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 = + +...
date post
03May2018Category
Documents
view
234download
3
Embed Size (px)
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)
Addition of two vectors:
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 XY 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 XYZ axes in S are parallel to the X
YZ 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
a a Adt dt dt
= + = +
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 frameS is not accelerating, then according to
Newtons Second Law, )'( AamamFrrrr
+== . So in the Sframe, 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 s1). 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 Xdirection and Ydirection, 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