Buoyancy and Flotation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. You must have heard the phrase: that is only the tip of the iceberg. So tell me, how much of a real iceberg sits above the sea? Half? One third? By the end of this lesson, you will work it out yourself, from just two numbers. Today: buoyancy and flotation. Questions on this come up often in JEE, so let's begin.

0:27Archimedes' principle: the upward push

So, lift a stone from the bottom of a bucket of water. Under water it feels light; out of the water, heavy. Why does it feel lighter under water? Because the water pushes it up! This is Archimedes' principle. Any body, wholly or partly immersed in a fluid, feels an upward force. We call it the buoyant force, or the upthrust. Okay? And how big is this push? The upthrust equals the weight of the fluid displaced, that is, the fluid the body pushes out of its way. Right? In symbols: F B equals rho l, times V sub, times g. F with a capital B as subscript is the buoyant force. V with the subscript s u b is the submerged volume, the part of the body under the liquid. And rho, the Greek letter, with a small curly l as subscript, is a density. So rho l times V sub is the mass of fluid pushed aside, and times g, the acceleration due to gravity, gives its weight. Clear? Now think carefully. Whose density goes into this formula: the body's, or the liquid's? The liquid's, exactly! The small l stands for liquid. This is a common trap: students put in the density of the body. Don't! The upthrust does not care what the body is made of, only how much liquid it pushes aside. Got it? And this upward force acts through a point called the centre of buoyancy. With me?

1:58Apparent weight and relative density

Now, back to the stone. Its weight, m g, mass times g, has not changed; the earth still pulls with m g. But the water pushes up with F B, the upthrust. So a spring balance reads less: the apparent weight equals m g minus F B, okay? So the stone seems to lose some weight in water. Tell me: how much weight does it lose? Exactly the upthrust! The loss of weight in the fluid equals F B, the buoyant force. Nothing vanished; the water simply holds up part of the stone. Right? And this gives a lovely trick. Relative density equals weight in air, over loss of weight in water. It is the body's density compared with water's, so it has no unit. Two spring balance readings, and you have it! With me? Why does it work? Weight in air is rho b V g: rho with a small b is the body's density, capital V its volume. The loss in water is rho w V g, with w for water: the same V, because the body is fully under water. V g cancels, leaving rho b over rho w. So remember: fully under water. Clear?

3:09Floating: a body displaces its own weight

Alright, now flotation. Take a piece of wood floating at rest on water. Its weight pulls it down, but it does not sink. So what balances its weight? The upthrust, yes! At rest the forces balance, so m g equals F B, the upthrust. In other words, a floating body displaces its own weight of fluid. Okay? Now write both sides. The weight is rho b times V times g, with capital V the whole volume of the body. The upthrust is rho l times V sub times g: only the part under the liquid counts. Right? Cancel g, and rearrange. V sub over V equals rho b over rho l. The fraction of the body under the liquid is just the ratio of the two densities. That's it! Got it? So, back to our iceberg. Ice has a density of nine hundred and seventeen kilograms per cubic metre; the power minus three means per cubic metre. Sea water: one thousand and twenty five. So, what fraction of the iceberg is above the water? Done? Let's see. The fraction under the sea is nine hundred and seventeen over one thousand and twenty five: about zero point eight nine five; the wavy equals sign means approximately. So the part above is one minus that: about ten point five percent. Only about one tenth! Truly the tip of the iceberg. But careful: the question asked for the part above. Stop at zero point eight nine five, and you lose the marks.

4:46Example 1: stone's volume and density from two weighings

Clear? Now, Example one. We have a stone that weighs twenty newtons in air, and fifteen newtons in water; capital N stands for newtons. Find its volume and its density. Take g as ten. Pause the video, and try it first. Done? Okay, let's check. In water, the stone loses twenty minus fifteen, that is five newtons. And the loss of weight is exactly the upthrust. So the upthrust is five newtons. Right? Now, the upthrust is rho of water, times V, times g. The stone sinks, so it is fully under water, and the submerged volume is its whole volume. Water's density is ten to the power three kilograms per cubic metre, so: five equals ten to the power three, times V, times ten. So V equals five times ten to the power minus four cubic metres; the double arrow means so. Okay? Now the density: mass over volume. So twenty over five times ten to the power minus four, right? No, no, careful! Twenty newtons is the weight, not the mass. Mass equals weight over g: twenty over ten, two kilograms. With me? So the density is two over five times ten to the power minus four. Four thousand kilograms per cubic metre. Got it? Now a quick check with our trick. Relative density is weight in air over loss of weight: twenty over five, that is four. So the stone is four times as dense as water: four thousand kilograms per cubic metre again. One line. Nice, right?

6:23Example 2: block floating between oil and water

Now a harder one. We have a block of density nine hundred kilograms per cubic metre. It floats where oil, of density eight hundred, meets water, of density one thousand, wholly immersed: partly in each. What fraction of it is in the water? Pause and try. Hint: two liquids push up here. Many students jump to our flotation formula. Block density over liquid density: nine hundred over one thousand, so nine tenths in the water, right? No, no, careful! That formula is for one liquid. Here the oil pushes up too. So let x be the fraction of the block in the water; then one minus x is in the oil. Okay? The block floats at rest, so its weight equals the total upthrust: the upthrust of the water, plus the upthrust of the oil. Each liquid pushes up with the weight of that liquid displaced. Right? With capital V for the block's volume: nine hundred V g equals one thousand times x V g, plus eight hundred times one minus x, times V g. Clear? Cancel V g everywhere. Nine hundred equals one thousand x, plus eight hundred times one minus x. With me? Open the bracket: nine hundred equals eight hundred plus two hundred x. So x is one hundred over two hundred, that is one half. Half the block is in the water, not nine tenths! Got it?

7:53Quick revision: upthrust, apparent weight, floating rule

So, quick revision. Upthrust equals the weight of the fluid displaced: F B equals rho l V sub g, with the liquid's density. Apparent weight is m g minus F B, so the loss of weight is the upthrust. Floating: V sub over V equals rho b over rho l; with two liquids, add the upthrusts. Now go and try the practice problems on this page. All the best!