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7/23/2019 Vector Mechanics for Engineers: Dynamics - Chapter 12 notes
http://slidepdf.com/reader/full/vector-mechanics-for-engineers-dynamics-chapter-12-notes 1/23
Chapter
12
Vector M echanics for Engineers: Dynamics
F. Beer & Al
RVCC
Kinetics of Particles:
Newton’s Second Law
ENGR 133-51 - Engineering Mechanics II : Dynamics
7/23/2019 Vector Mechanics for Engineers: Dynamics - Chapter 12 notes
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12- 2
Contents
• Newton’s Second Law of
Motion• Linear Momentum of a
Particle
• System of Units
• Equations of Motion
• Dynamic Equilibrium
• Angular Momentum of a Particle
• Equation of Motion in Radial & TransverseComponents
• Conservation of Angular Momentum
• Newton’s Law of Gravitation
7/23/2019 Vector Mechanics for Engineers: Dynamics - Chapter 12 notes
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12- 3
Introduction
Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or continue to move in a straight line
(Statics Ch. 2)
Newton’s Second Law: If the resultant force acting on a particle is
not zero, then the particle will have an acceleration such that
am F
(Dynamics)
Recalling Newton’s Laws:
Newton’s Third Law: The forces of action and reaction between two
particles have the same magnitude and line of action with oppositesense (Statics Ch. 6)
M m
F -F
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12- 4
Newton’s Second Law of Motion
• Newton’s Second Law: If the resul tant force acting on a
particle is not zero, the particle will have an acceleration
proportional to the magni tude of resul tant and in the
direction of the resul tant .
• Consider a particle subjected to constant forces,
m
a
F
a
F
a
F mass,constant
3
3
2
2
1
1
• When a particle of mass m is acted upon by a force
the acceleration of the particle must satisfy
, F
am F
• Acceleration must be evaluated with respect to a Newtonian frame of reference, i.e., one that is not
accelerating or rotating.
• If force acting on particle is zero, particle will not
accelerate (a =0), i.e., it will remain stationary or continue
on a straight line at constant velocity.
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Problem 1
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Systems of Units
• Of the units for the four primary dimensions (force,
mass, length, and time), three may be chosen arbitrarily.
The fourth must be compatible with Newton’s 2nd Law.
• 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,
22 s
mkg1
s
m1kg1 N1
• U.S. Customary Units: base units are the units of force(lb), length (m), and time (second). The unit of mass is
derived,
ft
slb1
sft1
lb1slug1
sft32.2
lb1lbm1
2
22
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Systems of Units
kg0.4536mass- pound1
kg14.59ft
slb1
slug1 Mass
N4.448lb1 Force
m0.3048ft1 Length
2
US CS → SI
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Equations of Motion
• Newton’s second law provides
am F
• Solution for particle motion is facilitated by resolving
vector equation into scalar component equations, e.g.,
for rectangular components,
z m F ym F xm F
ma F ma F ma F
k a jaiamk F j F i F
z y x
z z y y x x
z y x z y x
FBD
(Forces)
Kinetic Diagram
(motion)
7/23/2019 Vector Mechanics for Engineers: Dynamics - Chapter 12 notes
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Sample Problem 12.1
A 200-lb block rests on a horizontal
plane. Find the magnitude of the force
P required to give the block an accelera-
tion or 10 ft/s2
to the right. The coef-ficient of kinetic friction between the
block and plane is m k 0.25.
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Problem 3.a
12- 10
A 180-lb person stands in an elevator.
1) What is the apparent weight of the person while the elevator accelerate
upward with a constant value of 8 ft/s2? (FBD is the person)
2) What is the apparent weight if the elevator is accelerating downward with thesame value?
3) What is the tension in the rope of the elevator when it is moving downward?
(FBD is the person and the elevator)
4) For what value of acceleration of the elevator the person would appear
weightless? (FBD is the person)
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Problem 3.b
Hints:
1) Deceleration is not constant!
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Problem 4
Hints:
1) determine if A slips studying
static equilibrium on A (is the
force of friction necessary for
equilibrium exceeding themaximum frictional force?).
Only then you can consider if
Newton’s second law applies.
2) FBD for A with its own
coordinate system3) If finding that A slips, draw FBD
for B with its own coordinates
and apply also to B the dynamic
equilibrium equation
The two crates are released from rest. Their masses are mA = 40 kg and mB 30 kg, and the
coefficient of friction between crate A and the inclined surface are μs = 0.2 and μk = 0.15.
Are the two crates in static equilibrium? If not, what is the acceleration of the crates?
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Sample Problem 12.4
The 12-lb block B starts from rest and
slides on the 30-lb wedge A, which is
supported by a horizontal surface.
Neglecting friction, determine (a) the
acceleration of the wedge, and (b) the
acceleration of the block relative to the
wedge.
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Dynamic Equilibrium
• Alternate expression of Newton’s second law,
ector inertial vamam F
0
• With the inclusion of the inertial vector, the system
of forces acting on the particle is equivalent to
zero. The particle is in dynamic equil ibr ium .
• Methods developed for particles in static
equilibrium may be applied, e.g., coplanar forces
may be represented with a closed vector polygon.
• Inertia vectors are often called inertial forces as
they measure the resistance that particles offer to
changes in motion, i.e., changes in speed or
direction.
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@Hibbeler
Acceleration
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@Hibbeler
Deceleration
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Equations of Motion
• Newton’s second law provides
am F
• For tangential and normal components,
2v
m F dt
dvm F
ma F ma F
nt
nnt t
Recalling that tangential component of
acceleration reflects change of speed and
normal component reflects change of direction.
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Sample Problem 12.5
The bob of a 2-m pendulum describes
an arc of a circle in a vertical plane. If
the tension in the cord is 2.5 times the
weight of the bob for the positionshown, find the velocity and accel-
eration of the bob in that position.
A i l i ACB f 2 h h i C h i h d
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Problem 7
Hints:- What is the equation you are going to use to find v…?
- To get there you will need to find ρ as a function of L
and angles (L = total length of rope), then
- Draw the FBDs. To optimize your equations, is
convenient to choose x , y with x coinciding with n
- Apply Newton’s Law
A single wire ACB of 2 m passes through a ring at C that is attached to
a sphere. The sphere revolves at a constant speed v in the horizontal
circle shown. Knowing that ϴ1=60o and ϴ2=30o and that the tension is
the same in both portions of the wire, determine the speed v .
A i il i ’ li i d i f f ff i
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Problem 8A civil engineer’s preliminary design for a freeway off -ramp is
circular with radius R= 60m. If she assumes that the coefficient of
static friction between tires and road is at least μs=0.4, what is the
maximum speed at which vehicles can enter the ramp without loosing
traction?
Hints:
• The reason why we can turn following a curved path, is because of friction force
between the road and the tires (we cannot turn on a sheet of ice).
• This necessary force of friction is related to the normal acceleration of a circular motion
• Evaluating the maximum frictional force related to the static coefficient allows to
calculate the maximum velocity without slipping.
• Draw FBDs considering front view of the car
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Sample Problem 12.6
Determine the rated speed of a
highway curve of radius = 400 ft
banked through an angle q = 18o.
[The rated speed of a banked highway
curve is the speed at which a car
should travel if no lateral friction force
is to be exerted at its wheels]
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Problem 9
Hints:
1. The rated speed of a banked highway
curve is the speed at which a car should
travel if no lateral friction force is to be
exerted at its wheels.2. Consider that in a banked curve, the
normal acceleration is always horizontal.
3. Draw the FBD and analyze dynamic
equilibrium in a general case
(considering a frictional force that in (a)
is zero) with x // to the road.4. Specify each of the cases.
FBDF FBDK
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Homework:
Solving Problems on Your Ownpage ~704