Lesson 2: KINETICS OF
RECTILINEAR TRANSLATION
Prepared by:
JANE M. TECSON
Course Instructor
Where are we in the course?
Concept: what is dynamics?
Newton’s Laws of Motion of a Particle
1. A particle acted upon by a balanced force system has no acceleration
2. A particle acted upon by an unbalanced force system has an acceleration
in line with and directly proportional to the resultant of the force system.
F= ma ;
a= F/m
3. Action and reaction forces between two particles are always equal and
oppositely directed
FUNDAMENTAL EQUATION OF KINETICS OF A
PARTICLE
Resultant force = mass x acceleration
Fr=ma
Note: W = mg
;
m=W/g
So, (by substitution)
Fr=
𝑊
𝑔
(a)
Where:
Fr = resultant force
m=mass
W=weight
a= acceleration
g = acceleration due to gravity = 9.81m/s2 =32.2 ft/s2
EFFECTIVE FORCE ON A PARTICLE (INERTIAL
FORCE, Fr)
- Is the resultant force on the particle
Since
𝑊
(a)
𝑔
Fr=
𝑊
𝑔
(a);
is more convenient to be used as the effective force of a
particle in cases where the acceleration of the particle is known but
the actual force system producing this acceleration is not known.
D’ALEMBERT’S PRINCIPLE
-it states that the resultant of the external forces applied to a body (rigid or nonrigid) composed of a system of particles is equivalent to the vector sum of the
effective forces acting on all particles.
REF = F-f = ma =
𝑊
(a)
𝑔
Where:
REF - resultant of the external forces
F
– applied force
f
-frictional force
Problem 1
What horizontal force F can be applied to a 100 kg in a level surface with
coefficient of friction of 0.2, that will cause an acceleration of 2.5 m/s 2 ?
Given:
m = 100 kg
a = 2.5 m/s2
a
µ = 0.2
mass
Required:
F
Formula:
REF = F-f = ma
F = ma + f
Problem 1 (cont.)
Solving for frictional force (f):
f=µN
Solving for N:
∑Fy = 0 (upward forces = downward forces)
N = W = mg = 100 kg (9.81 m/s2) = 981 N
Therefore:
f=µN
f = 0.2 (981 N) = 196.2 N
a
mass
Problem 1 (cont.)
finally,
F
= ma + f
= 100 kg (2.5 m/s2) + 196.2 N
F = 446.2 N (Ans.)