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Chapter 4: The Laws of Motion
Forces There seem to be two kinds of forces in Nature: Contact forces and field forces.
A contact force is transferred from one object to another by physical contact.
A field force is transferred in space without physical contact.
• However, according to modern physics contact forces are field forces in disguise!• All the fundamental forces in Nature are all field forces.
Forces Examples of forces:
Forces
There are four know fundamental forces:
An example:Free neutron decay
Fundamental forces
Forces Examples of forces:
An example of weak interaction
Free neutron decay: n -> p + e-e
-
Newton’s First Law Galileo’s thought experiment
• Galileo thought about an object moving on a frictionless surface.• He posed a question: what will happen if no force is applied to the object?• He concluded that it’s not the nature of an object to stop, once set in motion, but rather to continue in its original state of motion.
Newton’s first law of motion
An object moves with a velocity that is constant in magnitude anddirection, unless acted on by a non-zero net force.
• External forces come from the object’s environment. If an object’s velocity is not changing in either magnitude or direction, then it’s acceleration and the net force acting on it must both be zero.• Internal forces originate within the object itself and cannot change the object’s velocity (although they can change the object’s rate of rotation).
Newton’s First Law Inertia
• It is much easier to throw a golf ball over a longer distance than a bowling ball – a bowling ball resists to move more than a golf ball.• Q: Why is that? A: Inertia.• The tendency of an object to continue in its original state of motion is called inertia.
Mass
• Mass is a measure of the object’s resistance to changes in its due to a force: more precisely inertial mass.
Newton’s First Law Examples of inertia
Ooops! Seat belt
motion of car
Newton’s Second Law Motion of an object under influence of net non-zero force
• How does an object under influence of net force?• Experiments show that, if you push an object with twice as strong force, the object is accelerated twice as much….
Newton’s second law
The acceleration a of an object is directly proportional to the netforce acting on it and inversely proportional to its mass where theconstant of proportionality is one.
m
Fa
inertial mass
net force
acceleration
zzyyxx maFmaFmaFamF ,,;
Newton’s Second Law Units of force and mass
• SI unit of force : newton (N)• SI unit of mass : kilogram (kg)• Newton’s second law: 1 N = 1 kg m/s2 =0.225 lb..
Newton’s Second Law Examples
• Example 4.1: Airboat
kg 1050.3 2m
N 1070.7 2
resistprop FFF
(a) Find the acceleration of the airboat.
22
2
m/s 20.2kg 1050.3
N 1070.7
m
FaFma net
net
(b) Starting from rest, find the time needed to reach a speed of 12.0 m/s.
s 5.45tm/s 0.12)m/s 20.2( 20 tvatv
const. force=const. accel.
(c) After reaching the speed, the engine is turned off and drifts to a stop over a distance of 50.0 m. Find the resistance force.
N 504)m/s kg)(-1.44 1050.3(
m/s 44.1m) 0.50(2m/s) 12(0222
2220
2
maF
aaxavv
resist
Newton’s Second Law Examples
• Example 4.2: Horses and a barge
x-components:kg 1000.2 3m
30.0o
-45.0o
N 1000.6 21 F
N 1000.6 22 F
N 1024.4cos 2222 FF x
N 1020.5cos 2111 FF x
y-components:
N 1024.4sin 2222 FF y
N 1000.3sin 2111 FF y
N 1044.9 221 xxx FFF
N 1024.1 221 yyy FFF
2m/s 472.0/ mFa xx
2m/s 0620.0/ mFa yy
46.7tan
m/s 476.0
1
222
x
y
yx
a
a
aaa
acceleration:
Newton’s Second Law Gravitational force
• Is the mutual attractive force between any two objects• Is described by Newton’s law of universal gravitation:
.Every particle in the Universe attracts every other particle witha force that is directly proportional to the product of the massesof the particles and inversely proportional to the square of thedistance between them.
221
r
mmGFg
G= 6.67 x10-11 Nm2/kg2
universal gravitation constant
Newton’s Second Law Weight
• Is the magnitude of the gravitational force acting on an object of mass m near Earth’s surface.
mgw gravity theofon accelerati :g
• SI unit : newton (N)
• Relation between g and G
2r
mMGmgw E
22E
EE
R
MG
r
MGg
ME: mass of EarthRE: radius of Earth
depends on environment
222 m/s 62.1,m/s 274,m/s 80.9 MoonSun ggg
Newton’s Third Law Newton’s 3rd law
• Forces in nature always exists in pairs.
If object 1 and object 2 interact, the force F12 exerted by object 1on object 2 is equal in magnitude but opposite in direction to theforce F21 exerted by object 2 on object 1.
action reaction
Newton’s Third Law An example of Newton’s 3rd law
• A TV on a table (TV at rest)
'
'
gg FF
nn
action reaction
normal force
gravitational force
nFam g
0
TV at rest
mgnmgFg ,
Newton’s 2nd law
Friction Friction and force
• Friction as a function of applied force (see the graph)
• Static region (Object at rest)
nf
Ff
ss
s
max
friction static
• Kinetic region (Object in motion)
friction kinetic : nf kk
normal forcecoefficientof friction
normal forcecoefficientof friction const.
Friction Coefficients of friction
Application of Newton’s Laws An example of Newton’s 3rd law
• A piece of rope at rest under constant tension along it
Newton’s 2nd law 00' amaTT No acceleration
• A crate pulled by a manForces onthe crate
Free-body diagram
mTa
mgnma
Tma
x
y
x
/
0
Newton’s 2nd law
'TT
Application of Newton’s Laws Objects in equilibrium
• Objects that are either at rest or moving with constant velocity are said to be in equilibrium.
Newton’s 2nd law
0a
)3D if 0(,0,00 zyx FFFF
Net force is zero.
Application of Newton’s Laws Examples
• Example 4.5: A traffic light at rest
Application of Newton’s Laws Examples
• Example 4.5: A traffic light at rest
T1x T2x
T1y
T2y
T3y
00 3 gy FTF
0N 1000.1
0.53sin)33.1(0.37sin
0N 1000.1
0.53sin0.37sin
33.10.53cos
0.37cos
0
0.53cos0.37cos
2
11
2
21
112
21
TT
TTF
TTT
TTF
y
x
N 1000.1 23 gFT
N 9.7933.1 N, 1.60 121 TTT
Application of Newton’s Laws Examples
• Example 4.6: Sled (at rest) on a frictionless hill
0 gFnTF
N 5.38
00.30sinN) 0.77(
sin0
T
T
mgTFx
N 7.66
00.30cosN) 0.77(
cos0
n
n
mgnFy
Application of Newton’s Laws Examples
• Example 4.8: The runaway car
=20.0o
(a) Determine the acceleration of the car.
nFFam g
0cos
sin
nmgF
mgFma
y
xx
2
2
m/s 35.3
0.20sin)m/s 80.9(sin
gax
(b) Time taken for the car to reach the bottom?
s 86.3
m 0.25)2/1( 2
t
tax x
(c) velocity at the bottom?
m/s 12.9s) 86.3)(m/s 35.3( 2 atv
Application of Newton’s Laws Examples
• Example 4.9: Weighing a fish in an elevator
(a) Find the weight when a>0.
)( gammgmaT
mgTFma
N 1.48
)m/s 80.9m/skg)(2.00 08.4( 22
T
(b) Find the weight when a<0.
N 8.31
)m/s 80.9m/skg)(-2.00 08.4( 22
T
a=2.00 m/s2 a=-2.00 m/s2
(c) Find the weight when the cable breaks (a=-g).
0)m/s 80.9m/skg)(-9.80 08.4( 22 T
Application of Newton’s Laws Examples
• Example 4.10: Atwood’s machine
gmTam 222
m2>m1
gmTam 212 12 aa
gmgmamm 12121 )(
gmTam 111
gmamT 111
gmm
mma
21
121
gmm
mmT
21
212
Application of Newton’s Laws Examples
• Example 4.11: A block on a ramp (static friction)
)3(0cos
)2( 0sin
)1(max
mgnF
nmgF
nf
y
sx
ss
s
mgn
tan)4)(2(
)4(cos)3(
Application of Newton’s Laws Examples
• Example 4.12: A sliding hockey puck (kinetic friction)
xavv 220
2
220
2
m/s 67.12
x
vva
v0=20.0 m/s -> v=0x=1.20x102 m
0 mgnFnF gy
mgn
mg
nfFma
k
kkx
170.0g
ak
Application of Newton’s Laws Examples
• Example 4.13: Connected objects
m1=4.00 kgm2=7.00 kg =0.300
(a) Find acceleration and tension I.
(2) 0
(1)
1
11
gmnF
amfTF
y
kx
For Object 1:
(3) )1( 111 amgmT k
(4) 12
222
am
amTgmFy
For Object 2:
(2)-(4)
12112 )( ammgmgm k
(1)) (from N 4.32,m/s 17.5 2
21
121
Tmm
gmgma k
Application of Newton’s Laws Examples
• Example 4.13: Connected objects
m1=4.00 kgm2=7.00 kg =0.300
(b) Find acceleration and tension II (system approach).For a system made ofObject 1+2:
gmgm
ngmamm
k
k
12
221 )(
21
12
mm
gmma k
Application of Newton’s Laws Examples
• Example 4.14: Two blocks and a cord
For top object 1:
mgn
nma s
1
1
0
ga
mgmamgn
s
s
1
For bottom object 2:
TmgMa s
mgTgM ss
N 5.51)( gMmT s
m=5.00 kgM=10.0 kgs=0.350
What is the max. force by the stringwithout causing the top object to slip?
