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### Transcript of Mechanics - hkage.org. Mechanics 1 Vector 1.1 Any vector x 0 y 0 z A A x A y A z 0 r r r r = + +...

• 1

Mechanics

1 Vector

1.1

Any vector 000 zAyAxAA zyxrrrr

++= (1.1)

000 )()()( zBAyBAxBABAC zzyyxxrrrrrr

+++++=+= (1.2)

Dot product of two vectors

zzyyxx BABABAABBA ++== cosrr

(1.3)

It is also referred to as the projection of Ar

on Br

, or vise versa.

Amplitude of the vector

AAAAAA zyx

rrr=++ 222|| (1.4)

1.2

Position vector of a particle: 000 zzyyxxrrrrr

++= (1.5).

If the particle is moving, then x, y, and z are function of time t.

Velocity:

000000 zvyvxvzdt

dzy

dt

dyx

dt

dx

dt

rdv zyx

rrrrrrr

r++=++= (1.6)

Acceleration

00002

2

02

2

02

2

000 zayaxazdt

zdy

dt

ydx

dt

xdz

dt

dvy

dt

dvx

dt

dv

dt

vda zyx

zyx rrrrrrrrrr

r++=++=++= (1.7)

1.3 Uniform circular motion

Take the circle in the X-Y plane, so z = 0,

)cos( += tRx , )sin( += tRy (1.8)

is the angular speed. is the initial phase. Both are constants.

Using the above definition of velocity (1.6),

00 )cos()sin( ytRxtRvrrr

+++ (1.9),

vr

is always perpendicular to rr

.

Its amplitude is Rv = (1.10)

The acceleration is:

rytRxtRarrrr 2

0

2

0

2 )sin()cos( =++ (1.11).

Its amplitude is R

vRa

22 == (1.12)

Ar

Br

Ar

Br

x

y

+= t

x

y

+= t

(x,y,z)

x

y

z

rr

o

(x,y,z)

x

y

z

rr

(x,y,z)

x

y

z (x,y,z)

x

y

z

rr

o

Cr

Br

Ar

Cr

Br

Ar

• 2

2 Relative Motion

A reference frame is needed to describe any motion of an object.

Consider two such reference frames S and S with their origins at

O and O, respectively. The X-Y-Z axes in S are parallel to the X-

Y-Z axes in S. One is moving relative the other.

Note: Rrrrrr

+= ' . (2.1)

So the velocity is: uvdt

Rd

dt

rd

dt

rdv

rrrrr

r '

'+=+== (2.2)

Similar for acceleration: '

'dv dv du

= + = +

r r rrr r

(2.3)

This is the classic theory of relativity. If ur

is constant, then 0A =r

, and 'aarr

= , i. e., Newtons Laws work in all inertia reference frames.

Properly choosing a reference frame can sometimes greatly simplify the problems.

3 Forces

Pull through a rope, push, contact forces (elastic force and friction force), air resistance, fluid

viscosity, surface tension of liquid and elastic membrane, gravity, electric and magnetic,

strong interaction, weak interaction. Only the last four are fundamental. All the others are the

net effect of the electric and magnetic force.

3.1 Tension

Pulling force (tension) in a thin and light rope:

Two forces, one on each end, act along the rope direction. The

two forces are of equal amplitude and opposite signs because

the rope is massless. It is also true for massless sticks.

3.2 Elastics

Elastic contact forces are due to the deformation of solids.

Usually the deformation is so small that it is not noticed. The

contact force is always perpendicular to the contact surface. In

the example both 1Fr

and 2Fr

are pointing at the center of the

sphere.

3.3 Friction

The friction force between two contact surfaces is caused by the relative motion or the

tendency of relative motion. Its amplitude is proportional to the elastic contact force, so a

friction coefficient can be defined.

When there is relative motion, the friction force is given by f =

kN, where k is the kinetic friction coefficient. The direction of the friction is always opposite to the direction of the

relative motion.

O

O

rr

'rr

Rr

O

O

rr

'rr

Rr

Tr

Tr

Tr

Tr

1Fr

2Fr

1Fr

2Fr

vr

Nr

fr

vr

Nr

fr

• 3

When there is no relative motion but a tendency for such

motion, like a block is being pushed by a force Fr

, the

amplitude of f is equal to F until it reaches the limiting value

of sN if F keeps increasing, where s is the static friction coefficient.

Once the block starts to move, the friction becomes f = kN. Usually k < s.

3.4 Viscosity

Air resistance and fluid viscous forces are proportional to the relative motion speed and the

contact area. A coefficient called viscosity is used in these cases.

3.5 Inertial force

Inertial force is a fake force which is present in a reference frame (say S) which itself is

accelerating. Recall that Aaarrr

+= ' . Assume frame-S is not accelerating, then according to

Newtons Second Law, )'( AamamFrrrr

+== . So in the S-frame, if one wants to correctly

apply Newtons Law, she will get 'amAmFrrr

= , i. e., there seems to be an additional force

AmFrr

=int (3.1)

acting upon the object.

Example 3.1

A block is attached by a spring to the wall and placed on the

smooth surface of a cart which is accelerating. According to

the ground frame, the force Fr

acting on the block by the spring

is keeping the block accelerating with the cart, so amFrr

= . In the reference frame on the cart, one sees the block at rest but

there is a force on the block by the spring. This force is

balanced by the inertial force amFrr

=int

3.6 Gravity

Between two point masses M and m, the force is

rr

GMmF

)r

2= (3.2)

The gravitation field due to M is rr

GMg

)r

2= (3.3).

The field due to a sphere at any position outside the

sphere is equal to that as if all the mass is concentrated at

the sphere center. (Newton spent nearly 10 years trying to

proof it.)

Potential energyr

GMU

= (3.4).

Nr

fr

Fr

Nr

fr

Fr

M

M

=

M

M

=

rr

Mm

rr

Mm

amFrr

=

Ground frame

ar

Cart frame

amFrr

=amFrr

=int

amFrr

=

Ground frame

ar

Cart frame

amFrr

=amFrr

=int

• 4

Application of superposition: Uniform density

larger sphere with a smaller spherical hole.

Near Earth surface, because the large radius

of Earth, the gravitation field of Earth can be

taken as constant, and its amplitude is

2

E

E

R

GMg = = 9.8 m/s2 (3.5).

Its direction is pointing towards the center of Earth, which in practice can be regarded as

downwards in most cases.

Example 3.2

The weight, or the force of the ground on a person on

different places on Earth. Take the radius of Earth as R, and

the rotation speed being (= 2/86400 s-1). On the Equator, we have mg - N = ma = m2R, so N = mg - m2R = m(g - 2R)

At the South Pole (or North Pole), we have N = mg

At latitude , by breaking down the forces along the X-direction and Y-direction, we have

mafNmg = sincos)( , and

0cossin)( =+ fNmg .

Solving the two equations, we get cos)( maNmg = , sinmaf = , where cos2 Ra = .

The negative sign of f means that its direction is the opposite of what we have guessed.

One can also break down the forces along the tangential and radial directions to obtain the

same answers. One can also take Earth as reference frame and introduce the inertia force to

account for the rotational acceleration.

3.7 Buoyancy

In a fluid (liquid or gas) of mass density at depth H, consider a column of it with cross section area A, then the

total mass of the column is AH, and the gravity on it is AHg. The gravity must be balanced by the supporting force from below, so the force of the column on the rest of the

liquid is F = AHg and pointing downwards. The pressure

=

1rr

2rr

1rr

2rr

O O O

-=

1rr

2rr

1rr

2rr

O O O

-

y

x

gmr

gmr

Nr

Nr

Nr

gmr f

rar

y

x

gmr

gmr

Nr

Nr

Nr

gmr f

rar

gmrHgmrH

• 5

P = F/A = Hg (3.6).

Now consider a very small cubic of fluid with all six side area of A at depth H. The force on

its upper surface is AHg and pointing down, the force on its lower surface is AHg but pointing upwards so the cubic is at rest. However, for the cubic not to be deformed by the two

forces on its upper and lower surfaces, the forces on its side surfaces must be of the same

magnitude. This leads to the conclusion that the pressure on any surface at depth H is AHg, and its direction is perpendicular to the surface. One can then easily prove that the net force of

the fluid (buoyancy) on a submerged body of volume V is equal Vg. (See the HKPhO 2003 paper.) The buoyancy force is acting on the center of mass of the submerged portion of the

object.

3.8 Torque

When two forces of equal amplitude and opposite directions

acting upon the two ends of a rod, the center of the rod remains

stationary but t