CHAPTER-13
KINETICS OF A PARTICLE:
Force & Acceleration
Book:
▪ Engineering Mechanics Dynamics, R. C. Hibbeler, 14th Ed
KINETICS OF A PARTICLE: FORCE & ACCELERATION
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Newton’s Second Law of Motion
•
Equation of Motion
•
Equation of Motion: Rectangular Components
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Equation of Motion: Normal & Tangential Components
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Equation of Motion: Cylindrical Components
NEWTON’S LAWS OF MOTION
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.
NEWTON’S LAWS OF MOTION
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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
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This equation is referred to as the equation of motion
Newton’s Law of Gravitational Attraction:
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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
KINETICS OF A PARTICLE: FORCE & ACCELERATION
•
Newton’s Second Law of Motion
•
Equation of Motion
•
Equation of Motion: Rectangular Components
•
Equation of Motion: Normal & Tangential Components
•
Equation of Motion: Cylindrical Components
EQUATION OF MOTION
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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
EQUATION OF MOTION FOR A SYSTEM OF PARTICLES
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The equation of motion for a system of particles can be written as:
ΣF = maG
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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.
KINETICS OF A PARTICLE: FORCE & ACCELERATION
•
Newton’s Second Law of Motion
•
Equation of Motion
•
Equation of Motion: Rectangular Components
•
Equation of Motion: Normal & Tangential Components
•
Equation of Motion: Cylindrical Components
EQUATION OF MOTION: Rectangular Coordinates
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When a particle is moving relative to an inertial x, y, z frame of reference, the forces
acting on the particle as well as its acceleration may be expressed in terms of their i, j,
and k components as:
∑F=ma
∑ Fx i + ∑ Fy j + ∑ Fz k = m( ax i + ay j + az k)
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For this equation to be satisfied, the respective i, j and k components on the left side
must be equal to the corresponding components on the right side.
EQUATION OF MOTION: Rectangular Coordinates
Examples:
13.1, 13.2, 13.3, 13.4,13.5
Fundamental Problems:
F13.1, F13.5
Practice Problems:
13.3, 13.6, 13.16, 13.18, 13.40
EXAMPLE 13-1
The 50kg crate shown rests on a horizontal plane for which the coefficient of kinetic
friction is 0.3 If the crate is subjected to a 400N towing force as shown, determine the
velocity of the crate in 3s starting from rest.
EXAMPLE 13-4
A smooth 2 kg collar shown is attached to a spring having a stiffness k = 3N/m and an
unstretched length of 0.75 m. If the collar is released from rest at A, determine its
acceleration and the normal force of the rod on the collar at the instant y = 1m.
PROBLEM 13-3
If the coefficient of kinetic friction between the 50-kg crate and the ground is µk = 0.3,
determine the distance the crate travels and its velocity when t=3 s. The crate starts from
rest, and P = 200 N.
PROBLEM 13-18
A 40-lb suitcase slides from rest 20 ft down the smooth ramp. Determine the point where it
strikes the ground at C. How long does it take to go from A to C?
KINETICS OF A PARTICLE: FORCE & ACCELERATION
•
Newton’s Second Law of Motion
•
Equation of Motion
•
Equation of Motion: Rectangular Components
•
Equation of Motion: Normal & Tangential Components
•
Equation of Motion: Cylindrical Components
EQUATION OF MOTION: Normal & Tangential Coordinates
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When a particle moves over a curved path which is known, the equation of motion for the
particle may be written in the tangential, normal and binormal directions. i.e.
∑F=ma
∑ Ft ut + ∑ Fn un + ∑ Fb ub = mat + man
Here ∑ Fn , ∑ Ft and ∑ Fb represent the sums of all the force components acting on the
particle in the normal, tangential and binormal directions respectively.
Since there is no motion of the particle in the binormal direction as the particle is
constrained to move along the path, so the above equation may be written as:
∑ Ft = m at
∑ Fn = m an
∑ Fb = 0
EQUATION OF MOTION: Normal & Tangential Coordinates
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Since at = dv/dt represents the time rate of change in magnitude of velocity, so if ∑Ft
acts in the direction of motion, the particle’s speed will increase, whereas, if it acts in
opposite direction, the particle will slow down.
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Similarly an = v2/ρ represents the time rate of change the velocity’s direction. Since this
vector always acts in the positive n direction i.e. towards the path’s center of curvature,
then ∑Fn, which causes an, also acts in this direction
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In particular, when the particle is constrained to travel in a circular path with constant
speed, there is a normal force exerted on the particle by the constraint. As this force is
always directed towards the center of the path, it is often referred to as centripetal force.
EQUATION OF MOTION: Normal & Tangential Coordinates
Examples:
13.6, 13.7, 13.8, 13.9
Fundamental Problems:
F13.7, F13.10
Practice Problems:
13.57, 13.60, 13.63, 13.71, 13.80
EXAMPLE 13-7
The 3kg disk is attached to the end of the cord as shown. The other end of the cord is attached
to a ball-and-socket joint attached at the center of the platform. If the platform is rotating
rapidly, and the disk is placed on it and released from rest as shown, determine the time it
takes for the disk to reach a speed great enough to break the cord. The maximum tension the
cord can sustain is 100N, and the coefficient of kinetic friction between the disk and the
platform is 0.1
EXAMPLE 13-9
The 60-kg skateboarder in Fig below coasts down the smooth circular track. If he starts from
rest when θ = 0°, determine the magnitude of the normal reaction the track exerts on him when
θ = 60°. Neglect his size for the calculation.
PROBLEM 13-60
At the instant θ = 60°, the boy’s center of mass G has a downward speed vG = 15 ft/s.
Determine the rate of increase in his speed and the tension in each of the two supporting
cords of the swing at this instant. The boy has a weight of 60 lb. Neglect his size and the
mass of the seat and cords.
PROBLEM 13-63
The pendulum bob B has a weight of 5 lb and is released from rest in the position shown, θ =
0°. Determine the tension in string BC just after the bob is released θ = 0°, and also at the
instant the bob reaches point D, θ =45°. Take r = 3 ft.
KINETICS OF A PARTICLE: FORCE & ACCELERATION
•
Newton’s Second Law of Motion
•
Equation of Motion
•
Equation of Motion: Rectangular Components
•
Equation of Motion: Normal & Tangential Components
•
Equation of Motion: Cylindrical Components
EQUATION OF MOTION: Cylindrical Coordinates
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When all the forces acting on a particle are resolved into cylindrical components i.e. along
the unit vector directions ur, uθ, uz the equation of motion may be expressed as:
∑F=ma
∑ Frur+ ∑ Fθuθ+ ∑ Fzuz = m ar ur + m aθ uθ + m az uz
Consequently, we may write the following three scalar equations of motion:
∑ Fr = m a r
∑ Fθ = m a θ
∑ Fz = m a z
If the particle is constrained to move only in the r-θ plane, then only the first two of the
above equations are used to specify the motion.
EQUATION OF MOTION: Cylindrical Coordinates
TANGENTIAL & NORMAL FORCES
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Problems involving cylindrical coordinates requires the determination of the resultant force
components ∑Fr, ∑Fθ, ∑Fz causing a particle to move with a known acceleration
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If however, the particle’s accelerated motion is not completely specified at the given instant, then
some information regarding the directions or magnitudes of the forces acting on the particle must
be known or computed in order to solve the equations of motions.
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Consider a force P causing the particle to move along a path defined by polar coordinates as
r=f(θ)
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The normal force N which the path exerts on the particle is always perpendicular to the tangent of
the path whereas the frictional force F always acts along the tangent in the opposite direction of
motion.
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The direction of N and F can be specified relative to the radial coordinate by computing the angle
Ψ (psi) which is defined between the extended radial line r=OP and the tangent to the curve.
EQUATION OF MOTION: Cylindrical Coordinates
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This angle can be obtained by noting that when the particle is displaced ds along the path,
the component of displacement in the radial direction is dr and the component of
displacement in the transverse direction is r dθ.
Since these components are mutually perpendicular, the angle Ψ can be determined from:
tan Ψ = r dθ / dr
or
tan Ψ = r / (dr/dθ)
If Ψ is calculated as a positive quantity, it is measured from the extended radial line to the
tangent in a counter clockwise sense or in the positive direction of θ.
If it is negative, it is measured in the opposite direction to positive θ.
EQUATION OF MOTION: Cylindrical Coordinates
Examples:
13.10, 13.11, 13.12
Fundamental Problems:
F13.14, F13.15
Practice Problems:
13.88, 13.91, 13.92, 13.107, 13.110
EXAMPLE 13-10
The smooth 0.5-kg double-collar in Fig. can freely slide on arm AB and the circular guide rod.
If the arm rotates with a constant angular velocity of 3 rad/s, determine the force the arm
exerts on the collar at the instant θ = 45°. Motion is in the horizontal plane.
EXAMPLE 13-12
A can C, having a mass of 0.5 kg, moves along a grooved horizontal slot as shown. The slot is
in the form of a spiral, which is defined by the equation r=(0.1θ) m, where θ is in radians. If
the arm OA is rotating at a constant rate of 4 rad/s in the horizontal plane, determine the force
it exerts on the can at the instant θ=π radians. Neglect friction and the size of the can.
PROBLEM 13-107
The forked rod is used to move the smooth 2-lb particle around the horizontal path in the
shape of a limaçon, r = (2 + cos θ) ft. If θ = (0.5t2) rad, where t is in seconds, determine the
force which the rod exerts on the particle at the instant t = 1 s. The fork and path contact the
particle on only one side.
PROBLEM 13-110
Rod OA rotates counterclockwise at a constant angular rate of 4 rad/s. The double collar B is
pin-connected together such that one collar slides over the rotating rod and the other collar
slides over the circular rod described by the equation r = (1.6 cos θ) m. If both collars have a
mass of 0.5 kg, determine the force which the circular rod exerts on one of the collars and the
force that OA exerts on the other collar at the instant θ = 45°. Motion is in the vertical plane.