The principle of flotation states that a floating body will displace its own
weight of the liquid in which it is floating. So, for a body of total volume V and density d floating on a liquid of density ρ, with volume of the body submerged,
Vsub:
Vdg = Vsubρg (since the body is at rest)
Vsub
V = dρ
Since d, the density of the body, is less than ρ, the density of the liquid
(condition required for floating) and d/ρ < 1, then Vsub = d/ρ V and the volume submerged will be the fraction (d/ρ) of the volume, V, of the floating body. It follows that if the fraction immersed is fimm then
fimm = dρ = (density of the body)
(density of the liquid)
When a body floats in a liquid, the fraction immersed,
fimm = density of the material of the body
density of the liquid
= ratio of the densities of the body and the liquid
We may summarise, then, by saying that if the body is placed in a liquid, and: (i) its density < the density of the liquid, the body will float;
(ii) its density = the density of the liquid, the body will remain at rest wherever it is placed, i.e. it will remain suspended in the liquid; (iii) its density > the density of the liquid, the body will sink.
Boats and rafts remain afloat because of the upthrust which supports them in water. In the case of rafts the density of the logs which are strapped together to make the raft must be less than that of water. This means that the density of the wood of the logs must be less than 1 g cm–3 or the relative density less than
1. This is the case for most woods found in the Caribbean. Even ‘greenheart’, one of Guyana’s densest woods, has been floated down rivers as rafts to saw- mills where they are sawn for use in the building trade. Purpleheart, too, should be able to float as a raft, its relative density being slightly below 1.
Where submarines are concerned, they are able to sink or float as desired because they can vary their overall density by using a ballast of air or water to give them a positive buoyancy or a negative buoyancy, as required. To float, their overall relative density must be less than 1; to sink, it must be greater than 1; and to remain suspended it must be equal to 1. The submarine is designed so that when the ballast tanks are full of air the submarine just floats. If the air in the ballast tanks is slowly replaced by sea-water the submarine becomes heavier and begins to sink – the more water in the tanks, the faster the sinking. To enable the submarine to rise again, its overall density is reduced by pumping water out of the ballast tanks and replacing this water with air. This causes the overall density of the submarine to be less than that of the sea- water and the submarine rises again.
floating ❯
principle of flotation ❯
The principle of flotation plays an important part in the dispersal of seeds by wind action and by water action.
Worked example 10.9
A boiling tube of cross-section 4.5 cm2 containing lead pellets floats in water
with 10.0 cm of its length immersed. The total mass of boiling tube and pellets is 200 g. What will be the depth of immersion of the tube if a further 5.0 g of pellets is added to the boiling tube?
Solution
Let the weight of the tube at first be W1 and the depth of immersion h1.
Let the second weight of the tube be W2 and the second depth of immersion be h2.
The equilibrium equations are (i) W1 = Ah1ρg and (ii) W2 = Ah2ρg
Subtracting the equations, we have W2 – W1 = (h2 – h1)Aρg and so h2 – h1 = (W2A – Wρg 1)
We recognise that the change in immersion depth = h2 – h1 and that the extra weight = W2 – W1.
So remembering that all units must be S.I. and substituting, we have
h2 – h1 = 5.0 × 10–3 × 10
4.5 × 10–4 × 103 × 10
= 0.0111 m
= 1.1 cm
Therefore the additional depth of immersion = 1.1 cm and the new depth of immersion will therefore be 11.1 cm.
the shape of a bowl, it is quite possible for that bowl to float on the water. As long as I can give the bowl such a shape that it can displace sufficient water for the displaced water to have the same weight as the bowl, the bowl will float. It seems then that the shape of the ‘hull’ of my bowl will be crucial to the floating or otherwise of the bowl. Is it a fact that flat -bottomed boats (like barges) will float more easily than boats with the conventional shape of a hull?
Chapter summary
• Pressure on a surface is the force acting at right angles to that surface over unit area. This area is the area of contact between the surface and the agency responsible for the force.
• The general formula for pressure is
pressure = area of contact between agency and surfaceforce acting perpendicular to the surface
• The S.I. unit of pressure is the newton per metre squared or the pascal (Pa). • A solid exerts a pressure on a surface when either the weight of the solid causes a
thrust to be exerted on that surface, or the solid transfers a thrust to the surface. • The pressure of a liquid is due to the perpendicular thrust exerted on a surface placed
in the liquid.
• The formula for the pressure due to a stationary liquid on a surface or at a point in that liquid is
pressure = (height of liquid above surface or point) × (density of the liquid) × (gravitational field strength)
Answers to ITQs
ITQ1 (a) 2000 N, (b) 3000 Pa
ITQ2 250 N
ITQ3 17.7 kPa
ITQ4 (i) Alter the angle between the glass tube and the thistle funnel between 0° and 90° by heating the glass. Use at least five different angles.
(ii) For each angle between glass tube and the funnel and for a given depth below the surface, measure the pressure with a water or oil manometer.
(iii) These values of the pressure should all be the same within the limits of
error, the errors being those involved in measuring the depth of the funnel (this could be a large error) and the head of liquid in the manometer.
ITQ5 (i) The R.D of mercury is large. So for the same pressure to be measured, the head will be shorter for mercury than for all other common liquids, since they all have a much smaller density than mercury.
ITQ6 (i) Select three or four liquids which can be obtained in large quantities (like water, kerosene, motor oil and one other), the densities of which are known.
Then either
A (ii) measure the pressure at widely different depths in each of the liquids
using a water or an oil manometer;
(iii) plot a graph of pressure against depth of the liquid for each liquid
used. Points on each graph should give a straight line, since for each set of measurements made the density of the liquid was constant. The slope of the graph would be the density of the liquid.
(iv) Different slopes would suggest that the density is important and that, for a given depth below the surface, the pressure of the liquid depends on the density of the liquid, but you would not know how.
Or
B (Which is the better method) Do you know why?
(ii) measure the pressure at given depths in the different liquids;
(iii) plot a graph of pressure against density.
(iv) If the graph is straight, the conclusion would be that the pressure in the liquid is proportional to the density if the depth in the liquid is kept constant.
ITQ7 R2 < R1 because of the upthrust on the cylinder and therefore on the newton-meter; K > K because of the downthrust in response to the initial
• The pressure of a fluid acts at all points in that fluid and in all directions about that point and towards the point. A fluid is any liquid or gas.
• The hydrostatic pressure at all points on a given horizontal level in a stationary, continuous liquid is constant.
• Pressure applied at any point in an enclosed liquid or gas is transmitted undiminished in value to all points within that liquid or gas. This is known as Pascal’s principle. • The fact that the pressure of a liquid or a gas varies with depth explains why a body
that is immersed partially or wholly in a fluid experiences an upward force, called an upthrust.
• Archimedes’ principle states that a body partially or completely immersed in a fluid experiences an upthrust that is equal to the weight of the fluid displaced by the body. • Archimedes’ principle leads to the principle of flotation. This principle states that a
floating body displaces its own weight of the liquid in which it is floating.
• Archimedes’ principle provides a very convenient method of finding experimentally the relative density of a solid body or of a liquid.
Examination-style questions
1 (i) The torr is the unit of pressure most often used for expressing the pressure of a partial vacuum. It may be defined as the pressure exerted by a column of mercury 1 mm high. Calculate an approximate value for the torr in pascals. Look up the origin of the unit ‘torr’. It is a contraction of the name of a past scientist. Who was he? What was his significant scientific contribution?
(ii) The bar is defined as 105 Pa. Calculate the ratio: 1 millibar/1 torr.
2 A writing desk stands on a base of external dimensions 80 cm × 70 cm and internal dimensions 70 cm × 50 cm (see diagram). The total weight of the desk and base is 400 N. Calculate the average pressure the desk exerts on a horizontal floor, (i) in N cm–2, and
(ii) in Pa.
3 A heavy lorry stands on six tyres, four at the back and two in front. The total weight of the lorry is 40 000 N. The area of contact between each rear tyre and the ground is 100 cm2 and
that between each front tyre and the ground is 80 cm2. Assuming that the pressure is the
same in all the tyres, calculate the value of this pressure. 4 Explain each of the following:
(i) If a drinking glass is filled to the brim with water and a flat sheet of stiff plastic is slid over the top and the glass inverted, the plastic sheet remains in place only if there is no air bubble in the water.
(ii) Heavy vehicles can move on soft ground without becoming stuck only if their tyres are very wide.
(iii) A sharp knife will cut hard objects more easily than a dull one.
(iv) A heavy haversack is more comfortable to carry over the shoulder if the straps are wide than if they are narrow.
5 The following is one method of comparing the densities of two immiscible liquids directly (see diagram). Stand a tall U-tube upright in a retort stand. Use a small funnel to pour the denser liquid into one limb of the U-tube to a height of about 15 cm. Now slightly incline the tube and, again using the funnel (having first washed and rinsed it), pour the other liquid down the other limb. Return the U-tube to its vertical position. When the liquids have settled, measure the vertical height of each liquid from the common junction. Denote these heights h1 and h2 as shown in the diagram.
(i) Using Pascal’s ‘law’ (pressures on the same horizontal level in a stationary liquid are the same), obtain an equation connecting the heights h1 and h2 and the corresponding densities ρ1 and ρ2.
(ii) Use this relation to find the density of an unknown oil, where h1 (for oil) = 33.4 cm and
h2 (for water) = 28.6 cm.
This method is called the ‘balancing columns’ method. It can be used to find the ratio of the densities of liquid 1 (like an oil) and liquid 2 (water). This ratio is then the relative density of liquid 1. This method can be used only if the liquids are immiscible.
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6 If the liquids in the previous question are miscible, a short length of mercury is introduced into the U-tube to separate them, as shown in the diagram.
The mercury menisci are both on the same horizontal. Use Pascal’s law to obtain an equation connecting the heights of the liquids and their densities.
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7 In another method, called Hare’s method, the U-tube is inverted and the ends are dipped into two liquids to be compared. A length of rubber tubing fitted to an opening at the bend in the tube is used to suck the two liquids up the tubes.
(i) Explain why the pressures due to these liquids at the levels of the menisci in the beakers (levels P and Q) are equal. Note that these levels need not be the same. (ii) To what agency are the pressures over the menisci in the two beakers due?
(iii) To what agency are the pressures at the same levels (i.e. the level of the meniscus in the beakers), but with the tubes, due?
(iv) Use Pascal’s law to write expressions for these pressures. (v) Hence obtain an equation connecting the pressures.
(vi) Now deduce an expression for the ratio of the densities of the liquids A and B.
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