### Chapter 5 Applications of Newton`s Law

```Chapter 5 Applications of
Newton’s Law

Sec. 5-1 Force Laws
Sec. 5-2 Tension and normal forces
Sec. 5-3 Friction forces ★
Sec. 5-4 The dynamics of uniform circular motion ★
Sec. 5-5 Time-dependent force
Sec. 5-6 Noninertial frames and pseudoforces ★
Sec. 5-7 Limitations of Newton’s law
Sec. 5-1 Force Laws
basic forces:
(1) the gravitational force
(2) the electromagnetic force
(3) the weak nuclear force, which causes
certain reactions among the fundamental
particles.
(4) the strong force, which operates among
the fundamental particles and is responsible
for binding the nucleus together.
Two protons in typical nucleus, for example, the
relative strength of these forces would be:
strong (relative strength = 1);
2
10
electromagnetic(
);
weak ( 109);
gravitational ( 1038).
In fact, everything we study about ordinary
mechanical systems involves only two force:
gravity and electromagnetism. Tension forces

Normal forces
Friction forces
Sec. 5-2 Tension and normal forces
(张力与压力)
(1) Tension force (such as in a stretched rope or
string), arises because each small element of the
string pulls on the element next to it.
m
If the mass of the rope is negligible, the values of
the force exerted on the two ends of the rope must
be nearly equal to each other.
(2) Normal force : Just like tension force,
the normal force is also contact force.

N

 N'
Both tension and normal forces originate
with the atoms of each body --- each atom
exerts a force on its neighbor. They belong
to electromagnetic forces.
Sample problem:
5-7 In a system, a block (of mass m1 =
9.5 Kg) slides on a frictionless plane
inclined at an angle   34 . The block is
attached by a string to a second block (of
mass m2=2.6 Kg). The system is released
from rest. Find the acceleration of the
blocks and the tension in the string.
m2
Sec. 5-3 Friction forces ★
Friction is the force that opposes (反抗) the
relative motion or the trend of relative motion of
two solid surfaces in contact
Friction
static friction
kinetic friction
sliding friction √
rolling friction
1) The forces of static friction (静摩擦力)
The frictional forces acting between surfaces at
rest with respect to each other.
Friction force can be measured by following expt.
Fig 5-12 Friction force

fk
fsMax

fs
rest
moving
The maximum force of static friction will be the same as
the smallest applied forces necessary to start motion.
The maximum force of static friction f s Max
between any pair of dry unlubricated surface
follows these two empirical laws :
(1) It is approximately independent of the area of
contact surfaces.
(2) It is proportional to the normal force
f s Max  s N
(5-7)
where N the magnitude of the normal force,
s the coefficient of static friction,
f s Max the maximum force of static friction.
f s  f s Max
 2)The force of kinetic friction(动摩擦力,
（滑动）):
f k  k N
(5-8)
where  k is the coefficient of kinetic friction.
Usually, for a given pair of surfaces  s   k.
The actual value of  s and  k depend on
the nature of both the surfaces in contact.
Table 5-1 some representative values of  s and  k .
Surface
s
k
1.0
0.8
Rubber on dry
concrete
Glass on glass
0.9 ~ 1.0
0.4
Steel on steel
0.6
0.6
Wood on wood
0.25 ~ 0.5
0.2
Waxed wood ski
on dry snow
0.04
0.004
Sample problem:
5-10 Repeat Sample Problem 5-7, taking
into account a frictional force between block
1 (m1) and the plane. Use the values s
=0.24 and  k =0.15. Find the acceleration of
the blocks and the tension in the string.
m1 = 9.5 Kg
m2=2.6 Kg
  34
m2
Sec. 5-4 The dynamics of uniform
circular motion
1) The conical pendulum(锥摆)
Fig 5-18
T

L
m
R
m

v
mg
Fig 5-18 shows a conical pendulum, as the mass m
is revolving in a horizontal circle with constant
speed v, the string L sweeps over the surface of an
imaginary cone.
Can we find the period of the motion?
T cos  mg
｛
2
T sin   mar  mv / R
(R  L sin )
(5-12)
(5-13)
v  Rg tan
If we let t represent the time for one complete
revolution of the body, then
2R
R
L cos
t
 2
 2
v
g tan
g
t is called the period of motion.
Fig 5-20
2) The banked curve

Let the block in Fig 5- v
20 represent an
a)
R
automobile or railway
c
car moving at constant

speed v on a level

v
b)
of curvature R.
Where does the centripetal force come from?

a): sidewise frictional
force exerted by the
b):
N sin   mv / R
2
N cos  mg
2
tan   v / Rg
Example 1
Example 2
See 动画库/力学夹
/2-02牛顿定律例题
.exe 例3
Problem:
A child whirls a stone in a horizontal
circle 1.9 m above the ground by means of
a string 1.4 m long. The string breaks, and
the stone flies off horizontally, striking the
ground 11m away. What was the centripetal
acceleration of the stone while in circular
motion?
Sec. 5-5 Time-dependent force
If the forces are dependent on time, we can
still use Newton’s laws to analyze the motion.
For simplicity, we assume here that the
forces and the motion are in one dimension,
which we take to be the x direction. Then
dv
Fx(t)  ma(t)  m
dt
Fx (t )
dvx 
dt
m
(5-18)

vx
v0 x
dvx  
t
0
v x (t )  v0 x
1 t
Fx (t )
dt or v x  v0 x  0 F(t)dt
m
m
1 t
  F (t )dt
m 0
(5-19)
wherev0 x is initial velocity, v is the velocity at
x
time t.
dx

v
In the same way with
x
dt , we have
t
x(t )  x0   v x (t )dt
0
(5-20)
If Fx is a constant, Eqs. 19 and 20 will reduce to the
formula we obtained for const. acceleration motion.
Discussion
  
Basic concepts in kinematic motion: a, v , r , t
How about the motion if the acceleration is a
function of position, such as a spring oscillator?
F   kx
v(t)? a(t)?
2
dv
d x
F  kx  m  m 2  mx
dt
dt
k
x  Acos(ωt  φ), ω 
m
A and φ are determined by initial conditions.
Sample problem 5-11
 A car of m=1260kg is moving at 105 km/h.
the driver begins to apply the brakes so that
the magnitude of the braking force increases
linearly with time at the rate of 3360N/s
(a) How much time passes before the car
comes to rest?
(b) how far does the car travel in the
process?
Solution:
(a) we choose the direction of the car’s velocity as
the positive x direction, then we can represent the
braking force as F  ct  3360t N
x
1 t
ct 2
and
vx (t） v0 x   (ct)dt  v0 x 
m 0
2m
Let this expression for vx equal to zero and solve
for t,
2v0 x m
2(29.2m / s)1260kg
t1 

 4.68s
c
5360N / s
The car comes to rest at
t1  4.68s
(b) According to Eq(5-20)
t
ct 2
ct 3
x  x0   (v0 x 
)dt  x0  v0t 
2m
6m
0
Evaluating the expression at t  t1 , x0  0
we obtain
(3360N / s)(4.68s)3
x(t1 )  (29.2m / s)(4.68s) 
 91.1m
6(1260kg)
Fig 5-22
x
v
t1
t (s)
ct 3
x  x 0  v0 t 
6m
2
a
t1
t (s)
t1
t (s)
ct
v x ( t） v0 x 
2m
ax (t )  ct / m
Sec. 5-6 Noninertial frames and
pseudoforces(赝力)
 1) How is the motion equation if a noninertial
frame is chosen?
Noninertial frame--- a frame that is accelerated
as viewed from an inertial frame.
See an example
Consider an observer s’ in
a van that is moving at
constant velocity. The van
contains a long airtrack
with a frictionless 0.25kg
glider at one end (Fig 522a).
The driver of the van
applies the brakes, and the
van begins to decelerate

with acceleration of a .
Fig 5-22
(a)
o
S

v
a 0
o
S’
a’

a0


a '  a
(b)

a'
Observer S’ sees the glider accelerate with
and can find no object in the environment of the
glider that exerted a force on it.
At same time S measures the van accelerate with
a . It is found that the relationship between the two

 .
accelerates is a
'  a
To preserve the applicability of Newton’s Second
law, S’ must assume that a force (a pseudo-force)

acts on the glider. According to S’, this force F '


must equal ma '   ma .
To apply classical mechanics in noninertial frames
( a ), we must introduce additional forces
known as ‘pseudoforces’ (sometimes called
‘inertial forces’（惯性力）).


F '  ma
2) Why is the force called ‘pseudoforce’?
Pseudoforces = non-Newtonian forces
Pseudoforces violate Newton’s third law. The
observer of s’ cannot find a reaction force exerted
by the glider on some other body.
Pseudoforces depends on the frames chosen.
 3) Pseudoforces are very real to those that
experience them.
To S, who is in inertial frame (ground), this
is quite natural; your body is simply trying
to obey Newton’s first law and move in a
straight line.
From your point of view in the noninertial
reference frame of the car, you must ascribe
your sliding motion to a pseudo-force pulling
you to the left. This type of pseudo-force is
called a centrifugal force(离心力), meaning a
force directed away from the center.
4) Is pseudoforce (inertial force) really pseudo?
A virtual experiment
• General Relativity:
found in 1915
• Principle of equivalence:
gravity ~ inertial force
g
Application of inertial forces:
See 动画库/力学夹/2-02牛顿定

Sec. 5-7 Limitations of Newton’s law
Special relativity teaches us that we can
not extrapolate the use of Newton’s laws to
particles moving at speeds comparable to
the speed of light. General relativity shows
that we can not use Newton’s laws in the
vicinity of extremely massive objects.
Quantum mechanics teaches us that we can
not extrapolate Newton’s laws to objects as
small as atoms.
```