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~
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
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
Static and Kinetic Frictional 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
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
Replacing an Engine (Equilibrium)
Another solution: Choose
Find: Tension forces
T1 and T2
Plane in Equilibrium
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
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 6.66).
(example 5.17)
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
Hauling a Crate with Acceleration
Exam Example 10: Blocks on the Inclines (problem 5.90)
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 kk
x
21
12212111121
))cos(cossin(sin)sincos(
mm
sgmmggsamTT k
kxxx
v
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)
(problem 5.62)
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)(
Centripetal Force
Rounding a flat curve (problem 5.48) Sources of the centripetal force
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 5.107)
Exam Example 12:
R
FF
gm
gm ca
ca
A pilot banks or tilts the plane at an angle θ to create the centripetal force Fc = L·sinθ
Lifting force
Rounding a Banked Curve
Example 5.22 (car racing):r = 316 m , θ = 31o
mphsmrgv 96/43tan
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