Chapters 5 & 6 – Laws of Motion
This is the first of two chapters on Newton’s laws of motion. Isaac Newton was one of
the greatest scientists of all time and his work provides the foundation of classical
mechanics. In addition to the laws of motion, Newton also discovered the law of
universal gravitation, which applies to planetary and satellite motion. He also invented
calculus.
Newton’s laws of motion are
1. A body at rest remains at rest and a body in motion remains in motion with
constant velocity unless acted on by a net external force. This is referred to as the
law of inertia.
2. Simply stated, F = ma. In this equation F is the vector sum of all the forces
acting on the object, m is the mass of the object, and a is the object’s acceleration.
3. For every action force, there is an oppositely directed reaction force of the same
magnitude. In equation form, F12 = -F21, where F12 is the force acting on object 1
by object 2 and F21 is the force on object 2 by object 1.
These equations involve the concepts of force, mass, and acceleration. We already know
about acceleration from previous chapters. Acceleration is the rate at which the velocity
changes, a = dv/dt. Qualitatively, a force is a push or a pull. It can come from direct
contact or it can be action at a distance (e.g., magnetic force, gravitational force). Force
can be measured in terms of the acceleration it gives to a standard mass using F = ma.
Mass can be measured relative to a standard mass by comparing the acceleration that a
fixed force gives it compared to the acceleration that the same force gives to the standard
mass. In the SI system, mass has units of kilograms (kg) and force has units of newtons
(N). 1 N = 1 kgm/s2.
First Law
The first law is simply a special case of the second law. If F = 0, then a = 0, which
means that v = constant (in magnitude and direction).
A ball going through the air does not maintain constant velocity since the force on the
ball is not zero. After the ball is thrown, the forces acting on it are gravity (down) and air
resistance (opposite to its velocity). (The force of the throw vanishes after the ball is
released.) If a ball were thrown in outer space far away from the pull of gravity of
celestial bodies, then it would continue moving indefinitely with constant speed without
changing direction.
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Second Law
The 2nd law says that an object will accelerate (speed up, slow down, or change direction)
if all the forces acting on the object don’t balance out. The acceleration will be in the
same direction as the net force. That is,
a = F/m
In terms of components, ax = Fx/m, ay = Fy/m, az = Fz/m.
Third Law
The 3rd law is very simple, but is easily
misunderstood. It says, for example, that if I
push on you then you push back with an
equal force. Our relative sizes and velocities
don’t affect this fact.
F12
1
F21 = -F12
2
Weight and Mass
The weight (W) of an object is the gravitational force acting on the object. If you drop an
object, then the force on the object is F = W and its acceleration is a = g. Thus, by
Newton’s 2nd law, F = ma, or W = mg. An object has the same mass regardless of
whether it is on earth, on the moon, or in outer space. Its weight depends on where it is.
On earth a 1 kg mass has a weight W = (1 kg)(9.8 m/s2) = 9.8 N. On the moon, where g =
1.6 m/s2, a 1 kg mass has a weight W = (1 kg)(1.6 m/s2) = 1.6 N. In outer space, far from
the earth or any other celestial body, the weight of a 1 kg mass is zero.
Examples:

If you weigh 150 lb, then the earth pulls down on you with a force of 150 lb. As a
consequence, you pull up on the earth with a force of 150 lb.

A Mack truck traveling at 60 mph makes a head-on collision with a motorcycle
traveling at 30 mph. The force exerted by the truck on the motorcycle is exactly the
same in magnitude as the force exerted by the motorcycle on the truck.
Equilibrium
If the sum of the forces on an object is zero, then it is in equilibrium (even if moving with
constant velocity). In 2-D,
Fx = 0, Fy = 0
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Example: Mass resting on a table.
n
The two forces acting on the mass are the upward force
exerted by the table, n, (n for ‘normal’ or perpendicular
direction) and the downward pull of gravity, mg. Thus,
Fy = n – mg = 0, or n = mg
mg
Example: Mass suspended by ropes. Find the tensions, T1, T2, and T3, in the ropes.
The tensions are the forces with which the ropes pull
on the knots which tie the ropes together. The mass is
in equilibrium, so
30o
50o
T2
T1
Fy = T3 – mg,
T3
2
so T3 = mg = (20 kg)(9.8 m/s ) = 196 N
m = 20 kg
The knot is in equilibrium. We show the forces acting
on the knot on a x-y coordinate system. Then
Fx = -T1cos(30o) + T2cos(50o) = 0
y
Or,
T2
T1
Fy = T1sin(30 ) + T2sin(50 ) – T3 = 0
o
o
30o
50o
0.866T2 = 0.643T1
x
0.5T1 + 0.766T2 = 196 N
T3
Combining these two eqs to solve for T1 and T2 –
0.5T1 + 0.766(0.643T1/0.866) = 196
1.069T1 = 196, or T1 = 183 N
Then, T2 = 0.643T1/0.866 = 0.643(183)/0.866, or T2 = 136 N
Example: A 4-kg block is pulled along a level
frictionless surface by a 20-N horizontal force.
What is the acceleration of the block?
A ‘freebody’ diagram is given to the right,
showing all the forces acting on the mass.
Applying the 2nd law,
m = 4 kg
n
F = 20 N
mg
3
Fx = max, or 20 N = (5 kg)ax, or
ax = 20/4 = 5 m/s2
What is the normal force exerted on the block by the table?
Fy = may = 0 (it can’t accelerate up or down)
Or, n – mg = 0, or n = mg = (4kg)(9.8 m/s2) = 39.2 N
Example:
The block in the example above is pulled with the 20-N force at an angle of 30o above the
surface.
The free body diagram is given to the right,
and the forces are also shown on an x-y
coordinate system. Now,
n
m = 4 kg
F = 20 N
30o
Fx = Fcos = max
20N cos(30o) = 17.3 N = (4 kg)ax
mg
ax = 17.3/4 = 4.3 m/s2
y
Fy = Fsin + n – mg = 0
n
F

n = mg - Fsin
x
= (4kg)(9.8 m/s2) – (20N)sin(30o)
mg
n = 29.2 N
Example: Inclined plane without friction.
y
n
n
mg cos
mg


mg sin
x
mg
Find the acceleration down the plane and the normal force exerted on the block by the
plane.
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For simplicity, we choose a coordinate system with the y-axis perpendicular to the plane
and resolve the forces into x- and y-components. This way, we know that ay = 0 and we
only have to find ax. Then,
Fy = n – mg cos = ay = 0, n = mg cos
Fx = mg sin = max, ax = g sin
So the inclination of the plane reduces the normal force below that for a level surface
(mg). If  = 30o, for example, then the normal force is 0.866 mg and the acceleration
down the incline is 4.9 m/s2.
Friction
Friction is a force due the interaction of an object with its environment that opposes its
motion. Common examples are air resistance and the horizontal force between two
objects in contact.
n
Static Friction
Fapplied
Consider a block resting on a horizontal
surface. If you try to slide the block one
fs
way the surface will exert an opposing force
in the opposite direction. (This is in
addition to the normal force exerted by the
surface on the block.) This static frictional
mg
force will increase as the push increases
until it starts moving. Empirically, it is
found that the maximum static frictional force is proportional to the normal contact force
between the block and the surface. That is,
f s , max   s n
where the proportionality constant s is called the coefficient of static friction.
Note: In the above diagram where the surface is horizontal and the applied force is
horizontal, n = mg. This will not be the case if the applied force is directed upward or
downward at an angle or if the surface is not horizontal.
Kinetic Friction
To a first approximation, a block sliding on a surface experiences a frictional force that is
independent of its speed and also is proportional to the normal force. That is,
f k  k n
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where s is the coefficient of kinetic friction. Typically, k < s. That is, it is easier to
keep an object sliding at constant speed than to start it moving. Also, typically s < 1 and
k < 1 (but not always).
n
Example:
F = 35 N
A 5-kg block is pulled along a horizontal
surface with a 35-N horizontal force. The
coefficient of kinetic friction between the
block and the surface is 0.4. Find the
acceleration of the block.
fk
mg
Fy = n – mg = 0, n = mg = (5 kg)(9.8 m/s2) = 49 N
fk = kn = kmg = 0.4(5 kg)(9.8 m/s2) = 19.6 N
Fx = F – fk = max
35 N – 19.6 N = (5 kg)ax, ax = 3.1 m/s2
n
Example:
F = 35 N
The block in the above example is pulled with a
force of 35 N at an angle of 15o above the
horizon. Find the acceleration.
15o
fk
mg
We will refer to the diagram to the right, where
we have resolved the applied force into x- and ycomponents.
Fy = n + Fsin – mg = 0,
n = mg - Fsin
= (5 kg)(9.8 m/s2) –(35 N)sin(15o) = 39.9 N
fk = kn = 0.4(39.9 N) = 16.0 N
y
n
fk
Fsin
Fcos
x
mg
Fx = Fcos – fk = max
(35 N)cos(15o) – 16.0 N = (5 kg)ax
ax = 3.56 m/s2
Note that the acceleration is actually larger than when pulled horizontally with the same
force. Does this make sense? Would this be the case if the angle were much larger than
15o?
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Example: Mass and pulley system.
Two masses are connected by a string draped over
a light, frictionless pulley, as shown to the right.
The weight of the hanging mass pulls the other
mass along the surface. Friction exists between
mass 1 and the surface. We want to find the
tension in the cord and the acceleration of the
masses.
m1
m2
We apply Newton’s 2nd law to each mass separately.
Mass 1:
mass 1
n
a
Fy = n – m1g = 0, n = m1g
fk
Fx = T – fk = m1a, or
T
m1g
T - km1g = m1a
Mass 2:
Fy = m2g – T = m2a (down is positive)
mass 2
There are two unknowns (T and a) and two equations
(boxed). If we add the two equations (left 1 + left 2 =
right 1 + right 2) we will get
T
a
m1g
m2g - km1g = (m1 + m2)a
or,
a
(m2   k m1 ) g
(m1  m2 )
Knowing a, we can find T using one of the boxed equations.
Uniform Circular Motion
For an object moving in a circle with constant speed, the acceleration is directed toward
the center of the circle and has magnitude
ac 
v2
r
Thus, according to Newton’s 2nd law the net force is given by
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v2
r
F  mac  m
Example: A ball attached to the end of a string moves in a circular path at constant speed
on a frictionless, horizontal table, with one end of the string fixed at the center of the
circle. The mass of the ball is 0.5 kg, the radius of the circle is 0.8 m, and the speed of
the ball is 10 m/s. What is the tension in the string?
Since the normal force exerted by the table on the ball and the weight of the ball cancel
out, the net force is just the tension in the string. Thus,
F  T  m
v2
(10m / s) 2
 (0.5kg)
 62.5N
r
0.8m
Example: If a mass hanging from a string swings in a circle, then the string sweeps out a
cone. This system is called a conical pendulum. (If it swings back and forth in a vertical
plane, then it is a simple pendulum.) What is the relationship between the speed of the
mass, the length of the string, and the angle that the string makes with the vertical in
order to have a conical pendulum?
L

T
Tcos

Tsin
r
mg
mg
Fx  max , or
v2
(1)
r
Fy  T cos   mg  0, or
T sin   m
T cos   mg
(2)
v2
rg
As a specific example, if the length of the pendulum is 1 m, what speed would be
required to have  = 30o? Solving the above equation for v and noting that r = Lsin , we
have
Combining eqs. (1) and (2), we get: tan  
v  rg tan   Lg sin  tan   (1m)(9.8m / s 2 ) sin( 30o ) tan( 30o )  1.68 m
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Example: How fast can a car go around a circular, flat track at constant speed without
skidding?
v2
Fx  f s  mac  m
r
Fy  n  mg  0  n  mg
n
fs
f s   s n   s mg, or
mg
v2
  s g , v   s gr
r
For example, if r = 200 m and s = 0.4, then vmax = 28 m/s (= 63 mph).
Example: What is the optimum angle of bank of a circular track? The optimum angle
would be such that no frictional force acts to cause the car to skid up the bank or slip
down the bank.
n

n cos
n sin

mg
mg
v2
r
Fy  n cos   mg  0  n cos   mg
Fx  n sin   m
From the above two equations we get
tan  
v2
gr
So, if v = 28 m/s and r = 200 m (the numbers for the flat track), then
tan  
(28m / s) 2
2
 0.4    21.8o
(9.8m / s )( 200m)
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n
Example: Ferris wheel.
The net force and the acceleration are always toward the
center of the circle.
mg
Bottom:
n  mg  m
v2
v2
v2
, n  m( g  )  mg (1  )
r
r
rg
n
Top:
v2
v2
v2
mg  n  m , n  m( g  )  mg (1  )
r
r
rg
mg
The ‘effective’ weight of the person on the Ferris wheel is the normal force (the force
exerted by the seat on the person). Thus, at the bottom Weff > mg and at the top Weff < mg.
For example, if r = 10 m and the time for a revolution = 30 s, then
v=
2π r 2π (10m)
=
= 2.09 m / s
T
30s
v2
(2.09m / s)2
=
= 0.045
rg (10m)(9.8m / s 2 )
Thus, a person’s effective weight would
be increased or decreased by 4.5%.
n
What would be the effective weight at
the side of the Ferris wheel?
n
In this case, the seat would tilt so that the
normal force (and the effective weight)
would be at an angle. You should be
able to show that
mg
v2
n  mg 1
rg
10
mg
Example: A ball on the end of a string hangs from the ceiling of a linearly accelerating
car.
Inertial system
Noninertial system
a
T
T

Ffictitious

mg
mg
To an observer standing at rest outside the car, the ball is accelerating and Newton’s 2 nd
law gives
Fx  T sin   ma
Fy  T cos   mg  0
This person is said to be in an ‘inertial’ coordinate system.
A person inside the car sees the ball at rest and in equilibrium. Thus, there appears to be
a fictitious force acting on the ball to make it hang at an angle. This fictitious force has
magnitude Ffic = ma and is directed to the left. He could also conclude that gravity
appears larger with g eff  g 2  a 2 and with the ‘down’ direction inclined at an angle .
Since Newton’s laws don’t hold in the accelerating system without invoking a fictitious
force, this system is called a ‘noninertial’ system.
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