Different Forces and Applications of Newton’s Laws
Types of Forces
Fundamental Forces
Non-fundamental Forces
Gravitational (long-range)
Weight, tidal forces
Strong: quark-gluon
( L ~ 10-13 cm)
Weak: lepton-quark
~ )L~10-16cm
( n → p + e-+
e
Electromagnetic:
charged particles
exchanging by photons
(long-range)
Nuclear forces
(residual, “Van der Waals”)
Nuclear β-decay
~
AZ → AZ+1 +e- +
e
Normal force and pressure
Tension force and shear force
Frictional forces
Propulsion force
Buoyant force
Electric and magnetic forces
Chemical bonds
...
Normal Force

FN

F
mg N

Fext
 
mg F

Fext

mg
N
F N  mg FN  mg  Fext FN  mg  Fext
Weight vs. mass (gravitational mass = inertial mass)
Apparent weight vs. true weight mg , g = 9.8 m/s2



FN  m a  m g
Note: weight varies with location on earth, moon,…
gmoon=1.6 m/s2

T1

T2
The Tension Force

mG g
Massless rope:
Massive rope:
-T1=T2=mGg

a
-T1=T2+mRg=(mG+mR)g>T2
Acceleration
 with massive rope
 and idealization of massless rope
TB
mB
mR
TR
X
TR= (mB+ mR)a > TB= mB a = TR – mR a

T
R
Pulley

TB
(massless & frictionless)
Static and Kinetic Frictional Forces
Fluid Resistance and Terminal Speed
Linear resistance at low speed
f=kv
Drag at high speed
f= D v2
due to turbulence
Newton’s second law:
ma = mg – kv
Terminal speed (a→0):
Baseball trajectory is greatly
affected by air drag !
vt = mg / k (for f=kv) ,
vt = (mg/D)1/2 (for f=Dv2)
v0=50m/s
Applying Newton’s Laws for

a  0


 
 
Fj  0 
Equilibrium: m a 


F

j
j


j



Nonequilibrium: a  0  m a   F j
j
 Fx  0


 F y  0


 Fz  0 
ma x   Fx 


 ma y   F y 


ma z   Fz 
Replacing an Engine (Equilibrium)
Find:
Tension forces
T1 and T2
Another solution: Choose
Plane in Equilibrium
Example 5.9: Passenger in an elevator
y

a
Find: (a) reaction forces to Fn and w;
(b) passenger mass m;
(c) acceleration ay .

FN

 FN
Data: FN= 620 N, w = 650 N
0

w  mg
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 ,

w
Center of the Earth
ay = (FN – w) / m = g (FN – w) / w =
= 9.8 m/s2 (620 N – 650 N)/650 N = - 0.45 m/s2
Exam Example 9:
(example 5.17) How to measure
friction by
meter and clock?
d) Find also the works
done on the block
by friction and by gravity
as well as the total work
done on the block if
its mass is m = 2 kg
(problem 6.66).
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
d) Work
Hauling a Crate with Acceleration

a

T2 
F
 N2
Exam Example 10: Blocks on the Inclines (problem 5.90)
Data: m1, m2, μk, α1, α2, vx<0
Find: (a) fk1x and fk2x ;
(b) T1x and T2x ;
(c) acceleration ax .
m1
F
v N1

W1 X
Solution:
Newton’s second law for
α1
T1
 X
f k1

W1

f
k
2
m
W
2
α2

W2
2X
X
block 1: FN1 = m1g cosα1 , m1ax= T1x+fk1x-m1g sinα1 (1)
block 2: FN2 = m2g cosα2 , m2ax= T2 x+fk2x+ m2g sinα2 (2)
(a) fk1x= sμkFN1= sμkm1g cosα1 ; fk2x= sμkFN2= sμkm2g cosα2; s = -vx/v
(c) T1x=-T2x, Eqs.(1)&(2)→a
x
m1 g ( s k cos  1  sin  1 )  m 2 g ( s k cos  2  sin  2 )

m1  m 2
(b) T1x  T2 x  m1 (a x  s k g cos 1  g sin 1 ) 
m1m2 g (sin 1  sin  2  s k (cos  2  cos 1 ))
m1  m2


T  T 
T
T

T
m
a Scaffold
Exam Example 11: Hoisting
a Y Data: m = 200 kg (problem 5.62)
Find:
0
(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
 

m a  5T  W
Solution
 Newton’s second law:
F (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


W  mg
(c) L = 5·10 m = 50 m (pulley’s geometry)
Dynamics of Circular Motion
Uniform circular motion:
speed


v  v  const , velocity v  const
360o
Dimensionless unit for an angle: 1 rad 
 57 o
2
2
2
2
R
θ
Period T=2πR/v , ac = v /R = 4π R/T
Cyclic frequency f=1/T , units: [f] = Hz = 1/s
Angular frequency ω = 2πf = 2π/T, units: [ω]=rad·Hz=rad/s
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
v dt  dx   v (t )dt  2R
0
Centripetal Force
Sources of the centripetal force
Rounding a flat curve (problem 5.48)
Data: L, β, m
Find:
(a) tension force F;
(b) speed v;
(c) period T.
Solution:
Newton’s second law
Exam Example 12:
The conical pendulum (example 5.20)
or a bead sliding
on a vertical hoop
(problem 5.107)

F

ac


 F j  m ac

F
j

R

 a
Two equations with two unknowns: F,v
mg
mg c
2
Centripetal force along x: F sin   ma c  mv / R
Equilibrium along y: F cos   mg  (a ) F  mg / cos 
(b )
v 2  R( F / m ) sin   Rg tan 
, R  L sin 
 v  Lg sin 2  / cos   sin  Lg / cos 
(c )
T  2R / v  2L /
Lg / cos  
T  2
L cos 
g
A pilot banks or tilts the plane
at an angle θ to create
the centripetal force
Lifting force
Fc = L·sinθ
Rounding a Banked Curve
Example 5.22 (car racing):
r = 316 m , θ = 31o
v  rg tan   43 m / s  96 mph
Uniform circular motion in a vertical circle
Find: Normal force nT
Newton’s second law
Top: nT – mg = -mac
v2
nT  m ( g  )
R
Note: If v2 >gR ,
the passenger
will be catapulted !
Bottom:
nB – mg = +mac
v2
nB  m ( g  )
R