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· @emperorhassy ·
<p class="MsoNormal"><span style="font-size: 1rem;">One of the applications of Bernoulli’s
principle is the fragrance spray. </span><span style="font-size: 1rem;">The diagram shows a hand-operated fragrance spray</span><span style="font-size: 1rem;">.
When you press the bulb, air is forced through the tube, which has a
constriction in it – that is, part of the tube is narrower than the rest. The
air travels faster in the narrow part and so its pressure decreases. Liquid is
drawn up the vertical tube, breaks into droplets as it meets the air jet and is
carried out of the nozzle in a fine spray.</span></p><p class="MsoNormal"></p><div style="text-align: center;"><img src="" data-filename="ciq0bnpuvqnt1rcwqzxo" style="font-size: 1rem; width: 527.5px;"><sup><a href="" target="_blank">A perfume bottle. Pixabay</a></sup></div><p></p><h2><span lang="">Aerofoils: the principle of flight<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">An aerofoil is the shaped wing of an
aircraft.&nbsp;The top surface of an aerofoil is convex, which
means that, while the aircraft is in flight, air travels faster over the top
surface than over the bottom. </span></p><p class="MsoNormal"><img src="" data-filename="sokiiyyltdjdeh5qvi67" style="width: 302px; float: right;" class="note-float-right"></p><div style="text-align: center;"><a href="" target="_blank"></a><a href="" target="_blank" style="background-color: rgb(255, 255, 255); font-size: 1rem;"><sup>Streamlines around a NACA 0012 airfoil at moderate angle of attack. Michael Belisle, Public Domai</sup>n</a><br></div><p></p><p class="MsoNormal"><span lang=""><br></span></p><p class="MsoNormal"><span lang="">This is because the air stream has to travel a
greater distance over the top surface in the same time – as with a liquid, the
air mass has to be continuous. The result is a smaller pressure on the upper
surface than on the lower, and the aerofoil experiences a net upward thrust –
called <b>lift</b>. The wing is usually angled to the direction of motion so
that it has to push against the air. This produces an additional force
(reaction) with an upward component to add to the lift. In general, the angle
the aerofoil makes with the forward direction of motion (angle of attack) is
large enough for this reaction force to be more effective in maintaining height
than the aerofoil’s Bernoulli effect.<o:p></o:p></span></p><h2><span lang="">STREAMLINES AND TURBULENT FLOW: FRICTION AND DRAG<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">The simple theory of fluid flow described
above assumes that the flow is steady, so that particles of the fluid flow in
smooth paths – as shown by the continuous streamlines made visible by
streamers. But above a certain speed the flow becomes chaotic and we have <b>turbulent</b>
flow. Energy is dissipated in sound and heating; and extra pressures are
produced which can affect the straightforward motion of a vehicle. There is often
turbulent flow when two air streams that have diverged over a moving object
join together again. This effect can be seen in the air behind a car being
tested in a wind tunnel. The disturbed turbulent air produces a low-pressure
region, which tends to pull the car backwards, contributing to the drag forces
that oppose the motion of any object through a fluid. The turbulent drag force
is reduced by vehicle design. The best shape is like an aerofoil, of course,
but drivers would worry if, at a certain speed, the car started flying! The
rear spoiler fitted to some cars,<b> </b>as shown in the figure below, is a way of
reducing the drag force by making the air flow less turbulent at the back of
the vehicle. Racing cyclists and downhill skiers wear specially shaped helmets
to reduce the effect of turbulence.</span></p><p class="MsoNormal" style="text-align: center; "><img src="" data-filename="r9wx6zlmzvxkaprzl8ea" style="width: 527.5px;"><span lang=""><o:p><br></o:p></span></p><p class="MsoNormal" style="text-align: center; "><a href="" target="_blank"><sup>A car having a rear spoiler. Pixabay</sup></a><span lang=""><o:p><br></o:p></span></p><p class="MsoNormal"><span lang="">Other drag forces act on a surface moving
through a fluid. For example, air has to be pushed away as a car drives through
it, and this produces a reaction force that acts on the car. Next, I will
discuss the frictional forces between the vehicle’s surface and the fluid.<o:p></o:p></span></p><h2><span lang="">FRICTION AND VISCOSITY<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">Layers of fluids move over each other very
easily; for this reason a fluid cannot resist a shearing stress. But fluids do
have a kind of internal friction called viscosity. This varies: moving a knife
blade through water is much easier than moving it through honey. Similarly,
viscous forces arise when adjacent layers of liquid move against each other. In
a simple model of what happens, we assume that when a fluid moves through a
pipe the fluid layer next to the wall of the pipe is at rest, and that the fastest
stream is at the centre of the pipe. There is a constant velocity gradient in
the fluid, </span><span lang="">Δv</span><span lang="">/</span><span lang="">Δ</span><span lang="">x, where x is
the distance measured radially in the pipe. A liquid that behaves like this is
called a <b>Newtonian</b> <b>liquid</b>. Friction between adjacent layers in
the fluid determines how fast the fluid can flow compare – pouring water and
treacle. A fluid has a <b>coefficient of viscosity</b> </span><b>η</b><span lang=""> (the Greek letter eta) which determines the size of the viscous force.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The resistive viscous force is greater for a
wider pipe. These factors are combined in <b>Newton’s law of viscosity</b>.
When F is the force due to viscosity acting on a fluid stream of
cross-sectional area A, against the direction of flow, then:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">F = </span><span lang="">η</span><span lang="">A (</span><span lang="">Δv/Δx)</span><span lang=""><o:p></o:p></span></h4><p class="MsoNormal"><span lang="">This force is also the drag force on the sides
of the container.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Viscosity has units of Ns m<sup>-2</sup> . Blood
is thicker than water i.e blood is more viscous than water. Viscosity varies
with temperature: as a general rule the coefficient decreases as temperature
increases, so that liquids are more ‘runny’ when hot.</span></p><h2><span lang="">STOKES’ LAW</span></h2><p><img src="" data-filename="oaz83xqxsvqv2m23ryun" style="width: 333px; float: right;" class="note-float-right"><sup><a href="" target="_blank">Creeping flow past a falling sphere in a fluid (e.g., a droplet of fog falling through the air): streamlines, drag force Fd and force by gravity Fg. Kraaiennest, CC BY-SA 3.0</a></sup></p><p class="MsoNormal"><span lang="">Viscous forces also affect the speed of
streamline motion of an object in fluid. For a sphere in streamline motion at a
speed v through a fluid, the Irish physicist George Stokes (1819-1903) proved
that it would experience a drag force:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">F = 6</span><span lang="">πη</span><span lang="">rv<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">Where </span><span lang="">η</span><span lang=""> is the
coefficient of viscosity of the fluid and r is the radius of the sphere. This
idea may be linked to the terminal speed of a falling sphere. In this case the
viscous drag on the sphere equals its weight:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">mg = 6</span><span lang="">πη</span><span lang="">rv<o:p></o:p></span></h4><h2><span lang="">EXAMPLE<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">Calculate the terminal speed of a) a lead
sphere, b) a hailstone, both having a radius of 2 mm. <o:p></o:p></span></p><p class="MsoNormal"><span lang="">[The density of lead is 1.14 </span><span lang="">×</span><span lang=""> 10 kgm<sup>-3</sup>; the density of ice is 9.2 </span><span lang="">×</span><span lang=""> 10<sup>2</sup> kgm<sup>-3</sup>.]<o:p></o:p></span></p><p class="MsoNormal" style="line-height:106%"><b>Solution</b><span lang="">: The condition for terminal speed is that mg = 6</span><span lang="">πη</span><span lang="">rv, so terminal speed v is:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">v = mg / 6πηr<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">The mass of the lead sphere is:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">m = volume </span><span lang="">×</span><span lang=""> density = 4/3 πr<sup>3 </sup></span><span lang="">× ρ</span><span lang=""><o:p></o:p></span></h4><p class="MsoNormal"><span lang="">So the equation can be simplified to:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">v = 2r<sup>2</sup></span><span lang="">ρ</span><span lang="">g / 9η<o:p></o:p></span></h4><h4 align="center" style="text-align:center"><span lang="">= 2 × 4 × 10<sup>-6</sup>
× 1.14 × 10<sup>4</sup> × 9.8 / 9 × 1.8 × 10<sup>-5</sup><o:p></o:p></span></h4><p class="MsoNormal"><span lang="">giving: <o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">v = 5.5 × 10<sup>3</sup>
ms<sup>-1</sup><o:p></o:p></span></h4><p class="MsoNormal" style="line-height:105%"><span lang="">The only difference
in the conditions of the problem is the density of the materials. Thus the ratio
of terminal speeds is the same as the ratio of densities. Ice is less dense
than lead, so it falls more slowly.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The terminal speed of the hailstone is:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">v = 5.5 </span><span lang="">×
10<sup>3</sup></span><span lang=""> </span><span lang="">× 9.2 × 10<sup>2</sup></span><span lang=""> / 1.14 </span><span lang="">× 10<sup>4</sup></span><span lang=""> = 4.4 </span><span lang="">×
10<sup>2</sup></span><span lang=""> ms<sup>-1</sup><o:p></o:p></span></h4><p class="MsoNormal"><span lang="">Note that in practice both spheres would fall
more slowly, since turbulence would set in at much lower speeds.<o:p></o:p></span></p><h2><span lang="">DRAG EFFECTS FOR A VEHICLE<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">Energy can be ‘saved’ in transportation by
reducing a variety of losses. Nowadays the drag factor is taken into account in
car design. This factor indicates how much energy a moving vehicle ‘loses’ as a
result of air friction. This kind of friction loss is due to the vehicle working
against a frictional <b>drag force F<sub>d</sub></b>. When v is the vehicle’s
speed, A its cross-sectional area and </span><span lang="">ρ t</span><span lang="">he density of
air, then:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">F<sub>d</sub> = ½ CA </span><span lang="">ρv<sup>2</sup></span><span lang=""><o:p></o:p></span></h4><p class="MsoNormal"><span lang="">Where C is its <b>drag</b> <b>coefficient</b>,
a constant which depends upon the shape of the vehicle. The car is also subject
to rolling friction, due to the contact of the tyres with the road surface.
This creates a force given by </span><i><span lang="">μ</span><span lang="">N</span></i><span lang=""> where N is the normal reaction between car and road. The coefficient of
rolling friction for a car is about 0.02. The normal reaction N gets slightly
less at higher speeds as the car behaves like an aerofoil (luckily not very
successfully at usual road speeds).<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Rolling friction is much more important than
drag for a car travelling at low speeds. However, drag increases as the square
of the speed. Drag and rolling friction become equal at a speed of about 18 ms<sup>-1</sup>
(40 mph) for a typical car, and eventually drag is the main source of
frictional loss.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The main energy losses from a car engine are not
due to these effects at all. Heat engines are inefficient, and dissipate almost
70 per cent of the energy that comes from the burning fuel. There are other friction
losses in the transmission, so that the energy available for moving the car is
only about 14 per cent of the energy supplied by the fuel.<o:p></o:p></span></p><h2><span lang="">EXAMPLE<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">How much power needs to be delivered to the
wheels of a car travelling at a typical motorway speed of 31 ms<sup>-1</sup>
(70 mph)?<o:p></o:p></span></p><p class="MsoNormal"><span lang="">(Data: At this speed, rolling friction
provides a force of 210 N; the car has a cross-sectional area of 1.8 m<sup>2</sup>;
the density of air is 1.3 kg m<sup>-3</sup>; the drag coefficient of the car is
0.4.)<o:p></o:p></span></p><p class="MsoNormal"><b>SOLUTION</b><span lang="">: &nbsp;The drag force on the car is Calculated from:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">F<sub>d</sub> = ½ CA </span><span lang="">ρv<sup>2</sup></span><span lang=""><o:p></o:p></span></h4><h4 align="center" style="text-align:center"><span lang="">= 0.5 × 0.4 × 1.8 ×
1.3 × (31)<sup>2</sup> N = 450 N<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">Drag force = 450 N. Thus the total frictional
force on the car is: <o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">210 N + 450 N = 600 N<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">This force does work that requires energy to
be supplied by the car’s engine.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The work done per second = force </span><span lang="">×</span><span lang=""> distance moved per second (speed)<o:p></o:p></span></p><p class="MsoNormal"><span lang="">that is: Power needed = work done per second =
frictional force </span><span lang="">×</span><span lang=""> speed<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">= 660 N </span><span lang="">×</span><span lang=""> 31 ms<sup>-1</sup> = 20.5 kW<o:p></o:p></span></h4><h2><span lang="">HOW FAST CAN A VEHICLE GO?<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">The speed of a train or a car – or a cyclist –
is the result of a battle between the driving force and resistive forces.
Newton’s laws act as the referee. The best way to tackle many problems is from the
point of view of <b>energy</b> and <b>power</b> (the rate of transferring
energy).<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The car in the figure below is travelling on
a horizontal road at a steady speed of 25 m/s with its engine delivering energy
at its maximum rate of 14 kW. What is this energy being used to do? It is, in
fact being used to do work against friction forces: inside the engine, the
transmission, the wheels and – mostly – against the forces of drag and viscous
air friction. </span></p><p class="MsoNormal" style="text-align: center; "><img src="" data-filename="rvxl3yzdifs849nysg4s" style="width: 527.5px;"><span lang=""><br></span></p><p class="MsoNormal" style="text-align: center; "><a href="" target="_blank"><sup>A racing car. Pixabay</sup></a><span lang=""><br></span></p><p class="MsoNormal"><span lang="">It is also pushing air out of its way and giving it some kinetic
energy. The car is not accelerating or going uphill so doesn’t gain potential
or kinetic energy. Suppose the total of the friction and other forces opposing
motion is F. When the car travels 25 metres it does work of 25F.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The car takes one second to travel 25 metres,
so the work done per second is also 25F. This is the power developed by the
engine of course, since all of it is being used to do work against friction.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">So, <i>25F = 14 kW</i> in this case. This
gives the total frictional force as 560 N i.e 14000 / 25.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The power generated by the engine is P. What I
have shown is that when the car is travelling at a steady speed on the flat<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">power = frictional
forces </span><span lang="">×</span><span lang=""> speed<o:p></o:p></span></h4><h4 align="center" style="text-align:center"><span lang="">P = Fv<o:p></o:p></span></h4><h3><span lang="">Going uphill<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">Suppose our car is now climbing a not very steep
hill that rises 1 metre for every 25 metres along the road. Some of the power
from the engine has to lift the car uphill and so give it potential energy. The
car can’t go as fast, because the engine has an extra task:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">P = Fv + mg</span><span lang=""> Δ</span><span lang="">h<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">Here, </span><span lang="">Δ</span><span lang="">h is the
increase in height of the car every second. This depends on the angle of slope
and the speed of the car: </span><span lang="">Δ</span><span lang="">h = v sin </span><span lang="">θ</span><span lang=""> &nbsp;which shows that this is simply
v/25 in this example.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The car’s speed drops:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">Fv + mg</span><span lang=""> Δ</span><span lang="">h = P<o:p></o:p></span></h4><h4 align="center" style="text-align:center"><span lang="">(F + mg/25) v = P<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">If the car’s mass is 800 kg and if (not too
likely) F stays the same at 560 N, then you should be able to check that the
car’s speed drops to just under 16 ms<sup>-1</sup>.<o:p></o:p></span></p><h2><span lang="">FREE-BODY DIAGRAMS<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">A good way to test your understanding of what
is happening when bodies are acted upon by forces is to draw a simple diagram
showing where and in which direction the forces act. </span></p><p class="MsoNormal"><img src="" data-filename="j6ilay1vuxnn8l2gk08z" style="width: 187px; float: right;" class="note-float-right"><sup><a href="" target="_blank">Free Body Diagram. AndrewDressel at English Wikipedia, CC BY-SA 3.0</a></sup></p><p class="MsoNormal"><span lang="">To do this you need to
have a good understanding of Newton’s laws of motion and be able to apply them to
the situation the body finds itself in. For most everyday situations the forces
acting will be caused by gravity (or some other applied force), friction and
any force of contact between the object and another object. This contact force
is often called a <i>reaction</i> force – but this can be a confusing name
because it may not be the same force as that discussed above in connection with
Newton’s third law. The figure below shows some simple situations with the forces
involved drawn as vectors from the points where they act.<o:p></o:p></span></p><h2><span lang="">CONCLUSION<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">After reading all these series of posts starting
from&nbsp;</span><a href="!/@emperorhassy/newton-on-the-move-g-1571423646" target="_blank">NEWTON ON THE MOVE: Going Around the Bend</a><span lang="">, </span><a href="!/@emperorhassy/newton-on-the-move-p-1572374061" target="_blank">Practical Applications of Collisions</a><span lang="">,&nbsp;</span><a href="!/@emperorhassy/fluid-pressure-float-1572701505" target="_blank">Fluid Pressure</a><span lang=""> and this, you should have improved your understanding of
Newton’s laws of motion as applied to transport and moving fluids. You should
have also understood the role of friction in transportation, ability to apply
the laws of static and sliding friction and coefficient of friction. I also
mentioned about the ideas of viscosity, streamline flow and drag factor, and
relating them to Stokes’ law and terminal speed. I discussed the physics of
rotating bodies and used the ideas of angular velocity, angular acceleration,
torque, rotational inertia and rotational kinetic energy.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">I demonstrated how to apply the laws of
conservation of momentum and conservation of energy as appropriate to elastic
and inelastic collisions. The concept of pressure in fluids and relating it to
practical situations involving hydraulics and flotation. The Archimedes’
principle and the applications of Bernoulli’s principle to explain how fluid
pressure and speed are related when a fluid moves relative to an object such as
an aerofoil.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Finally, I explained how the physics in this
series of post is important in society and practical applications, such as
collisions, road safety, wheels and transport in general.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Till next time, I still remain my humble self,
@emperorhassy.<o:p></o:p></span></p><h3><span lang="">Thanks for reading.<o:p></o:p></span></h3><h2><span lang="">REFERENCES<o:p></o:p></span></h2><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href="'_law" target="_blank">'_law</a><span lang="">&nbsp;</span></p><p>

</p><p class="MsoNormal"><a href="" target="_blank"></a><span lang="">&nbsp;</span></p><p class="MsoNormal"><a href=""></a><a href="" target="_blank"></a></p><p class="MsoNormal"><a href="" target="_blank"></a><br></p><p class="MsoNormal"><a href="" target="_blank"></a><br></p><p class="MsoNormal"><a href="" target="_blank"></a><br></p><p class="MsoNormal"><a href="" target="_blank"></a><br></p><p class="MsoNormal"><br></p>
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last_update2019-11-08 18:10:06
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