Fluid Mechanics - Hydrostatics Sections 11 – 5 and 6

A closed system If you take a liquid and place it in a system that is CLOSED like plumbing for

example or a car’s brake line, the PRESSURE is the same everywhere.

Since this is true, if you apply a force at one part of the system the pressure

is the same at the other end of the system. The force, on the other hand MAY

or MAY NOT equal the initial force applied. It depends on the AREA.

You can take advantage of the fact that the

pressure is the same in a closed system as

it has MANY applications.

The idea behind this is called

PASCAL’S PRINCIPLE.

Pascal’s Law

Pascal’s Law: An external pressure applied to an

enclosed fluid is transmitted uniformly throughout

the volume of the liquid.

Fout Fin Aout Ain

Pressure in = Pressure out

in out

in out

F F

A A

Pascal’s Principle

Another Example - Brakes

shoepadbrake

shoepadbrake

pedalbrake

pedalbrake

A

F

A

F

PP

/

/

21

In the case of a car's brake

pads, you have a small initial

force applied by you on the

brake pedal.

This transfers via a brake line,

which had a small cylindrical

area.

The brake fluid then enters a

chamber with more AREA

allowing a LARGE FORCE to

be applied on the brake

shoes, which in turn slow the

car down.

Example 1. The smaller and larger pistons of a hydraulic

press have diameters of 4 cm and 12 cm. What input force

is required to lift a 4000 N weight with the output piston?

Fout Fin Aoutt Ain ; in out out in

in

in out out

F F F AF

A A A

2

2

(4000 N)( )(2 cm)

(6 cm)inF

2; 2

DR Area R

F = 444 N

Rin= 2 cm; Rout = 6 cm

Example 2 - The drawing shows a hydraulic chamber with a

spring, k = 1600 N/m, attached to the input piston, area

15 cm2 and a rock of mass 40.0 kg resting on the output

plunger, area 65 cm2. The piston and plunger are nearly at the

same height, and each has a negligible mass. By how much is

the spring compressed from its unstrained position?

Solution - Since the piston and the plunger are at the same height,

Equation 11.5, , applies, and we can find an

expression for the force exerted on the spring. Then F = kx, can be

used to determine the amount of compression of the spring.

)A/A(FF 1212

kx A

A F

1

2 1

From the drawing in the text, we see that the force on the right

piston must be equal in magnitude to the weight of the rock, or

, Therefore, mgF 1 kx A

A mg

1

2

Solving for x, we obtain

cm 5.65 m 0.0565 cm 65

cm 15 x

N/m 1600

)m/s kg)(9.8 (40.0

A

A

k

mg x

2

22

1

2

Buoyancy When an object is immersed in a fluid, such as a liquid, it

is buoyed UPWARD by a force called the BUOYANT

FORCE. When the object is placed in fluid is DISPLACES a certain amount of fluid. If the object is completely submerged, the VOLUME of the OBJECT is EQUAL to the VOLUME of FLUID it displaces.

Archimedes’ Principle " An object is buoyed up by a force equal

to the weight of the fluid displaced."

In the figure, we see that the difference

between the weight in AIR and the

weight in WATER is 3 lbs. This is the

buoyant force that acts upward to

cancel out part of the force. If you were

to weight the water displaced it also

would weigh 3 lbs.

Archimedes’ Principle

Fluidobject

FluidB

FLUIDB

VV

VgF

VmmgF

)(

)(

Archimedes’ Principle

• An object that is completely or partially submerged in

a fluid experiences an upward buoyant force equal to

the weight of the fluid displaced.

2 lb

2 lb

The buoyant force is due to the displaced fluid. The block material doesn’t matter.

Calculating Buoyant Force

FB = f gVf

Buoyant Force:

h1

mg

Area

h2

FB

The buoyant force FB is due to the

difference of pressure ΔP between the

top and bottom surfaces of the

submerged block.

2 1 2 1; ( )BB

FP P P F A P P

A

2 1 2 1( ) ( )B f fF A P P A gh gh

2 1 2 1( ) ( ); ( )B f fF g A h h V A h h

Vf is volume of fluid displaced.

Example 3 - A 2-kg brass block is attached to a string

and submerged underwater. Find the buoyant force and

the tension in the rope.

All forces are balanced:

mg

FB = gV

T

Force diagram

FB + T = mg FB = wgVw

3

2 kg;

8700 kg/m

b bb b

b b

m mV

V

Vb = Vw = 2.30 x 10-4 m3

Fb = (1000 kg/m3)(9.8 m/s2)(2.3 x 10-4 m3)

FB = 2.25 N

Example (Cont.): A 2-kg brass block is attached to a string

and submerged underwater. Now find the the tension in the

rope.

mg

FB = gV

T

Force diagram

FB + T = mg T = mg - FB

FB = 2.25 N

T = (2 kg)(9.8 m/s2) - 2.25 N

T = 19.6 N - 2.25 N

T = 17.3 N

This force is sometimes referred to as the apparent weight.

Floating objects: When an object floats, partially submerged, the buoyant force

exactly balances the weight of the object.

FB

mg

FB = f gVf mx g = xVx g

ρf gVf = ρxVx g, eliminate gravity

ρf Vf = ρxVx Floating Objects:

If Vf is volume of displaced water Vwd,

the relative density of an object x is

given by:

Relative Density:

x wdr

w x

V

V

Example 4 - A student floats in a salt lake

with one-third of his body above the surface.

If the density of his body is 970 kg/m3,

what is the density of the lake water?

1/3

2/3

Assume the student’s volume is 3 m3.

Vs = 3 m3; Vwd = 2 m3; s = 970 kg/m3

ρw Vwd = ρsVs

3

w3

32 m;

3 m 2

s wd s

w s

V

V

3

w

3 3(970 kg/m )

2 2

s

ρw = 1460 kg/m3

Example 5 - A frog is a hemispherical pod finds that he just floats

without sinking in a blue-green sea (density is 1.35 g/cm3). If the

pod has a radius of 6 cm and has negligible mass, what is the mass

of the frog?

Without sinking, the buoyant force pushing up on the frog and pod

is equal to the weight of the frog and the pod. Since the pod has

negligible mass, we will ignore it. Therefore, the buoyant force is

equal to the weight of the frog.

g 610.73 )5)(.(6cm) 3

4( )g/cm (1.35 V

g

Vg m

Vg, mg

33

Assignment – Chapter 11

• Pages 352 - 353,

• #33, 34, 40, 43, 45

Problem Solving Strategy

1. Draw a figure. Identify givens and what is to be

found. Use consistent units for P, V, A, and r.

2. Use absolute pressure Pabs unless problem

involves a difference of pressure DP.

3. The difference in pressure DP is determined by

the density and depth of the fluid:

2 1

m F; = ; P =

V AP P gh

Problem Strategy (Cont.)

4. Archimedes’ Principle: A submerged or floating

object experiences an buoyant force equal to the

weight of the displaced fluid:

B f f fF m g gV

5. Remember: m, r and V refer to the displaced

fluid. The buoyant force has nothing to do with

the mass or density of the object in the fluid. (If

the object is completely submerged, then its

volume is equal to that of the fluid displaced.)

Problem Strategy (Cont.)

6. For a floating object, FB is

equal to the weight of that

object; i.e., the weight of the

object is equal to the weight of

the displaced fluid:

x or x f x f fm g m g V V

FB

mg