Kinetics of Particles:
Newton’s Second Law
KINETICS
Newton’s Method
Work and
Energy Method
Impulse and
momentum
Kinetics - is a branch of dynamics that deals with the
relationship between the change in motion of a body and
the forces that cause this change.
FIRST LAW:
A particle originally at rest, or moving in a straight line with a constant velocity,
will remain in this state provided the particle is not subjected to an unbalanced
force
SECOND LAW:
A particle acted upon by an unbalanced force F experiences an acceleration a
that has the same direction as the force and a magnitude that is directly
proportional to the force.
THIRD LAW:
The mutual forces of action and reaction between two particles are equal,
opposite and collinear.
The first and third laws are extensively used during statics
However, Newton’s second law of motion forms the basis for most of the

dynamics concepts, since this law relates the accelerated motion of the particle
to the forces that act on it.
If the mass of the particle is ‘m’, Newton’s second law of motion may be

written in mathematical form as:
F = ma
This equation is referred to as the equation of motion

Newton’s Law of Gravitational Attraction:

Newton’s law of Gravitational Attraction may be expressed mathematically as:
F = [G m1 m2]/r2
where
F = Force of attraction between two particles
G = Universal constant of gravitation, 66.73x10-12 m3/kg.s2
m1 m2 = mass of each of the two particles
r = distance between the centers of the two particles

• If the resultant force acting on a particle is not zero,
the particle will have an acceleration proportional to the
magnitude of resultant and in the direction of the
resultant.


F  ma
• If particle is subjected
to several forces:


 F  ma
• We must use a Newtonian frame of reference, i.e., one that is not
accelerating or rotating.
• If no force acts on particle, particle will not accelerate, i.e., it will remain
stationary or continue on a straight line at constant velocity.
5
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
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
The equation of motion is:
F = ma
Consider a particle P which has a mass m and is subjected to the action of two
forces F1 and F2
We can graphically account for the magnitude and direction of each force acting
on the particle by drawing the particle’s free body diagram
Since the resultant of these forces produces the vector ma, its magnitude and
direction can be represented graphically on the kinetic diagram

The equation of motion for a system of particles can be
written as:
ΣF = maG

i.e. the sum of the external forces acting on the system of
particles is equal to the total mass of the particles times the
acceleration of its center of mass G.





Gravitational Force
Normal Force
Friction Force
Tension Force
Spring Force



The magnitude of the gravitational force acting on
an object of mass m near the Earth’s surface is
called the weight w of the object: w = mg
g can also be found from the Law of Universal
Gravitation
Weight has a unit of Newton
mM
w  Fg  mg
Fg  G 2
R
M
g  G 2  9.8 m/s 2
R

Weight depends upon location
R = 6,400 km




Mass is sometimes confused with weight.
Mass is a measure of the amount of matter
in an object; weight is the measure of the
gravitational force exerted on an object.
The force of gravity on a person or object at
the surface of a planet is known as weight.
So, when you step on a bathroom scale, you
are determining the gravitational force
Earth is exerting on you.



Force from a solid
surface which
keeps object from
falling through
Direction: always
perpendicular to
the surface
Magnitude:
depends on
situation
w  Fg  mg
N  Fg  ma y
N  mg  ma y
N  mg
•
•
•
•
μ – Greek letter “Mu”
Coefficient of friction.
– μs – coefficient of static friction
– μk – coefficient of kinetic friction
– depends on materials used and their surface
conditions
Coefficient of friction: Decimal between 0.0 and 1.0,
unitless.
Formula for solving friction:
Ff = μ FN



A taut rope exerts
forces on whatever
holds its ends
Direction: always
along the cord (rope,
cable, string ……) and
away from the object
Magnitude: depend on
situation
T1
T2
T1 = T = T2
• International System of Units (SI Units): base units
are the units of length (m), mass (kg), and time
(second). The unit of force is derived,
kg  m
 m
1 N  1 kg 1 2   1 2
 s 
s
• U.S. Customary Units: base units are the units of
force (lb), length (ft), and time (second). The unit of
mass is derived,
1lb
lb  s2
1slug 
1
2
1ft s
ft
1.
2.
3.
4.
Draw a free-body diagram by establishing a coordinate
system and showing all external forces applied to the
particle. Resolve forces into their appropriate
components.
Draw the kinetic diagram, showing the particles inertial
force, ma. Resolve this vector into its appropriate
components.
Apply the equations of motions in their scalar
components and solve these equations for the
unknowns.
The second law only provides solutions for forces and
accelerations. If velocity or position have to be found,
kinematic equations are used once the acceleration is
found from the equation of motion.
Sample Problem 1
The crate has a mass of 50 kg. If the crate is subjected
to a 400[N] towing force as shown, determine the
velocity of the crate in 3[s] starting from rest.
ms= 0.5, mk= 0.3,
SOLUTION:
Free body diagram
Kinetics diagram
400sin30
400cos30
Equations of Motion :

SOLUTION:
Free body diagram
Kinetics diagram
Kinematics : The acceleration is constant, P is constant
An 80-kg block rests on a horizontal plane. Find the
magnitude of the force P required to give the block an
acceleration of 2.5 m/s2 to the right. The coefficient of
kinetic friction between the block and plane is mk = 0.25.
SOLUTION:
FBD
Equations of Motion :
Kinetics diagram
 Fx  ma :
P cos 30   0.25 N   80  2.5 
Psin30
Pcos30
 200
 Fy  0 :
N  P sin 30  785  0
W  mg  80  9.81  785 N Solve for P and N
F  mk N  0.25 N
N  P sin 30  785
P cos30  0.25  P sin 30  785   200
P  534.7 N
N  1052.4 N
Sample Problem 3
The two blocks shown start from rest. The
horizontal plane and the pulley are frictionless,
and the pulley is assumed to be of negligible
mass. Determine the acceleration of each block
and the tension in the cord.
SOLUTION:
xA
O
xB
x
• Kinematic relationship: If A moves xA to the
right, B moves down 0.5 xA
xB  12 x A
aB  12 a A
y
FBD
Kinetics diagram
Considering Block A:
Equations of Motion :
F
x
 mA a A
T1  100  a A
(1)
FBD
Kinetics diagram
Considering Block B:
Equations of Motion :
F
y
 mB aB
m B g  T2  m B a B
300  9.81  T2   300  a B
T2  2940-  300  a B
FBD
Kinetics diagram
(2)
Considering Pulley C:
Equations of Motion :
F
y
 mC aC
T2  2T1  0
(3)
T1  100 aA
(1)
T2  2940  (300)aB
(2)
T 2  2 T1  0
(3)
Combining eq. (1) and eq. (2) in eq. (3), we have;
2940-  300  aB  2T1  0
2940-  300  aB  200a A  0
Combine kinematic relationships with equations of motion
to solve for accelerations and cord tension.
x B  12 x A
a B  12 a A
2940-  300  aB  2  200aB  0
aB  4.2 m / s 2
a A  8.4 m / s 2
Solving for T1
T1  100 aA
T1  100 (8 . 4 )
T1  840 N
(1)
Solving for T2
T 2  2 T1  0
T 2  2 T1
T 2  2 ( 840 )
T 2  1680 N
(3)