The document discusses Newton's laws of motion, focusing on the relationship between force, mass, and acceleration. It outlines the first three laws, introduces concepts of static and dynamic equilibrium, and explains how to apply these principles using equations of motion in various scenarios. Additionally, it includes examples involving friction and the calculation of velocities in dynamic systems.

1. F E B R U A R Y 2 6 , 2 0 1 4
KINETICS AND PARTICLES:
FORCE and ACCELERATION

2. 2 Main Factors that affect motion
in an Object
1. The Forces acting on an object
2. The Mass of an object

3. Mass – amount of matter in an object
- property of an object that specifies how
much resistance an object exhibits to
change in velocity
Acceleration – is the rate of change of the
velocity
Force – push or pull of an Object
– according to Isaac Newton, these are
what causes any change in the
velocity of an object at the same
time, it causes acceleration.

4. Newton's Laws
Newton described force as the ability to cause a mass to
accelerate.
First law: When the sum of the forces acting on a particle is
zero, its velocity is constant. In particular, if the particle is
initially stationary, it will remain stationary. Or if it moves
with constant speed in a single direction.
Second law: The rate of change of linear momentum of an
object is directly proportional to the applied force F and the
object moves in the direction in which force F is applied. If
the mass is constant, the sum of the forces is equal to the
product of the mass of the particle and its acceleration.
F = ma

5. Newton's Laws
Third law: The forces exerted by two particles on each other
are equal in magnitude and opposite in direction.
F2 = −F1
4th Law : Newton’s Gravitational Attraction
This law governs the gravitational attraction between
any two particles/bodies.

6. Newton's 2nd Law of Motion
The force and acceleration are directly
proportional, the constant of proportionality can be
determined in the ratio m = F1 / a1,
on the other hand if another force of different magnitude
say, (F2) is acted on an object, it will create another
acceleration say, a2 such that m = F2 / a2…
Thus, m serves as the constant of proportionality.
Equation of Motion
F = ma

7. In the case that 2 or more forces acting on a particle, the
resultant force is determined by a vector summation of all
the forces.
FR = F
Generally, the equation of motion is written as
F = ma
Static Equilibrium:
F = 0 FR = F = 0 thus , a = 0
Dynamic Equilibrium:
FR = F = ma F – ma = 0

8. Equations of Motion: Rectangular Coordinates
In a x,y & z frame of reference,
the forces acting on a particle can be expressed in
terms of i, j, k components, so we have
F = ma
Fx i + Fy j + Fz k = m ( ax i + ay j + az k )
To satisfy the given condition with its respective
i, j, k components:
Fx = max Fy = may Fz = maz

9. F R I CT I O N :
If the particle contacts on rough surface,
it is necessary to use the frictional equation, which relates
the coefficient of Kinetic Friction µk to the magnitude of
the Frictional Force (Ff) and Normal Force (N) acting on
the surface of contact,
Ff = µk N
Where: Ff - frictional force acting opposite
the subjected Force
µk - coefficient of kinetic friction
N - normal force
- force acting perpendicular to the point of
contact.

10. The 50 kg crate rests on a horizontal plane for which the
coefficient of kinetic friction is µk = 0.3 . If the crate is
subjected to a 400-N towing force 30⁰ from the horizontal,
determine the velocity of the crate in 5 sec starting from
rest.
EXAMPLE # 1

11. The 100 kg block A is released from rest. If the
mass of the pulleys and the cord are neglected,
determine the speed of the 20 kg block B in 2
seconds.
EXAMPLE # 2

12. At a given instant the 10 kg block A is moving downward
with a speed of 6 m/s. determine its speed 2 sec later.
Block B has a weight of 4 kg, and the coefficient of kinetic
friction between it and the horizontal plane is µk = 0.2 .
Neglect the mass of the pulleys and cord.
EXAMPLE # 3

13. Determine the time needed to pull the cord at B down 4 m
starting from rest when a force of 10 kg is applied to the
cord. Block A weighs 20 kg. Neglect the mass of the
pulleys and cords.
EXAMPLE # 4