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PHYSICS 125
Lecture Notes Book 2
Prepared by Kai Wong
Table of Contents
6. Newton’s Laws of Motion ………………………………………………………. 2
7. Weight and Normal Force ……………………………………………………… 9
8. Friction …………………………………………………………………………… 18
9. Tension …………………………………………………………………………… 26
10. Centripetal Acceleration ………………………………………………………. 32
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6. Newton’s Laws of Motion
Newton’s First Law
When we ask what causes an object to move, we have in mind an object that is initially at
rest and is made to move. A force is clearly needed. For example, we need to give a chair
a push to start it moving on the floor. When we start a car, the force is the friction from
the road surface. This force is forward because the wheels begin to spin in a direction
such that the tires tend to slip backward relative to ground. As there is a change of
velocity (from zero to nonzero), the object experiences acceleration. Therefore, a force
can cause acceleration.
Once the object is moving, is a force necessary to maintain its motion? The answer seems
to be yes when we consider moving a chair on the floor. The chair would stop if we don’t
continue to push. But what if the floor is very slippery? In this case the chair can go a
long way on the floor after our hands are removed from it. If we examine the motion of
an ice hockey puck, after being hit, it can maintain its velocity without further help for a
very long way indeed. The commonality of these two examples is the lack of friction, and
therefore a lack of force on the object. Without force, an object at rest cannot move, and
an object that is already moving would continue to do so with the same velocity. This is
the essence of Newton’s First Law of Motion, which can be thus stated :
An object at rest will remain at rest, and an object moving will continue to move with the
same velocity, unless compelled by forces.
The tendency of objects to continue with the same velocity is called inertia. When you
are driving and suddenly slam on the brake, you get thrown forward because of inertia:
you keep moving with the original speed of the car until perhaps hitting the windshield.
Similarly, inertia explains why you are thrown backward when you suddenly accelerate
the car, because you are maintaining the original speed while the car is speeding away
from you. But for the friction from your seat and the back of the seat, you would have
been left behind on the road!
Newton’s Second Law
The first law deals with situations without forces. What happens if there is a force? We
have already seen that it causes acceleration. Consider the case of a single force acting on
an object. Experiments show
1. The stronger the force, the larger the acceleration. In fact, they are proportional:
when the force doubles, so does acceleration. This is captured by the relation
aF
2
2. If the same force is applied to objects of different masses, the more the mass of
the object, the less its acceleration. In fact, they are inversely proportional: when
the mass doubles, the acceleration is halved. This is expressed by the equation
a
1
m
Because of this relation, we say that an object’s mass is a measure of its inertia:
the larger the mass, the harder it is to accelerate, so that its tendency to stay with the
same velocity is the stronger.
Combining the two proportionality relations, we get
ak
F
m
(3)
where k is some constant.
Defining the force unit newton. How do we determine the constant k ? Let’s
remember that we have units for m and a (kg and m/s2 respectively in the SI system),
and so can assign numbers to these quantities. But up to now, we do not have a unit
for forces. In fact, we can use Eq.(3) to define a unit for force in the following way.
Take a 1-kg object, apply a force to it and adjust the strength of the force until the
object is observed to move with an acceleration of 1 m/s2. We call the strength of
such a force 1 newton (1N for short). Substituting the values for m, a, F in this
scenario into Eq.(3):
1 m / s2  k
1N
1 kg
k 1
and newton can be identified with the compound unit:
N  kg  m / s 2
If we use Newton as the unit for forces, kg as unit for mass, and m/s2 as unit for
acceleration, Eq.(3) simplifies to
a
F
m
which is equivalent to
F  ma
3
which is arguably the most famous equation in physics.
So far we have ignored the vector nature of forces and acceleration. It is not hard to
see that a force produces acceleration in the same direction. So we should write


F  ma
Then in general there could be more than one force acting on an object. Experiments
show that, in their ability to cause acceleration, multiple forces are equivalent to a
single force that is equal to the (vector) sum of the forces. Thus arrive at the final
form of Newton’s second law expressed as an equation


F  ma
For calculation, it is more convenient to use the component form, which in 2dimension becomes
Fx  ma x
Fy  ma y
How to use Newton’s second law
We suggest the following procedure to do problems involving the 2nd law.
1. Draw a diagram (called the fee body diagram) in which the object is represented
by a point, and the forces acting on it are arrows coming out of it.
2. Choose x and y axes for reference (Just choose x axis if the problem is 1-D)
3. Use Eq.(4) as templates and fill in quantities that are known. This requires
examining each force vector in the diagram and calculate its x and y components.
4. Solve the equations to find the unknowns.
Example: A 2-kg block on a table is pushed with a 10N force. The friction force is
6N opposing motion. Find the acceleration.
Solution: The free body diagram is as shown. Note that we represent the push as a
pull to conform with the rule that forces originate from the object.
4
6N
2kg
10N
x
Filling in the template
Fx  max
10   6  2a x
 ax 
4
 2m / s 2
2
Example: A 5-kg block is pushed with a 20N force on a rough table. It is observed to
travel forward with acceleration 3m/s2. Find the friction force.
The free body diagram is
F
5kg
20N
x
Since we expect the direction of the friction force to be opposite to the 20N force, we
draw its arrow as indicated. It is labeled as F. This unknown value represents the
magnitude of the friction force, and so is expected to be a positive number.
Filling in the template
Fx  max becomes
20   F   5  3
since the x-component of the friction force is –F. Solving the equation gives
F  20  15  5N
Example: How large a force is required to stop a 800-kg car in a distance of 50m if the
car is traveling at 30m/s?
Solution: The free body diagram is as shown:
5
F
x
First we find the acceleration:
vox  30m / s
vx  0
x  50m
 ax 
v x2  v02 0  30 2

 9m / s 2
2x
2  50
Next we apply Newton’s second law:
Fx  max
 F  800   9
 F  7200N
The British unit for force is lb. The conversion factor is
1 lb  4.445 N
The above force is therefore
7200 N  7200 
1
lb  1620 lb
4.445
Example: The action of rowing exerts a force of 150N due East on a 250-kg boat with a
100-kg passenger. Wind exerts a 120N force in the direction 60ºW of N. Find the
magnitude and direction of the acceleration of the boat.
Solution:
With the free body diagram as shown
6
y
60º
150N
120N
x
350kg
\
The x-component of Newton’s law is
150  120 sin 60   350a x
 ax 
150  120 sin 60 
 0.13m / s 2
350
The y-component of Newton’s law is
120 cos 60  350a y
ay 
120 cos 60 
 0.17m / s
350
The magnitude of acceleration is

a  0.132  0.17 2  0.21m / s 2
The direction is   tan 1
0.17
 53 N of E
0.13
7
Newton’s Third Law
This law has to do with the nature of forces. A force on an object must come from
another object. If you call one the agent, the other may be called a receiver. But the role
of agent and receiver are reversible. The agent also gets a force from the receiver that is
equal and opposite in direction. If you hit a table with your fist, your fist also receive a hit
by the table. In a tug of war, if team A is pulled by team B with a certain force, then team
B also experiences the same force in the opposite direction, even though team A is not
making any apparent effort, and is in fact losing the game. Summarizing these situation,
Newton’s 3rd law can be thus stated:
If object A acts on object B with a force, then object B also acts on object A with a force
of equal magnitude but opposite in direction.
If we call the pair of forces action and reaction, we can state the law succinctly:
Action and reaction are equal and opposite.
It is important to know that action and reaction do not cancel each other out, because they
act on different objects. Consider two persons sitting on swivel chairs, pulling each other
with a rope. (Or, only one person does the pulling). If the masses of the persons are
m A and mB respectively, then a free body diagram on the two persons is as shown:
aA
mA
F
F
aB
mB
x
in which we have identify the x-axis and the accelerations of the two. Applying Newton’s
second law to person A:
F  mA a A
To person B:
 F  mB aB
 mA a A  mB aB
So they move toward each other. If person A has more mass, he or she would experience
a larger acceleration.
Consider the scenario of a donkey pulling a cart. Both move forward with an
acceleration. How is this possible in view of the fact that the donkey is pulled backward
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with a force by the cart that is of equal magnitude to the one with which it pulls the cart?
The answer lies in the presence of friction with ground (at least for the donkey). The
force diagram is the following:
P
P
F
We see that the net force on the donkey can be forward if friction F is more than the pull
P, resulting in a forward acceleration for the donkey. The cart also accelerates forward
because P is the only force and is forward.
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7. Weight and Normal Force
Newton’s law of universal gravitation
Between two point masses m1 and m2 , that are separated from each other by a distance
d , there is a force of attraction with a magnitude given by
F G
m1m2
d2
d
F
F
m2
m1
The constant G , known as the gravitational constant, is given by
G  6.67  10 11 N  m 2 / kg 2
and is very small. Therefore the gravitational attraction between ordinary objects is tiny.
However, it becomes very important fro astronomical objects.
The gravitational force on an object due to the earth is called its weight. To calculate this
force on an object, whose mass is denoted by m , we need the following data:
Mass of earth M  6.0  10 24 kg
Radius of earth R  6.4  10 6 m
Additionally, we need to know the height h (altitude) of the object above ground. The
earth is not a point mass. But it can be shown that for the purpose of calculating
gravitational force due to a sphere, all the masses inside the sphere can be imagined to
reside on its center, effectively making the sphere into a point mass at the center. Thus,
we can write
d  Rh
10
h
R
For most ordinary situations, the height is very small compared with the radius of the
earth. For example, for a jet air plane at a cruising altitude of 30,000ft, or 10,000m,
h
10 4

 1.5  10 3
R 6.4  10 6
In such cases, we can use the approximation
dR
The weight of the object is therefore
M
6.0  10 24
11
W  G 2 m  6.67  10 
m  9.8m newton
2
R
6.4  10 6


If the object is not supported, it will experience a downward acceleration, because when
we apply Newton’s second law, we obtain
y
ay
W
 W  ma y
ay  
W
9.8m

 9.8 m / s 2
m
m
11
independently of the mass. This is Galileo’s observation. Thus, Newton laws of motion
together with Newton’s law of universal gravitation provide an explanation of
Galileo’s observation that all objects fall with the same acceleration, independently of
their masses. We have also obtained a formula for the acceleration due to gravity:
g
GM
R2
The same formula applies to the acceleration due to gravity on the surface of the moon, if
we take M to be the mass of the moon and R to be its radius. The value of g on the moon
turns out be approximately 1/6 that on the earth.
When an object is at a very high altitude, the acceleration due to gravity becomes smaller.
For example, Hubble’s telescope (a satellite) is 600km above the surface of the earth. At
this location,
d  R  h  6.4  10 6  6.0  10 5  7.0  10 6 m
and the weight (gravitational attraction due to the earth) of an object of mass m at that
location is
W G
M
6.0  10 24
11
m

6
.
67

10

m  8.2m newton
2
d2
7.0  10 6


But as long as we stay close to the surface of the earth, we can write
W  mg
for the relation between mass and weight.
It is important to note the difference between mass and weight. Mass is a measure of

inertia, and occurs in Newton’s second law in the combination ma . Weight is a special
kind of force, namely, the gravitational attractive force due to the earth, and as such,


occurs on the left side of  F  ma . Weight has no meaning in outer space, but mass
does.
Example: What is the mass of an object weighing 2000N?
W 2000
Solution: m 

 204kg
g
9.8
Example: What is the mass of an object weighing 200lb?
Solution: Since 1 lb  4.45 N
m
W 200  4.45

 90.8kg
g
9.80
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Normal Force
A block resting on a table does not fall in spite of its weight. The forces on the block
must therefore balance out (add up to be zero). The free body diagram would look like:
FN
W=mg
where FN is a force on the block due to the table. This force is perpendicular to the
surface of the block, and is called the normal force for this reason. The word normal is a
mathematical (rather than psychological) term meaning the direction perpendicular to a
plane. The origin of the normal force in this case comes from the slight deformation of
the table, resulting in a push on the block because of elasticity.
The following diagrams identify the normal force in two other situations.
1. A block on an incline plane:
2. A magnet stuck to a refrigerator door:
13
For the block resting on a horizontal table, applying Newton’s second law, since the
block does not accelerate, we expect
FN  W
The normal force can be changed if we apply a lift force to the block. The free body
diagram in this case is
FN
L
W=mg
Newton’s second law says in this case
L  FN  W
 FN  W  L
and so is less than the weight.
Consider the case of two blocks resting on a table, with one on top of the other. Suppose
the weight of the top block is 20N and that of the bottom block is 30N. What is the
normal force
(1) On the top block due to the bottom block?
(2) On the bottom block due to the table?
First we draw the free body diagram of the top block in steps:
Step 1: Identify its weight:
20N
14
Step 2: Since it is not moving, there must be a normal force of 20N upward from the
bottom block. This answers the first question:
20N
20N
Step 3: Next we construct the free body diagram for the bottom block. We already
know two forces on the block: its weight and the reaction to the normal on the top
block:
30N
20N
Step 4: With only these two forces, the bottom block cannot remain at rest. Therefore
we expect a upward force of 50N for balance, and this must come from the table, and
is the normal force on the block due to the table:
50N
30N
20N
15
The reaction to the 50N force is a downward force of 50N on the table. Notice that we
could have gotten the same result by considering the two blocks as one object that
weighs 50N.
The Elevator Problem
The normal force is also affected by the state of motion.
Consider a person riding an elevator. If the elevator is at rest, or moving with constant
velocity up or down, its acceleration is zero. This also applies to the person because
he is moving together with the elevator. In this case, the normal force is equal to his
weight. If the elevator moves upward with increasing speed, its acceleration a is
upward, so is that of the person. The free body diagram for the person is as shown:
y
FN
a
mg
Newton’s second law yields
FN  mg  ma
 FN  mg  ma  mg  a 
Thus the normal force exceeds his weight. The normal force needs to overcome the
weight with enough excess to cause the upward acceleration. In this case, the person
feels heavier. For this reason, we call the normal force the apparent weight.
When the elevator slows down on its upward journey, the accelerator is downward.
The free body diagram becomes
16
y
FN
a
mg
and Newton’s second law gives
FN  mg  ma
 FN  mg  ma  mg  a 
and is less than the weight. The person feels lighter.
In the last case, if a  g , we have FN  0 . The person feels no push from the floor.
Indeed he will lose contact with the floor. Of course, for this to happen, the elevator
has to be pulled downward with a sufficient force, which is unlikely. But this scenario
occurs whenever you push something up on your palm and let it fly in the air.
During the downward journey of the elevator, when it speeds up, acceleration is
downward, and the normal force is also less than the weight. Again, the normal force
would vanish if a  g . In this case, the elevator and the person are free falling
together. The person might as well be outside the elevator and still be together with
the elevator all the time. This condition is known as weightlessness. The condition
also applies to astronauts in a space station encircling the earth. An astronaut can
apparently float inside the space station because he is experiencing the same
downward acceleration due to the gravitation attraction of the earth as the space
station. Neither objects actually fall to the surface of the earth in spite of the
acceleration because they have sideward velocities like a projectile.
17
8. Friction
Friction force is due to the roughness of surfaces. As one surface glides past or has the
tendency to glide past another, the roughness causes resistance. The friction that occurs
without actual gliding is called static friction. When gliding occurs, it is called kinetic
friction.
The nature of friction can be demonstrated by pulling a block on a table. Let P be the
magnitude of the pull force. When P is small, the block does not move. The friction force
is static in this case, and its magnitude FS must equal P , but it points in the opposite
direction:
FS
P
As P increases, so does FS. Therefore static friction can change. However, there will
come a point when further increase of P causes the block to start moving. When this
happens, the friction force reaches its largest possible value, which we denote by Fmax.
Once the block starts moving, the friction force is due to kinetic friction. Experiments
show that its value is less than the maximum static friction Fmax. Thus it takes less force
to keep an object moving than to start it from rest. It is also found that the value of kinetic
friction remains the same no matter what value the pull P assumes. The following sketch
illustrates what happens in this experiment:
F
Fmax
FS
FK
P
The maximum static friction Fmax is found to be proportional to the normal force:
Fmax   S FN
The constant of proportionality, denoted by  S , is called the coefficient of static friction,
and depends on the roughness of the surfaces involved. In the example of pulling a block
on a table, the complete free body diagram is as shown:
18
y
FN
FS
P
x
W=mg
Since the block is not moving in the y-direction, we have
FN  mg
Thus the maximum static friction is
Fmax   S mg
and increases with the mass.
We can also increase Fmax by pushing down on the block with a force of magnitude P’. In
this case the free body diagram is
y
FN
FS
P
x
P’
W
Force balance in the y-direction gives
FN  W  P  0
 FN  mg  P
Fmax   S FN   S mg  P
The reason why it is so hard to remove a nail from the wall is because the normal force
acting on the nail from the wall is large.
19
The kinetic friction force is also proportional to the normal force:
FK   K FN
The coefficient of kinetic friction  K is always less than  S .
Example: A block on rough table is given a push and immediately acquires a velocity of
4m/s2. Given that the coefficient of kinetic friction is 0.25, how far can the block travel
until it stops? Note that we are not given the mass of the block.
Solution: Kinetic friction applies because gliding occurs. The free body diagram is as
shown:
y
FK=μKFN
FN
x
mg
First we need to find the acceleration. Filling in the template Fx  max leads to
 FK  ma x
(1)
On the other hand, the template Fy  ma y gives
FN  mg  0
because a y  0 . Therefore FN  mg and
FK   S FN   S mg
Substituting into Eq.(1):
  S mg  max
 a x    S g  0.25  9.8  2.5m / s 2
20
The mass m drops out of the equation.
Next, for the motion, we now have the data:
a  2.5m / s 2
v 0  4m / s
v0
Using the equation v 2  v02  2ax ,
x 
v 2  v02
0  42

 3.2m
2a
2 2.5
Static friction can help propel motion as the following example shows.
Example What is the maximum acceleration of a car if the coefficient of static friction is
0.5?
Solution: When a car speeds up, the spinning tires have a tendency to slide backward on
the road surface. The sliding does not occur unless there is slipping. With no slipping, the
friction is static, and the direction is forward (to oppose the tendency to slip). The free
body diagram is therefore as shown:
amax
FN=mg
Fmax=μSFN=μSmg
mg
We have used the fact that the acceleration is maximum when the static friction is
maximum, and indicated the dependence of the maximum friction on  S , FN and
therefore mg . Newton’s 2nd law in the x-direction becomes
 S mg  mamax
 a max   S g  0.5  9.8  4.9m / s 2
A heavy truck and a light sedan have the same maximum acceleration if the coefficient of
static friction is the same. Clearly, the tires of a race car should be designed to achieve a
large value for  S .
21
Example: Consider a 50-kg crate resting on a flatbed truck. What is the friction force on
the crate when (A) the truck is moving with constant velocity, and (B) when it is
speeding up forward with acceleration 2.5 m/s2 ?
Solution: Focus attention on the crate. The free body diagram for it is as shown.
a
FN=mg
FS
mg
=
From the x-component of Newton’s second law,
FS  ma
In case (A), a  0 . The friction is zero.
In case (B), FS  50  2.5  125N
Example: It is observed in the previous example that the crate starts to slip backward on
the flatbed as soon as the acceleration of the truck reaches 6 m/s2. What is the coefficient
of static friction?
Solution: The free body diagram remains the same, except that FS is replaced by Fmax
and a  6m / s 2 . The forward static friction force on the crate reaches the largest possible
value to produce a 6m/s2 acceleration of the crate. Further increase of acceleration of the
truck will result in the crate slipping.
S 
Fmax ma a
6

 
 0.61
FN
mg g 9.8
Can you answer the question: When the flatbed truck slows down, what is the direction of
the friction force on the crate?
Example: A block of mass m sits on an inclined plane that makes an angle  with the
horizontal. Find the acceleration down the incline if it is frictionless.
Solution: The free body diagram together with the choice of x and y axes is as shown
22
y
FN
x
mg
θ
To find the x component of the weight mg, we complete the triangle as shown:
y
θ
x
θ
mg
The angle identified inside the triangle is equal to θ because it is complementary (adding
up to 90º) to the same angle (the one between the black and the red arrow down the
incline) as the angle between the incline and the horizontal. The x-component is indicated
by the red arrow down the incline, and the y-component is the other red arrow.
 x component of weight  mg sin 
y component of weight  mg cos 
The x-component of Newton’s 2nd law is
mg sin   max
Therefore the acceleration down the incline is given by
a x  g sin 
and is always less than 1 because sine of any angle lies between 0 and 1. When
  90  , we have sin   1 , so that a x  g , which is to be expected because then the
23
inline is vertical, and the object is free falling. When   0 , we have sin   0 , so that
a x  0 , which is again expected because the incline is flat.
From the y-component of Newton’s 2nd law, using the fact that a y  0 as the y-coordinate
of the object does not change (the objects stays on the incline), we find
FN  mg cos   0
 FN  mg cos 
and is always less than the weight because cosine also lies between 0 and 1.
If the incline in the above example has friction, the object can stay still for a while as the
inclination is increased, starting from being horizontal. In this case, the free body diagram
includes a friction force, which comes from static friction as long as the object is not
moving.
y
FN
x
FS
mg
θ
The friction force can be found from the special case of Newton’s 2nd law when there is
no acceleration. In this case, the net force is zero. The x-component of Newton’s 2nd law
becomes
mg sin   FS  0
 FS  mg sin 
Since FS has no y-component, we still have
FN  mg cos 
(2)
as before.
24
As  increases, so does FS . Since FS has a maximum called Fmax (which is equal to
 S FN ), the equilibrium cannot be satisfied when the angle  is too large. The largest
possible  can be called the critical angle  crit because once it is exceeded, the object
will start to move. We therefore expect
Fmax  mg sin  crit
  S FN  mg sin  crit
Using Eq.(2) for FN , we find
 S mg cos crit  mg sin  crit
leading finally to tan  crit   S
Thus, if we measure the critical angle and find it to be 30º, we would have
 S  tan 30   0.58
25
9. Tension
When you stretch a rubber band with both hands, it gets longer, and pulls back on your
hands. The same thing happens with a string, although the extension is too small to be
seen. In these cases, the rubber band and the string are said to be in tension. The
following diagram shows the forces on either ends of the string, which are equal if the
string is not moving. If the left end of the string is tied to the wall, and only the right end
is pulled by your hand, the diagram is the same, except that the agent of the pull force on
the left is the wall instead of your left hand.
Let’s mentally divide the string into a left and a right parts. Focusing attention on the
right part, since it is not moving, and is pulled to the right with a force P, there must be a
force T pulling it to the left as shown:
T
P
P
T
Thus we write
T P
This force T is due to the left part of the string. The reaction of this force is an equal and
opposite force on the left part as shown. The force T is called the tension in the string.
From this we see that the pull force P has been transmitted undiminished along the length
of the string.
If the string is accelerating to the right with acceleration a, as would happen when you are
pulling a block and cause it to accelerate, then the forces on the right part cannot balance.
A net force is required to cause its acceleration. If the mass of the right part is m ,
Newton’s second law requires
P  T  ma
T  P  ma
There is therefore a reduction in tension as we move away from the point of contact of
the pull force P, resulting in less than perfect transmission. In most circumstances,
(except for a rope) the mass of the string is very small compared with the objects being
26
pulled. We can then neglect the mass of the string, and again use the relation T=P even
when there is acceleration.
Consider a small part of the string, for instance a part cut out of the right part of the string
in the preceding example.
T
P
T
P
T
\
In addition to the tension on its right end as previously identified, there must be a tension
force T on its left hand for equilibrium, and this force is due to the remaining left part of
the string. Thus we see that every portion of the string is being stretched by tension.
Example When a hanger is hung from a clothe line which is frictionless, you might notice
that it goes to the middle of the line. Suppose the mass of the hanger is 0.5 kg, the length
of the clothe line is 120cm, and it sacks a distance of 3cm. What is the tension on the
cloth line?
Solution: We expect the tension on either side of the clothe line be the same for
symmetry, but this can actually be demonstrated. If TL and TR are the tensions on the left
and right parts of the line, we can draw the free body diagram together with the
configuration of the clothe line as shown:
TL
TR
θ
θ
5N
At equilibrium, we require
Fx  0
which becomes
TR cos  TL cos  0
27
TL  TR  T
To find the tension, we need to use
F y  0
which gives
T sin   T sin   5  0
5
or T 
2 sin 
From the triangle identified in the diagram, knowing the length of half of the clothe line
to be 60cm, and the sack is 3cm, we find
sin  
T 
3
1

60 20
5
1
2
20
 50 N
Example: A 5-kg block is connected to a 3-kg block by a string with both sitting on the
frictionless table. The 3-kg block is pulled by a force of 16N in the direction away from
the other block. Find the accelerations of the blocks and the tension in the string. (Take
g=10m/s2)
Solution: We expect both blocks to have the same acceleration as they move together as
long as the string is taut. A schematic diagram and the free body diagrams for both blocks
are as shown
3kg
5kg
50N
30N
T
50N
y
T
16N
30N
a
28
x
Applying the x-component of Newton’s 2nd law, we find,
T  5a
for the left block,
16  T  3a
for the right block
Adding these equations (sum of left sides=sum of right sides)
16  8a
 a  2m / s 2
T  5a  10N
We could have gotten the answer for a by considering the two as one single object with
mass 5+3=8kg, and use
a
F 16

 2m / s 2
m 8
This way of thinking regards the two as forming a system, acted upon by external forces.
The tension is an internal force that need not concern us if acceleration is all we want to
find. However, finding the tension itself requires us to examine the components of the
system and apply Newton’s law to each component.
Example: Three blocks are connected by two strings as shown. The front block is pulled
so that the whole assembly accelerates to the right. Which string has more tension?
T2
T1
Answer: T1  T2 because T1 causes the acceleration of two blocks while T2 causes only
one. With equal acceleration, the force required is proportional to the mass.
Example: The same two connected blocks as the previous example are positioned so that
the 5-kg block sits on a frictionless table, while the string passes over a pulley and the 3kg hangs freely from the edge of the table. Find the acceleration and the tension in the
string.
29
50N
T
a
T
50N
30N
Solution: We have combined the configuration and free body diagrams into a single
picture. We have also drawn the direction of acceleration as making a turn from pointing
to the right to pointing downward for the 5-kg and the 3-kg blocks respectively.
Newton’s 2nd law for the 5-kg block is (rightward is positive)
T  5a
Newton’s 2nd law for the 3-kg block is (downward is positive)
30  T  3a
Adding the two equations:
30  8a
 a  3.75m / s 2
T  5a  5  3.75  18.75N
Note that the tension is less than the weight of the hanging mass. The difference is
responsible for the downward acceleration.
Example (Attwood’s machine) Two unequal masses connected with a light string are
hung from either side of a light pulley. Find the acceleration of the masses.
Solution:
30
a
a
T
T
m2 g
m1 g
m1 g
The configuration and the free body diagrams for the two masses are as shown. The two
masses share the same magnitude of acceleration but in opposite direction.
The downward component of Newton’s second law for the right mass is
m1 g  T  m1a
The upward component of Newton’s second law for the left mass is
T  m2 g  m2 a
Adding the two equations:
m1 g  m2 g  m1a  m2 a
a 
m1  m2
g
m1  m2
This can form the basis for an experimental determination of g . For example, suppose
m1  0.051kg , m2  0.050kg , and it takes 5.5s for m1 to fall a distance of 1.5m to the
floor, starting from rest. We can find a from the data v0  0 x  1.5m t  5.5s . Using
1
the equation x  v0 t  at 2
2
1
2
1.5  0  a5.5
 a  0.1m / s 2
2
m  m2
0.051  0.050
g 1
a
 0.1  10m / s 2
m1  m2
0.001
31
10. Centripetal Acceleration and Centripetal Force
An object traveling along a circle with constant speed is said to execute a uniform
circular motion. The time it takes to go once around the circle is called the period. Let
r  radius of circle
v  speed
T  period
During the time interval T , the distance traveled equals the circumference of the circle,
which is 2r . Therefore
v
dis tan ce 2r

time
T
Example A gramophone record rotates at 33-1/3 rpm (revolutions per minute). Find the
velocity of a scratch mark which is at a distance of 5cm from the center.
Solution First find the period, which is the time taken to make one revolution.
3
min 
 60s  1.8s
1
100
33
3
The radius is r  5cm  5  10 2 m
2r 2 0.05
v 

 0.17m / s 2
T
1.8
When an object is executing a uniform circular motion, the direction of velocity changes
even though its magnitude does not. The object therefore experiences acceleration. To
find the acceleration, let’s first look at a motion graph, which is snap shots of the object
v2
T
1
v1
at equal time interval.
32
  
To find the change of velocity v  v2  v1 when the object is at the point P, we reverse
the vector v1 to obtain –v1 and add it to v2. This leads to a v that is directed toward the
center:
v2
-v1
v
r
r

 v
Since acceleration is by definition a 
, its direction is toward the center. For this
t
reason it is called a centripetal acceleration. To find its magnitude, we note that the little
black triangle and the blue triangle are similar triangles because they are both isosceles
and share a common angle. Their lengths are therefore proportional. The base of the blue
triangle is the displacement traveled in the time interval t , which is equal to vt .
Therefore we have
 v v t

v
r

v v 2

t
r
So the magnitude of the centripetal acceleration is
a
v2
r
According to Newton’s second law, there must a net force pointing toward the center that
equals ma . We call this net force the centripetal force, whose magnitude is therefore
given by
33
F
mv 2
r
The role of centripetal force can be played by any or combination of the types of forces
that we have considered, which include gravitational force, normal force, friction force
and tension force.
When a ball is twirled on the end of string to move in a horizontal circle, tension force
plays the role of centripetal force. The forces acting on the ball, the string, and the hand
are as shown, where action-reaction pairs are color-coded.
hand
v2
a
ball
r
string
We see that the string is under tension, and the hand experiences a pull toward the ball.
The tension is given by
mv 2
T
r
As the speed of the ball goes up, so does the tension force. When the latter reaches the
maximum sustainable tension, the string will break.
Example 2-kg object is twirled at the end of a string and moves in a horizontal circle of
radius 0.2m. If the maximum tension sustainable tension is 5N, what is the maximum
speed of the object without breaking the string?
Solution: We use the equation
Tmax 
2
mvmax
r
from which
vmax 
Tmax r
5  0.2

 0.71m / s
m
2
34
Example If in the previous example, the speed of the object is maintained at 0.5 m/s, but
the string is continuously shortened. At what length will the string break?
Solution: At constant speed, the shorter the radius of the circle, the stronger is the tension
force that plays the role of centripetal force. The string breaks when the tension is at its
maximum. The smallest radius rmin without breaking therefore satisfies the equation
Tmax
mv 2

rmin
 rmin 
mv 2 2  0.5 2

 0.1m
Tmax
5
For a car turning on a flat road surface, static friction plays the role of centripetal force:
mv 2
FS 
r
The free body diagram is as shown, where the center of the circle is on the left.
FN
a
FS
mg
As the car speeds up while maintaining the same radius for the turn, the static friction
increases to match the demand from increasing centripetal acceleration. When it reaches
the maximum possible value, the car is at the point of skidding. Since maximum static
friction is
Fmax  S FN   S mg
We now have
 S mg 
2
mvmax
r
35
 vmax   S gr
Note that the result does not depend on the mass of the car. To make it easier to turn, the
roughness of the surface should be increased.
Example: What is the minimum value of the coefficient of static friction in order that a
car can safely turn in a circle of 10m while traveling at 30mph?
Solution:
30mph=13.4m/s
From the reasoning above,
S g 
v2
r
v2
13.4 2
 S 

 1.8
gr 9.8  10
To increase the maximum speed for cars to turn without slipping, the road surface can be
banked, as is often done for the racing tracks. As shown in the free body diagram below
for a car turning in a circle whose center is to the left, both static friction force FS and the
normal force FN have components in the direction pointing to the axis of rotation.
a
FN
To axis of
rotation
FS
mg
Turning is also possible without friction. You can no doubt appreciate that a marble can
travel in a horizontal circle on the inside of a smooth conical surface. In this case, the
component of the normal force on the marble in the direction pointing to the axis of
rotation plays the role of centripetal force.
What is playing the role of centripetal force when an airplane turns? To answer this
question, we first note that to stay in midair, there must be a lift force that balances the
36
weight of the plane. The origin of this lift force is the slightly lower air pressure on top of
the wings than below them, generated by the design of the shape of the wing section.
When the wings are level, the lift force is vertical. When the wings are tilted, the lift force
remains perpendicular to the wings, and is no longer vertical. Instead it has component in
the direction pointing to the axis of rotation.
Satellites travel in circles around the earth. The centripetal force for satellites is due to
the gravitational attraction from the earth, or the weight of the satellite in short. Using
Newton’s law of universal gravitation, we can therefore write
Weight = centripetal force
GMm mv 2

r
r2
where
M  mass of the earth
m  mass of the satellite
r  radius of satellie orbit
v  speed of the satellite
For near earth orbits, we can replace r by the radius of the earth, which we denote by R .
We can therefore write
mv 2
mg 
R
It follows that the speed of such satellites is
v  gR  9.8  6.4  10 6  8000m / s
If a projectile is thrown horizontally with this speed, it will never fall to the floor, but will
instead become a satellite, provided of course its motion is unimpeded by buildings and
mountains, and air resistance can be neglected.
When the orbits have high altitude, we can no longer set r equal the earth’s radius. In this
case the speed is give by
v
GM
r
and gets smaller the higher the altitude of the satellite.
37
How does the period pf the satellite depend on the radius of its orbit? To answer this
question, we first square both sides of the above equation to get
v2 
GM
r
and the make the substitution
v
2r
T
The result is
GM
 2r 

 
r
 T 
2
Collecting r on one side and T on the other, we find
T2 
4 2 3
r
GM
This means T 2 is proportional r 3 :
T 2  r3
Therefore, the higher the altitude the longer is the period. The same argument can be
applied to the solar system. In this case, the planets are the satellites, and M is the mass of
the sun. The proportionality between the square of the period of the planets and the cube
of radius of their orbits is known as one the Kepler’s laws. It was deduced by Kepler
from observational data on planet motions. We see that the law can be explained by
Newton’s law of universal gravitation.
The proportionality relation can be written in a number of ways when we compare two
orbits, labeled as 1 and 2:
 T2

 T1
2

r
   2

 r1



3
T2  r2

T1  r1



32
r2  T2

r1  T1



23
Satellites that have the same period as the earth’s rotation around its axis and whose
orbits are on the equatorial plane are called geosynchronous satellites. They stay in one
spot in the sky if observed from anywhere on the earth. If we compare such satellites with
the moon, labeling the latter as 1 and the satellite as 2, we have
38
T2
1d
1


T1 27 d 27
where we have taken the moon’s period around the earth to be approximately 27 days.

r2  1 
 
r1  27 
23
2
1
1
  
9
 3
Thus, a geosynchronous satellite is 1/9 of the way toward the moon. If we take the
average distance between the moon and the earth to be 3.85  10 8 m , the radius of the orbit
of a geosynchronous satellite is 3.85  10 8 / 9  4.28  10 7 m . Dividing this by the radius
of the earth 4.28  10 7 / 6.4  10 6  6.7 . The altitude of such satellites are approximately
six times the earth’s radius.
39
40