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Škola: Gymnázium, Brno, Slovanské náměstí 7Šablona: III/2 – Inovace a zkvalitnění výuky prostřednictvím

ICTNázev projektu: Inovace výuky na GSN prostřednictvím ICTČíslo projektu: CZ.1.07/1.5.00/34.0940Autor: RNDr. Bc. Miroslav HruškaTematická oblast: Gravitational field and motion of

bodies (gravitační pole a pohyb těles)Název DUMu: Hydrostatics – revisionKód: VY_32_INOVACE_FY.4.17Datum: 22. 4. 2014Cílová skupina: Žáci středních škol s CLIL výukou fyziky v

angličtiněKlíčová slova: Hydrostatics, hydrostatic pressure, Pascal’s law,

Archimedes’ principle Anotace: Prezentace procvičuje principy Pascalova a

Archimedova zákona

Gravitational field and motion of bodies

17. Hydrostatics – revision

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Hydrostatics

P a sca l'sla w

A rch im e d e s'p rin c ip le

L a w s o fh yd ros ta tics

Which law deals with pressure, and which with force?

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The cause of the ……………. in a fluid can be: • an ……………… force acting on the fluid• the forces of ………… acting on individual …………….. of the fluid•

The pressure ………… by an ………….. force acting on a liquid in a closed ……………… is at all ………. of the liquid the same

Fill in the gaps I

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The cause of the pressure in a fluid can be: • an external force acting on the fluid• the forces of gravity acting on individual particles of the fluid•

The pressure caused by an external force acting on a liquid in a closed container is at all points of the liquid the same

Fill in the gaps I – solution

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Exercise I A force of 10 N acts

on the small piston of a hydraulic device of radius 2 cm. The large piston has got a radius of 20 cm. Find the magnitude of the force acting on the large piston?

F1 = 10 N, r1 = 2 cm, r2 = 20 cm F2 = ?

1F

2F

2S1Sp

pp

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Solution I The area of the small piston is

S1 = π × 22 cm2

The area of the large piston isS2 = π × 202 cm2

The force acting on the large piston is

N1000440010

1

212

S

SFF

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We refer to the …………… caused by the forces of …………… acting on individual particles of a liquid as the ……………… pressure

Any object ………………. in a liquid is …………… up by a force equal to the ………… of the liquid …………… by the object

Fill in the gaps II

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We refer to the pressure caused by the forces of gravity acting on individual particles of a liquid as the hydrostatic pressure

Any object immersed in a liquid is buoyed up by a force equal to the weight of the liquid displaced by the object

Fill in the gaps II – solution

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Exercise II Find the hydrostatic pressure in water

at a depth of 5 m.

h = 5 m, ρ = 1000 kg/m3, g = 10 m/s ph = ? ph = hρg = 5 × 1000 × 10 Pa = 50 kPa

The hydrostatic pressure is 50 kPa.

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Find the net force actingon an iron cube of volume 100 cm3 completely immersed in water (ρt = 7860 kg/m3).

The buoyant force Fvz

The force of gravity FG

Exercise III

gVF kvz gVmgF tG

vzF

GF

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V = 100 cm3 = 0.0001 m3

ρT = 7860 kg/m3

g = 10 m/s2

ρk = 1000 kg/m3

The buoyant force Fvz

The force of gravity FG

Solution III

gVF kvz gVmgF TG

vzF

GF

Since ρT > ρk ,

F = FG – Fvz = 7.86 N – 1.00 N = 6.86 N

That is all for now

Thank you for your attention

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Materiál je určen pro bezplatné používání pro potřeby výuky

a vzdělávání na všech typech škol a školských zařízeních. Jakékoliv další využití podléhá autorskému zákonu.

Zdroje: Bednařík, M., Široká, M. Fyzika pro gymnázia, Mechanika.

Dotisk 3. vyd. Praha: Prometheus, 2004. 288 s. ISBN 80-7196-176-0

Hornby, A. S. Oxford Advanced Learner’s Dictionary of Current English. 5. vyd. Oxford: Oxford University Press, 1995. 1428 s. ISBN 0-19-431423-5