Ch.4 Forces And Newton’s Laws
What is a Force?
A Push or a Pull
Two types of forces: Contact and Non contact
Non Contact - Action at a distant force
Is the Force a Vector or a Scalar ? Units and Dimensions ?
Mass – Measure of Mass is inertia
(How does it differ from Weight ?)
Higher mass – harder to change the motion, i.e. higher inertia
1
Newton’s First Law Of Motion
• An object continues in a state of rest or in a
state of motion at a constant speed along a
straight line unless compelled to change that
state by a net force.
• What is the net force?
If so, why will a ball rolling on a flat surface
stop?
2
• Vector sum of all
the forces acting on
the object is called
the net force.
( i.e. both magnitude
and direction needs
to be considered)
What is the net force on the car ?
3
Inertial Reference Frame
Newton’s laws are valid in an inertial frame,
i.e. the acceleration of the frame should be
zero (Constant Velocity).
Ideally we do not have such a frame, but earth
is a very good approximation.
4
Newton’s Second Law
• First law: no net force
no change in velocity
• Second law: What happens if there is an external
force? (net)
Acceleration of the object
 Net Force
1
Acceleration inversely proportional to mass 
m
5
F

a
m
or
 F  ma
Only two factors determine the
acceleration---Force and Mass
Force Units Newton (N) = kg. m/s2
What is 1 Newton ?
6
Newton’s Second Law Of Motion
When a net external force  F acts on an object of
mass m, the acceleration a results is directly
proportional to the net force and has a magnitude that
is inversely proportional to the mass. The direction of
the acceleration is the same as the direction of the net
force.
F

a
m
or
 F  ma
SI Unit of Force: kg.m/s2 = newton (N)
7
Units for Mass, Acceleration & Force
System
Mass
Acceleration
Force
SI
kilogram(kg) meter/second2(m/s2) newton(N)
CGS
gram(g)
centimeter/sescond2 dyne(dyn)
(cm/s2)
BE
slug(sl)
foot/second2(ft/s2)
pound(lb)
8
Example 1. Pushing A Car
Frictional force = 560 N
Push = 275 N+ 395 N = 670 N
Net force = 670 N – 560 N, N= 110 N
F = ma , 110 N = 1850 kg*a
a = 110 N/1850 kg = 0.059 m/s2
(what is the direction ?)
9
Forces In 2-D
Fx = max
Fy = may
Two forces are applied onto an object. 1.) 12N towards
North, and 2.) 5N towards East, what is the direction of
the motion of the object?
Resultant force =
12N
12 2  52  13
tan θ = 12/5 =2.4
θ
5N
θ = tan-1(12/5) = 67.38 0
10
1300 kg raft
Example 2.
P= Force by the man
In 65 seconds,
where will the boat
be if v0x = 0.15 m/s ?
(a)
A= Force of wind
ax = ?, ay = ?
(d)
11
Resultant Force And Acceleration
Force
X- component
Y- Component
P
17 N
0N
A
15cos67 N
15sin67 N
Resultant
F
x
 17  15 cos 67  23N ,
F


23N

 0.018 m / s 2 ,
m 1300 kg
x
ax
 F  15 sin 67  0  14N
F
14 N

a 

 0.011 m / s
y
y
y
m
1300 kg
2
12
Displacement Of The Boat In 65 sec
For x: v0x = 0.15 m/s, ax = 0.018 m/s2
t = 65 sec, x = ?
x = v0x t + (½)axt2
= 0.15*65 +(1/2)*0.018*(65)2 = 48 m
For y: v0y = 0, ay = 0.011 m/s2, t = 65 sec
y = v0yt + (½) ay t2
= 0*65 + (½)*0.011*(65)2 = 23 m
13
Which of these will cause the
acceleration to be doubled ?
a) All of the forces acting on the object
doubles
b) The net force acting on the object
doubles
c) Both the net force and the mass doubles
d) The mass of the object is reduced by a
factor of two.
a, b, d
14
Newton’s Third Law of Motion
• Whenever one body exerts a force on a
second body, the second body will exert
an oppositely directed force of equal
magnitude on the body.
• (Action reaction law)
Acting on which object ?
15
MS = 11000 kg
mA = 92 kg
Force by Astronaut on Space Ship = P
Force on Astronaut by Space ship = P
What will be the acceleration of the two objects ?
For the astronaut,
If P = 36 N,
mA=92
-36 N = 92 aA , i.e. aA = -0.39 m/s2
36 N = 11000 kg as ,
i.e.
as = 0.0033 m/s2
16
Types Of Forces
Fundamental Forces
• Gravitational force
• Strong Nuclear
• Electroweak
Electromagnetic force
All other forces are Non Fundamental
(can be explained by a fundamental force)
Maxwell, Unification theory
17
Gravitational Force
• Newton’s law of Universal gravitation
m1m2
F G 2
r
r
center to center
G = 6.673 x 10 –11 N m2/kg2
Always attractive , Planetary motion, Satellites
18
Weight Of An Object
• Mass m on the surface of
the earth
W = mg
Me
Radius of earth = Re
m1m2
F G 2
r
W G
mM e
2
 mg
Re
So:
Me
g G
2
Re
(6.67  10 11 N  m 2 / kg 2 )(5.98  10 24 kg)

(6.38  10 6 m) 2
Distance ~ Re (approx)
= 9.799 m/s2
19
Hubble Telescope
• The mass of Hubble telescope is 11600 kg.
(Re = 6.38*106 m, Me = 5.98*1024 kg)
• (i) What is the weight when resting on earth
surface,
M e M h 6.67 1011  5.98 1024 11600
W G

2
6 2
r
(6.38 10 )
= 1.14*105 N
• (ii) and in orbit 598 km above
(r = Re +598 km)
R=6.38*106m+598*103m=6.98*106m
20
Check Your Understanding 2
m2>m1, and the net gravitational force acting on the
third object is zero. Which of the drawings correctly
represents the locations of the object?
(d)
21
The Normal Force & Newton’s 3rd Law
Table exert a force on the block: FN
Net Force on the block: FN – W
Normal Force: The force component
that a surface exerts on the object
(Normal to the contact surface)
How can the table exert a force ?
(Consider the springs in a mattress)
22
What happens if I push or pull on the
block ?
What happens to the normal force if the rope has a pull = 15N?
23
Balancing Act
24
Apparent Weight
• How do you feel when the
elevator suddenly starts
going up?
FN
Feel heavy?
FN – W = ma
FN = Apparent weight
W = True weight
25
Elevator Accelerating Up With “a”
Applying 2nd law vertically
upward
FN – W = ma
FN = W+ma
Apparent weight is larger than
the true weight (W = mg)
g = 9.8 m/s2
26
Elevator Accelerating Down With “a”
FN – mg
=
ma
( if downward a = negative)
FN = mg +ma
What if a = -g ?
FN=0
27
Static And Kinetic Friction
Two surfaces touching
each other
Contact surface is not
smooth
In addition to the normal
force, there is a force
parallel to the surface
FRICTIONAL FORCE
28
Static Friction (fs)
• As F increases, fs
increases from zero up
to fs Max
• fs

fs Max
• Just before motion
fsMax = μs FN
(μs= Coefficient of static friction)
Units and Dimensions of μs?
29
Force Needed To Start The Sled
The maximum force needed to
just begin to move = need to
overcome the max frictional force
fsMax
F
i.e. F = fsMax = μs FN = μs mg
If μs = 0.35,
m = 38 kg
fsMax = 0.35 * 38 * 9.8 = 130 N
What happens once it just starts moving ?
Easier to move, i.e. needs force going down
i.e. fs changes to fk
Kinetic Friction
30
Kinetic Friction
• fk = μkFN
(μk
Coefficient of kinetic friction)
μk is usually less than μs
Example 10: Sled riding : How far does the sled
go before stopping ?
μk = 0.05
31
fk = μk FN = μk mg
Net force on the sled = kinetic frictional force
 f k  u k mg
ak 

 u k g
m
m
Acceleration
What is the meaning of the negative sign?
Does the acceleration depend on the mass ?
Stopping distance x?
v2 = v02 +2ax
v v
0  (4.0) 2
x

 16.3m
2(u k g )  2(0.05)  9.8
2
x
2
0x
32
Tension
Mass-less and non-stretch rope
Force gets applied to the box
undiminished.
33
Traction Of The Foot
• T1 and T2 keep the pulley on the foot at rest. i.e.
pulley is at equilibrium (Net Force = 0)
• Find the  Fx and  Fy
34
F
y
 T1 sin 35  T2 sin 35  0
F
x
 T1 cos 35  T2 cos 35  F  0
Let T1=T2=T, then F  2T cos 35
F  2T cos 35  2mg cos 35
 2(2.2kg)(9.80m / s ) cos 35  35N
2
35
F
y
 T1 sin 35  T2 sin 35  0
T1 =T2 ( Same rope)
F
x
 F  T1 cos 35  T2 cos 35
 F  2T cos 35 (Say T1=T2 = T)
Since
F
x
0
T = 2.2 g (why?)
F  2T cos 35
F = 35 N
36
Replacing The Engine
37
Force
x Component
y Component
T1
-T1 sin 10.00
+T1 cos 10.00
T2
+T2 sin 80.00
-T2 cos 80.00
W
0
-W
F
x
 T1 sin 10.0  T2 sin 80.0  0
F
y
 T1 cos10.0  T2 cos 80.0  W  0
38
 sin 80.0 
T1  
T2
 sin 10.0 
 sin 80.0 

T2 cos10.0  T2 cos 80.0  W  0
 sin 10.0 
T2 
W
 sin 80.0 

 cos10.0  cos 80.0
 sin 10.0 
Setting W=3150N, T2=582N
T1=3.30*103N
39
Equilibrium At Constant Velocity
40
Force
x Component y Component
W
-W sin 30.00
-W cos 30.00
L
0
+L
T
+T
0
R
-R
0
F
x
 W sin 30.0  T  R  0
F
y
 W cos 30.0  L  0
L  W cos 30.0  (86500 N ) cos 30.0  74900 N
41
Check Your Understanding 4
Which of the following could lead to equilibrium?
a) Three forces act on the object. The forces all point
along the same line but may have different
directions.
b) Two perpendicular forces act on the object.
c) A single force acts on the object.
d) In none of the situations described in (a), (b) and
(c) could the object possibly be in equilibrium.
Answer: (a)
42
Non-equilibrium Applications
• Non zero net force
• Net force – x component, y component
• Acceleration in x direction and in y
direction
43
Towing A Super Tanker
ax=2.0*10-3
D=75.0*103N, m=1.50*108N, R=40.0*103N
44
F
y
F
x
Force
x Component
y Component
T1
+T1cos 30.00
+T1sin 30.00
T2
+T2cos 30.00
-T2sin 30.00
D
+D
0
R
-R
0
 T1 sin 30.0  T2 sin 30.0  0
 T1 cos 30.0  T2 cos 30.0  D  R  max
45
ax=2.0*10-3
max  R  D
T
2 cos 30.0
3
(1.50  10 kg)( 2.00  10 m / s )  40.0  10 N  75.0  10 N

2 cos 30.0
8
2
3
3
=1.53*105N
46
Hauling A Trailer
47
F
x
 T  m2ax  (27000kg)(0.78m / s )  21000N
2
F
x
  D  T   m1a x
If the drawbar has no mass
T=T`
D  m1a x  T   (8500kg)(0.78m / s )  21000 N
2
= 28000N
48
The Motion Of A Water Skier
What is the net force on skier in each picture?
49
(a) The skier is floating
motionless in the water.
The skier is floating
motionless in the water, so
her velocity and acceleration
are both zero. Therefore, the
net force acting on her is zero,
and she is in equilibrium.
50
(b) The skier is being pulled out of the water and
up onto the skis.
As the skier is being pulled up and out of the
water, her velocity is increasing. Thus she is
accelerating, and the net force acting on her is
not zero. The skier is not in equilibrium. The
direction of the net force is shown in the figure
above.
51
(c) The skier is moving at a constant speed along
a straight line.
The skier is now moving at a constant speed along
a straight line, so her velocity is constant. Since
her velocity is constant, her acceleration is zero.
Thus, the net force acting on her is zero, and she is
again in equilibrium, even though she is moving. 52
(d) The skier has let go of the
tow rope and is slowing down.
After the skier lets go of the tow
rope, her speed decreases, so she
is decelerating. Thus, the net
force acting on her is not zero,
and she is not in equilibrium.
The direction of the net force is
shown in the figure.
53
Hauling A Crate
54
MAX
F


mg
sin
10
.
0



F

ma
 x
s N
a
 mg sin 10.0   s FN

m
MAX
F
y
 mg cos10.0  FN  0
or
a
MAX
FN  mg cos10.0
  g sin 10.0   s g cos 10.0
 (9.80m / s 2 ) sin 10.0  (0.350)(9.80m / s 2 ) cos10.0
=1.68m/s2
55
Accelerating Blocks
56
For mass m1
F
x
 W1 sin 30.0  T  m1a
What is the mass m1 in y direction?
For mass m2
F
y
 T  W2  m2 (a)
What is the mass m2 in x direction?
m1=8.00 kg,
m2=22.0 kg,
T=T’
T=86.3 N, a=5.89 m/s2
57
Hoisting A Scaffold
m=155 kg, T=540 N
F
y
 T  T  T  W  ma y
3T  W 3(540 N )  1520 N
ay 

 0.65m / s 2
m
155kg
58
Check Your Understanding 5
Two boxes have masses m1 and m2, and m2 is greater than
m1. The boxes are being pushed across a frictionless
horizontal surface. As the drawing shows, there are two
possible arrangements, and the pushing force is the same in
each. In which arrangement does the force that the left box
applies to the right box have a greater magnitude, or is the
magnitude the same in both cases?
Answer: (a)
59
Velocity, Acceleration And
Newton’s Second Law Of Motion
If the craft is moving with constant velocity, are
there other forces?
v=850m/s
F
x
 F1  F2  F3  0
F3  ( F1  F2 )  (3000 N  5000 N )  8000 N
60
The Importance Of Mass
?
gmoon=1.60m/s2, T=24N, μk=0.20, mgearth=88 N
61
ax
F


x
m
 T   k FN

m
 T   k mg moon
ax 
m
Wearth  mgearth
Wearth
88N
m

 9.0kg
2
g earth 9.80m / s
 T   k mg moon 24 N  (0.20)(9.0kg)(1.60m / s 2 )

 2.3m / s 2
ax 
9.0kg
m
62
Conceptual Questions 1.
Why do you lunge forward when a car suddenly stops ?
Pressed backward when the car suddenly accelerates ?
REASONING AND SOLUTION When the car comes to a sudden
halt, the upper part of the body continues forward (as predicted
by Newton's first law) if the force exerted by the lower back
muscles is not great enough to give the upper body the same
deceleration as the car. The lower portion of the body is held in
place by the force of friction exerted by the car seat and the floor.
When the car rapidly accelerates, the upper part of the body tries
to remain at a constant velocity (again as predicted by Newton's
first law). If the force provided by the lower back muscles is not
great enough to give the upper body the same acceleration as the
car, the upper body appears to be pressed backward against the
seat as the car moves forward.
63
Conceptual Question 6.
Father and daughter on ice pushing each other, who has the
higher acceleration ?
REASONING AND SOLUTION
Since the father and the daughter are standing on ice skates, there is virtually no
friction between their bodies and the ground. We can assume, therefore, that the
only horizontal force that acts on the daughter is due to the father, and similarly,
the only horizontal force that acts on the father is due to the daughter.
a.) According to Newton's third law, when they push off against each other, the
force exerted on the father by the daughter must be equal in magnitude and
opposite in direction to the force exerted on the daughter by the father. In other
words, both the father and the daughter experience pushing forces of equal
magnitude.
b.) According to Newton's second law,  F  ma Therefore, .a   F / m The
magnitude of the net force on the father is the same as the magnitude of the
net force on the daughter, so we can conclude that, since the daughter has the
smaller mass, she will acquire the larger acceleration.
64
Problem
4) A 5.0 kg projectile accelerates from rest to 4.0*103
m/s . The net force on the projectile is 4.9 * 105 N.
What is the time for the projectile to come to the
speed?
F=4.9*105N, M=5.0kg, v0=0, vf=4.0*103
F = ma
v = v0 + at ,
4.9*105 = 5 a , a = 9.8*104 m/s2
v  v0 4.0  10 3
2
t

 4.08  10 sec
4
a
9.8  10
65
Problem
8)An arrow starting from rest leaves the bow with a
speed of 25.0 m/s. If the average force on the
arrow is doubled what will be the speed?
(Say mass m, travels a distance x before leaving the bow)
initial velocity v0=0, find velocity v1
v12 = vo2 + 2a1x (first case)
v22 = vo2 + 2a2x (second case)
2
1
2
2
v
a1

a2
v
66
v1
a1

v2
a2
If the net force is doubled in
case 2, i.e.
F=ma
F=ma1
2F=ma2
F/m=a1
2a1=a2
F = ma
a2 = 2a1
v1
a1

v2
2a1
v1 2  v 2
v 2  25 2  35.4m / s
67
Problem
15)A duck of mass 2.5 kg has a force of 0.1 N due east. Water
exerts a force 0.2 N 52o south of east . Velocity of the duck
is 0.11 m/s due east. Find the displacement of the duck in
3.0s while the forces are active.
x component of force = 0.1 + 0.2 cos 52 = Rx=0.2231N
v0= 0.11 m/s
y component of force = - 0.2 sin 52 = Ry= -0.1576 N
0.1 N
52o
Rx = 2.5 ax
ax = Rx /2.5 = 0.08924m/s2
Ry = 2.5 ay
ay = Ry/2.5 = -0.06304m/s2
0.2 N
68
Start
x=?

y=?
x = vox t + ½ ax t2
= 0.11 * 3 + ½ ax 32 =0.73158m
End
y = voy t + ½ ay t2
= 0 + ½ ay * 9= -0.28368m
Displaceme nt  x 2  y 2  0.7847m
y
1 y
tan    0.3878   tan
 21.19
x
x
69
Problem
20)REASONING AND SOLUTION The forces that act
on the rock are shown at the right. Newton's second
law (with the direction of motion as positive) is R
F  mg – R  ma
Solving for the acceleration a gives
mg
2
2
45
kg
9.80
m/s
N


 250
mg – R 45kg 9.8m / s  –250

N
a


m
45kgkg
45
2m/s 2
4.2
 4.2m / s
70