is going to be equal to this area times this distance. Let's see what we can do. Pressure is a scalar?!
So why do people seem to represent pressure in diagrams with arrows as if it were a vector with a particular direction? Even though pressure is not a vector and has no direction in and of itself, the force exerted by the pressure on the surface of a particular object is a vector.
So when people draw diagrams with pressure pointing in specific directions, those arrows can be thought of as representative of the direction of the forces on those surfaces exerted by the pressure from the fluid. If there were no surface upon which the pressure could exert a force, it would make no sense to draw a direction for the force at that point inside the water. On the left hand side of the diagram below there are water molecules and pressure, but no well defined direction of force.
The right hand side of the diagram below shows the well defined directions of forces on an ice cream cone submerged in the water. While we're on the topic, we might as well make it clear that the force exerted on a surface by fluid pressure is always directed inwards and perpendicular at a right angle to the surface.
At this point, if you've been paying close attention you might wonder "Hey, there's air above the water right? Shouldn't the weight of the column of air above the column of water also contribute to the total pressure at the top of the can of beans?