Different Forces and Applications of Newton’s Laws Types of Forces Fundamental...

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ferent Forces and Applications of Newton’s L Types of Forces Fundamental Forces Non-fundamental Forces Gravitational (long-range) Weight, tidal forces Strong: quark-gluon ( L ~ 10 -13 cm) Nuclear forces (residual, “Van der Waals”) Weak: lepton-quark ( n → p + e - + )L~10 -16 cm Nuclear β-decay A Z → A Z+1 +e - + Electromagnetic: charged particles exchanging by photons Normal force and pressure Tension force and shear force Frictional forces e ~ e ~

Transcript of Different Forces and Applications of Newton’s Laws Types of Forces Fundamental...

Page 1: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Different Forces and Applications of Newton’s LawsTypes of Forces

Fundamental Forces Non-fundamental Forces

Gravitational (long-range) Weight, tidal forces

Strong: quark-gluon

( L ~ 10-13 cm)

Nuclear forces

(residual, “Van der Waals”)

Weak: lepton-quark

( n → p + e-+ )L~10-16cm

Nuclear β-decay

AZ → AZ+1 +e- +

Electromagnetic:

charged particles

exchanging by photons

(long-range)

Normal force and pressure

Tension force and shear force

Frictional forces

Propulsion force

Buoyant force

Electric and magnetic forces

Chemical bonds . . .

e~ e~

Page 2: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Normal Force

gm

gm

gm

NF

NF

NFextF

extF

extN FmgF extN FmgF mgFN Weight vs. mass (gravitational mass = inertial mass)

Apparent weight vs. true weight mg , g = 9.8 m/s2

gmamFN

Note: weight varies with location on earth, moon,…

gmoon=1.6 m/s2

Page 3: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Pulley (massless & frictionless)

The Tension Force

Massless rope:

-T1=T2=mGg

1T

2T

gmG

Massive rope: -T1=T2+mRg=(mG+mR)g>T2

Acceleration with massive rope and idealization of massless rope

mB mR

BT

RTa

X

TR= (mB+ mR)a > TB= mB a = TR – mR a RT BT

Page 4: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Static and Kinetic Frictional Forces

Page 5: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Fluid Resistance and Terminal Speed

Linear resistance at low speed f = k v

Drag at high speed f= D v2 due to turbulence

Newton’s second law: ma = mg – kv

Terminal speed (a→0):

vt = mg / k (for f=kv) ,

vt = (mg/D)1/2 (for f=Dv2)

Baseball trajectory is greatlyaffected by air drag !

v0=50m/s

Page 6: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Applying Newton’s Laws for

Equilibrium:

0

0

0

00

z

y

x

jj

jj

F

F

F

FFam

a

Nonequilibrium:

zz

yy

xx

jj

Fma

Fma

Fma

Fama

0

Page 7: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Replacing an Engine (Equilibrium)

Another solution: Choose

Find: Tension forces

T1 and T2

Page 8: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Plane in Equilibrium

Page 9: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Example 5.9: Passenger in an elevator

y

0

a

NF

gmw

NF

Center of the Earth

w

Data: FN= 620 N, w = 650 N

Find: (a) reaction forces to Fn and w; (b) passenger mass m; (c) acceleration ay .

Solution:(a)Normal force –FN exerted on the floor and gravitational force –w exerted on the earth.

(b) m = w / g = 650 N / 9.8 m/s2 = 64 kg

(c) Newton’s second law: may = FN – w ,

ay = (FN – w) / m = g (FN – w) / w = = 9.8 m/s2 (620 N – 650 N)/650 N = - 0.45 m/s2

Page 10: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

How to measurefriction by meter and clock?

Exam Example 9:

d) Find also the works done on the block by friction and by gravityas well as the total work done on the block if its mass is m = 2 kg (problem 6.68).

Page 11: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

d) Work done by friction: Wf = -fkL = -μk FN L = -L μk mg cosθmax = -9 J ; work done by gravity: Wg = mgH = 10 J ;

total work: W = mv||2 /2 = 2 kg (1m/s)2 /2 = 1 J = Wg + Wf = 10 J + 9 J = 1 J

Page 12: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Hauling a Crate with Acceleration

Page 13: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Exam Example 10: Blocks on the Inclines (problem 5.92)

m1

m2

X

X

α1 α2

1W

1NF

2NF

2W

XW1

XW2

1kf

2kf

Data: m1, m2, μk, α1, α2, vx<0 a

Solution:Newton’s second law for

block 1: FN1 = m1g cosα1 , m1ax= T1x+fk1x-m1g sinα1 (1)

block 2: FN2 = m2g cosα2 , m2ax= T2 x+fk2x+ m2g sinα2 (2)

Find: (a) fk1x and fk2x ;(b) T1x and T2x ;(c) acceleration ax .

1T 2T

(a) fk1x= sμkFN1= sμkm1g cosα1 ; fk2x= sμkFN2= sμkm2g cosα2; s = -vx/v (c) T1x=-T2x, Eqs.(1)&(2)→

(b)

21

222111 )sincos()sincos(

mm

sgmsgma kkx

21

12212111121

))cos(cossin(sin)sincos(

mm

sgmmggsamTT k

kxxx

v

Page 14: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Exam Example 11: Hoisting a Scaffold

Ya

T

TT

TT

0

m F

gmW

Data: m = 200 kg Find: (a) a force F to keep scaffold in rest;(b) an acceleration ay if Fy = - 400 N;(c) a length of rope in a scaffold that would allow it to go downward by 10 m

SolutionNewton’s second law: WTam

5

(a) Newton’s third law: Fy = - Ty , in rest ay = 0→ F(a=0)= W/5= mg/5 =392 N

(b) ay= (5T-mg)/m = 5 (-Fy)/m – g = 0.2 m/s2

(c) L = 5·10 m = 50 m (pulley’s geometry)

Page 15: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Dynamics of Circular Motion

θ

Uniform circular motion:

constvvelocityconstvvspeed

,

Period T=2πR/v , ac = v2/R = 4π2R/T2 Cyclic frequency f=1/T , units: [f] = Hz = 1/sAngular frequency ω = 2πf = 2π/T, units: [ω]=rad·Hz=rad/s

Roo

rad 572

3601

Dimensionless unit for an angle:

Example: 100 revolutions per second ↔ f=1/T=100 Hz or T=1s/100=0.01 s

Non-uniform circular motion:equation for a duration of one revolution T

T

Rdttvdxdtv0

2)(

Page 16: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Centripetal Force

Rounding a flat curve (problem 5.44) Sources of the centripetal force

Page 17: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Data: L, β, m Find: (a) tension force F;(b) speed v;(c) period T.

Solution:Newton’s second law

j

cj amF

Centripetal force along x: RmvmaF c /sin 2Equilibrium along y: cos/)(cos mgFamgF

cos/sincos/sin

sin,tansin)/()(

2

2

LgLgv

LRRgmFRvb

g

LTLgLvRTc

cos2cos//2/2)(

Two equations with two unknowns: F,v

The conical pendulum (example 5.20)

or a bead sliding on a vertical hoop (problem 5.119)

Exam Example 12:

R

FF

gm

gm ca

ca

Page 18: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

A pilot banks or tilts the plane at an angle θ to create the centripetal force Fc = L·sinθ

Lifting force

Page 19: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Rounding a Banked Curve

Example 5.22 (car racing):r = 316 m , θ = 31o

mphsmrgv 96/43tan

Page 20: Different Forces and Applications of Newton’s Laws Types of Forces Fundamental ForcesNon-fundamental Forces Gravitational (long-range)Weight, tidal forces.

Uniform circular motion in a vertical circle

Newton’s second lawTop: nT – mg = -mac

Bottom: nB – mg = +mac

)(2

R

vgmnT

)(2

R

vgmnB

Note: If v2 >gR , the passenger will be catapulted !

Find: Normal force nT