Chapter 5 Applications of Chapter 5 Applications of Newton’s LawNewton’s Law

Sec. 5-1 Force LawsSec. 5-1 Force LawsSec. 5-2 Tension and normal forcesSec. 5-2 Tension and normal forcesSec. 5-3 Friction forcesSec. 5-3 Friction forcesSec. 5-4 The dynamics of uniform circular motionSec. 5-4 The dynamics of uniform circular motionSec. 5-5 Time-dependent forceSec. 5-5 Time-dependent forceSec. 5-6 Noninertial frames and pseudoforces Sec. 5-6 Noninertial frames and pseudoforces Sec. 5-7 Limitations of Newton’s lawSec. 5-7 Limitations of Newton’s law

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Sec. 5-1 Force Laws Sec. 5-1 Force Laws

Physicists have traditionally identified four Physicists have traditionally identified four basic forces:basic forces:

(1) (1) the gravitational forcethe gravitational force (2) (2) the electromagnetic forcethe electromagnetic force (3) (3) the weak nuclear forcethe weak nuclear force, which causes , which causes

certain radioactive decay processes and certain radioactive decay processes and certain reactions among the fundamental certain reactions among the fundamental particles. particles.

(4) (4) the strong forcethe strong force, which operates among , which operates among the fundamental particles and is responsible the fundamental particles and is responsible for binding the nucleus together. for binding the nucleus together.

Two protons in typical nucleusTwo protons in typical nucleus, for example, the , for example, the relative strength of these forces would be:relative strength of these forces would be:

strong (relative strength = ); strong (relative strength = );

electromagnetic( );electromagnetic( );

weak ( ); gravitational ( ). weak ( ); gravitational ( ).

210

3810910

1

In fact, everything we study about In fact, everything we study about ordinary ordinary mechanical systems involves only two force: mechanical systems involves only two force: gravitygravity and and electromagnetismelectromagnetism..

Tension forcesTension forces

Friction forcesFriction forces

Normal forcesNormal forces

Sec. 5-2 Tension and normal forces Sec. 5-2 Tension and normal forces ((张力与压力张力与压力 ))

(1) Tension force(1) Tension force (such as in a stretched rope or (such as in a stretched rope or string), arises because each small element of the string), arises because each small element of the string pulls on the element next to it.string pulls on the element next to it.

m

If the mass of the rope is negligible, the values of If the mass of the rope is negligible, the values of the force exerted on the two ends of the rope must the force exerted on the two ends of the rope must be nearly equal to each other. be nearly equal to each other.

(2) Normal force(2) Normal force : : Just like tension force, Just like tension force, the normal force is also contact force. the normal force is also contact force.

Both tension and normal forcesBoth tension and normal forces originate originate with the atoms of each body --- each atom with the atoms of each body --- each atom exerts a force on its neighbor. They belong exerts a force on its neighbor. They belong to to electromagneticelectromagnetic forces. forces.

'N

N

5-7 In a system, a block (of mass 5-7 In a system, a block (of mass m1 = 9.5 Kg) slides on a frictionless plane ) slides on a frictionless plane inclined at an angle . The block is inclined at an angle . The block is attached by a string to a second block (of attached by a string to a second block (of mass mass m2=2.6 Kg). The system is released ). The system is released from rest. Find the acceleration of the . Find the acceleration of the blocks and the tension in the string.blocks and the tension in the string.

Sample problem:Sample problem:

34

m 1

m2

Sec. 5-3 Friction forcesSec. 5-3 Friction forces ★★ FrictionFriction is the force that opposes ( is the force that opposes ( 反抗反抗 ) the ) the rrelative motionelative motion or the or the trend of relative motiontrend of relative motion of t of two solid surfaces in contactwo solid surfaces in contact

Frictionstatic friction

kinetic frictionsliding friction

rolling friction

√√

1) The forces of static friction (1) The forces of static friction ( 静摩擦静摩擦力力 ))

The frictional forces acting between surfaces The frictional forces acting between surfaces at at

restrest with respect to each other. with respect to each other.

The The maximum force of static frictionmaximum force of static friction will be the same as will be the same as the the smallest applied forcessmallest applied forces necessary to start motion. necessary to start motion.

Fig 5-12 Friction forceFig 5-12 Friction force

fsMax

rest moving

Friction force can be measured by following expt. Friction force can be measured by following expt.

kf

sf

The The maximum force of static frictionmaximum force of static friction betwee between any pair of dry unlubricated surface follows thesn any pair of dry unlubricated surface follows these two empirical laws :e two empirical laws :

(1) It is approximately (1) It is approximately independent of the areaindependent of the area of c of contact surfaces.ontact surfaces.

(2) It is (2) It is proportional to the normal forceproportional to the normal force

(5-7)(5-7)

where where N the magnitude of the normal force, the magnitude of the normal force,

the coefficient of static friction,the coefficient of static friction,

the maximum force of static friction.the maximum force of static friction.s

s sMaxf N

sMaxf

sMaxf

s sMaxf f

2)The force of kinetic friction(2)The force of kinetic friction( 动摩擦力动摩擦力 ,,（滑动）（滑动） ):):

(5-8)(5-8)

where is the coefficient of kinetic friction. where is the coefficient of kinetic friction.

Nf kk

k

kks

sUsually, for a given pair of surfaces . Usually, for a given pair of surfaces . The actual value of and depend on The actual value of and depend on the nature of both the surfaces in contact.the nature of both the surfaces in contact.

SurfaceSurface

Rubber on dry Rubber on dry concreteconcrete

1.01.0 0.80.8

Glass on glassGlass on glass 0.9 0.9 ~~ 1.0 1.0 0.40.4

Steel on steelSteel on steel 0.60.6 0.60.6

Wood on woodWood on wood 0.25 0.25 ~ 0.5~ 0.5 0.2 0.2

Waxed wood ski Waxed wood ski on dry snowon dry snow

0.040.04 0.0040.004

s

k

Table 5-1 some representative values of and .Table 5-1 some representative values of and .k

s

Sample problem:Sample problem:

5-10 Repeat Sample Problem 5-7, taking 5-10 Repeat Sample Problem 5-7, taking into account a frictional force between block into account a frictional force between block 1 (m1 (m11) and the plane. Use the values ) and the plane. Use the values

=0.24 and =0.15. =0.24 and =0.15. Find the acceleration of Find the acceleration of the blocks and the tension in the stringthe blocks and the tension in the string..

sk

m 1

m2

m1 = 9.5 Kg

m2=2.6 Kg

34

Sec. 5-4 The dynamics of uniform Sec. 5-4 The dynamics of uniform circular motioncircular motion

1)1) The conical pendulum(The conical pendulum( 锥摆锥摆 ))

mg

m

T

L

m

v

R

Fig 5-18

Fig 5-18 shows a conical pendulum, as the mass m is revolving in a 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 horizontal circle with constant speed v, the string L sweeps over the surface of an imaginary cone. surface of an imaginary cone.

Can we find the period of the motion?Can we find the period of the motion?

)sinLR(

mgT cos

RmvmaT r /sin 2

(5-12)(5-12)

(5-13)(5-13)｛

If we let If we let tt represent the time for one complete represent the time for one complete revolution of the body, then revolution of the body, then

tt is called the period of motion. is called the period of motion.

tanRgv

g

L

g

R

v

Rt

cos2

tan2

2

2) The banked curve2) The banked curve

Let the block in Fig 5-Let the block in Fig 5-20 represent an 20 represent an automobile or railway automobile or railway car moving at constant car moving at constant speed speed v on a level on a level roadbed around a roadbed around a curve having a radius curve having a radius of curvature R. of curvature R.

Fig 5-20 Fig 5-20

cR

v

v

Where does the centripetal force come from?Where does the centripetal force come from?a):a): sidewise frictional sidewise frictional forceforce exerted by the exerted by the road on the tires.road on the tires.

RmvN /sin 2

a)

b)

b):b):mgN cos

Rgv /tan 2

Example 1 Example 1 Example 2 Example 2

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Problem:Problem:

A child whirls a stone in a horizontal A child whirls a stone in a horizontal circle 1.9 m above the ground by means of circle 1.9 m above the ground by means of a string 1.4 m long. The string breaks, and a string 1.4 m long. The string breaks, and the stone flies off horizontally, striking the the stone flies off horizontally, striking the ground 11m away. What was the centripetal ground 11m away. What was the centripetal acceleration of the stone while in circular acceleration of the stone while in circular motion?motion?

Sec. 5-5 Time-dependent forceSec. 5-5 Time-dependent force

For simplicity, we assume here that the forcFor simplicity, we assume here that the forces and the motion are in one dimension, whies and the motion are in one dimension, which we take to be the x direction. Then ch we take to be the x direction. Then

(5-18)(5-18)dt

dvmma(t)(t)Fx

dtm

tFdv x

x

)(

If the forces are dependent on time,If the forces are dependent on time, we can we can still use Newton’s laws to analyze the motion.still use Newton’s laws to analyze the motion.

xv0

dtm

tFdv

tx

v

v x

x

x

0

)(0

t

xx dttFm

vtv00 )(

1)(

t

xx F(t)dtm

vv00

1

xv

(5-19)(5-19)

where is initial velocity, is the velocity at where is initial velocity, is the velocity at time time t..

If is a constant, Eqs. 19 and 20 will reduce to the If is a constant, Eqs. 19 and 20 will reduce to the formula we obtained for const. acceleration motion. formula we obtained for const. acceleration motion.

xF

t

x dttvxtx0

0 )()( (5-20)(5-20)

In the same way with , we haveIn the same way with , we havedt

dxvx

oror

Discussion

Basic concepts in kinematic motion: Basic concepts in kinematic motion: t,rva

,,

How about the motion if the acceleration is a How about the motion if the acceleration is a function of position, such as a spring oscillator?function of position, such as a spring oscillator?

kxF ?(t)?(t) av

xmdt

xdm

dt

dvmkxF

2

2

m

ktAcosx ω,φ)(ω

AA and are determined by initial conditions. and are determined by initial conditions. φ

Sample problem 5-11Sample problem 5-11 A car of m=1260kg is moving at 105 km/h. A car of m=1260kg is moving at 105 km/h.

the driver begins to apply the brakes so that the driver begins to apply the brakes so that the magnitude of the the magnitude of the braking forcebraking force increases increases linearly with time at the rate of 3360N/slinearly with time at the rate of 3360N/s

(a) How much time passes before the car (a) How much time passes before the car comes to rest?comes to rest?

(b) how far does the car travel in the (b) how far does the car travel in the process?process?

Solution:Solution: (a) we choose the direction of the car’s velocity as the (a) we choose the direction of the car’s velocity as the

positive x direction, then we can represent the braking positive x direction, then we can represent the braking force as force as

and and

Let this expression for equal to zero and solve for t,Let this expression for equal to zero and solve for t,

The car comes to rest at The car comes to rest at

Nt3360ctFx

ssN

kgsm

c

mvt x 68.4

/5360

1260)/2.29(22 01

st 68.41

xvm

ctvdtct

mvtv x

t

xx 2)(

1(

2

000 ）

(b) According to Eq(5-20) (b) According to Eq(5-20)

Evaluating the expression at , Evaluating the expression at ,

we obtain we obtain

00 x1tt

mkg

ssNssmtx 1.91

)1260(6

)68.4)(/3360()68.4)(/2.29()(

3

1

m

cttvxdt

m

ctvxx

t

x 6)

2(

3

00

0

2

00

Fig 5-22 Fig 5-22

t (s)

t (s)

t (s)

x

v

a

1t

1t

1t

m2

ctt(v

2

x0x v）

m6

cttvxx

3

00

mta /ct)(x

Sec. 5-6 Noninertial frames and pseSec. 5-6 Noninertial frames and pseudoforces(udoforces( 赝力赝力 ))

Noninertial frameNoninertial frame--- a frame that is --- a frame that is acceleratedaccelerated as viewed from an inert as viewed from an inertial frame.ial frame.

1) How is the 1) How is the motion equationmotion equation if a if a noninertial framenoninertial frame is chosen? is chosen?

Consider an observer Consider an observer s’s’ in in a van that is moving at a van that is moving at conconstant velocitystant velocity. The van cont. The van contains ains a long airtrack with a fa long airtrack with a frictionlessrictionless 0.25kg glider0.25kg glider at at one end (Fig 5-22a).one end (Fig 5-22a).

The driver of the van The driver of the van appliapplies the brakeses the brakes, and the van , and the van begins to decelerate with abegins to decelerate with acceleration of . cceleration of .

Fig 5-22Fig 5-22

oooo

SSSS

ooooS’S’S’S’

aa '

(a)(a)

(b)(b)

vvvv

See an exampleSee an example

a’a’a’a’

0a 0a

0a 0a

a

To preserve the applicability of Newton’s Second To preserve the applicability of Newton’s Second law, S’ must assume that a force (law, S’ must assume that a force (a pseudo-forcea pseudo-force) ) acts on the glider. According to S’, this force acts on the glider. According to S’, this force must equal .must equal .

Observer S’ sees the glider Observer S’ sees the glider accelerateaccelerate with withand can find and can find no object in the environment of the no object in the environment of the glider that exerted a force on itglider that exerted a force on it..

'F

amam '

'a

At same time S measures the van At same time S measures the van accelerateaccelerate with with . It is found that the relationship between the two . It is found that the relationship between the two accelerates is .accelerates is .a

aa '

Pseudoforces violatePseudoforces violate Newton’s third lawNewton’s third law. The obse. The observer of s’ cannot find a reaction force exerted by thrver of s’ cannot find a reaction force exerted by the glider on some other body. e glider on some other body.

2)2) Why is theWhy is the force called ‘pseudoforce’? force called ‘pseudoforce’?

To apply classical mechanics in noninertial frames To apply classical mechanics in noninertial frames ( ), we must introduce ( ), we must introduce additional forcesadditional forces know known as ‘n as ‘pseudoforcespseudoforces’ (sometimes called ’ (sometimes called ‘‘inertial forcinertial forceses’’（惯性力）（惯性力） ).).

amF

'

a

Pseudoforces = non-Newtonian forcesPseudoforces = non-Newtonian forces

Pseudoforces depends on the frames chosen.Pseudoforces depends on the frames chosen.

3) Pseudoforces are 3) Pseudoforces are very realvery real to those that to those that experience them. experience them.

From your point of viewFrom your point of view in the in the noninertial noninertial referreference frame of the car, you must ascribe your sence frame of the car, you must ascribe your sliding motion to a pseudo-force pulling you to tliding motion to a pseudo-force pulling you to the left. This type of pseudo-force is called a he left. This type of pseudo-force is called a cecentrifugal force(ntrifugal force( 离心力离心力 )), , meaning a force directmeaning a force directed away from the centered away from the center..

To To SS, who is in , who is in inertial frameinertial frame (ground), this (ground), this is quite natural; your body is simply trying is quite natural; your body is simply trying to obey Newton’s first law and move in a to obey Newton’s first law and move in a straight line.straight line.

gg

4) Is pseudoforce (inertial force) really pseudo?4) Is pseudoforce (inertial force) really pseudo?

A virtual experimentA virtual experiment

• General General Relativity: Relativity: foundfound in 19 in 191155

• Principle of equivalence:Principle of equivalence: gravity ~ inertial forcegravity ~ inertial force

Application of inertial forces: Application of inertial forces:

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Sec. 5-7 Limitations of Newton’s lawSec. 5-7 Limitations of Newton’s law

Special relativitySpecial relativity teaches us that we can teaches us that we can not extrapolate the use of Newton’s laws to not extrapolate the use of Newton’s laws to particles moving at speeds comparable to particles moving at speeds comparable to the speed of light.the speed of light. General relativityGeneral relativity shows shows that we can not use Newton’s laws in the that we can not use Newton’s laws in the vicinity of extremely massive objectsvicinity of extremely massive objects.. Quantum mechanicsQuantum mechanics teaches us that we can teaches us that we can not extrapolate Newton’s laws to objects as not extrapolate Newton’s laws to objects as small as atomssmall as atoms..