Ninth Edition
CHAPTER
12
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Kinetics of Particles:
Newton’s Second Law
Texas Tech University
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Ninth
Edition
Vector Mechanics for Engineers: Dynamics
Introduction
• Newton’s first and third laws are sufficient for the study of bodies
at rest (statics) or bodies in motion with no acceleration.
• When a body accelerates (changes in velocity magnitude or
direction), Newton’s second law is required to relate the motion of the
body to the forces acting on it.
• Newton’s second law:
- A particle will have an acceleration proportional to the magnitude
of the resultant force acting on it and in the direction of the resultant
force.
- The resultant of the forces acting on a particle is equal to the rate of
change of linear momentum of the particle.
- The sum of the moments about O of the forces acting on a particle is
equal to the rate of change of angular momentum of the particle
about O.
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Vector Mechanics for Engineers: Dynamics
Newton’s Second Law of Motion
• Newton’s Second Law: 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.
• Consider a particle subjected to constant forces,
F1 F2 F3
=
=
= = constant = mass, m
a1 a2 a3
• When a particle of mass m is acted upon by a force F ,
the acceleration of the particle must satisfy
F = ma
• 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, i.e., it will remain stationary or continue on a
straight line at constant velocity.
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Vector Mechanics for Engineers: Dynamics
Linear Momentum of a Particle
• Replacing the acceleration by the derivative of the
velocity yields
dv
F = m
dt
d
dL
= (m v ) =
dt
dt
L = linear momentum of the particle
• Linear Momentum Conservation Principle:
If the resultant force on a particle is zero, the linear
momentum of the particle remains constant in both
magnitude and direction.
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Vector Mechanics for Engineers: Dynamics
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,
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 (m), and time (second). The unit of mass is
derived,
1 lb
1 lb
lb s 2
1 lbm =
1slug =
=1
2
2
ft
32.2 ft s
1 ft s
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Vector Mechanics for Engineers: Dynamics
Equations of Motion
• Newton’s second law provides
F
=
m
a
• Solution for particle motion is facilitated by resolving
vector equation into scalar component equations, e.g.,
for rectangular components,
(Fx i + Fy j + Fz k ) = m(a x i + a y j + a z k )
Fx = ma x Fy = ma y Fz = ma z
Fx = mx Fy = my Fz = mz
• For tangential and normal components,
F t = mat
F n = man
dv
F
=
m
t
dt
v2
Fn = m r
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Vector Mechanics for Engineers: Dynamics
Rectangular Components
Rectangular Components
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Vector Mechanics for Engineers: Dynamics
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 acceleration of 10 ft/s2 to the right. The
coefficient of kinetic friction between the block and plane is k = 0.25.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.1
SOLUTION:
• Resolve the equation of motion for the block
into two rectangular component equations.
Fx = ma :
(
y
O
)(
P cos 30 − 0.25 N = 6.21 lb s 2 ft 10 ft s 2
)
= 62.1 lb
x
W
200 lb
m= =
g 32.2 ft s 2
lb s 2
= 6.21
ft
F = k N
= 0.25 N
Fy = 0 :
N − P sin 30 − 200 lb = 0
• Unknowns consist of the applied force P and
the normal reaction N from the plane. The two
equations may be solved for these unknowns.
N = P sin 30 + 200 lb
P cos 30 − 0.25( P sin 30 + 200 lb ) = 62.1 lb
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P = 151 lb
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.3
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.
The kinematic relationships for the dependent
motions and accelerations of the blocks are:
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y B = 12 x A
a B = 12 a A
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.3
O
x
• Write equations of motion for blocks and pulley.
y
Fx = m A a A :
T1 = (100 kg )a A
Fy = m B a B :
m B g − T2 = m B a B
(300 kg )(9.81 m s 2 )− T2 = (300 kg )a B
T2 = 2940 N - (300 kg )a B
Fy = mC aC = 0 :
T2 − 2T1 = 0
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.3
• Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
O
x
y
y B = 12 x A
a B = 12 a A
T1 = (100 kg )a A
T2 = 2940 N - (300 kg )a B
(
= 2940 N - (300 kg ) 12 a A
)
T2 − 2T1 = 0
2940 N − (150 kg )a A − 2(100 kg )a A = 0
a A = 8.40 m s 2
a B = 12 a A = 4.20 m s 2
T1 = (100 kg )a A = 840 N
T2 = 2T1 = 1680 N
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Vector Mechanics for Engineers: Dynamics
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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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.4
SOLUTION:
• The block is constrained to slide down the
wedge. Therefore, their motions are dependent.
aB = a A + aB A
• Write equations of motion for wedge and block.
Fx = m A a A :
N1 sin 30 = m A a A
y
0.5 N1 = (W A g )a A
x
Fx = mB a x = mB (a A cos 30 − a B A ) :
− WB sin 30 = (WB g )(a A cos 30 − a B A )
a B A = a A cos 30 + g sin 30
Fy = mB a y = mB (− a A sin 30) :
N1 − WB cos 30 = −(WB g )a A sin 30
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.4
• Solve for the accelerations.
0.5 N1 = (W A g )a A
N1 − WB cos 30 = −(WB g )a A sin 30
2(W A g )a A − WB cos 30 = −(WB g )a A sin 30
aA =
gWB cos 30
2W A + WB sin 30
(
32.2 ft s 2 )(12 lb ) cos 30
aA =
2(30 lb ) + (12 lb ) sin 30
a A = 5.07 ft s 2
a B A = a A cos 30 + g sin 30
(
)
(
)
a B A = 5.07 ft s 2 cos 30 + 32.2 ft s 2 sin 30
a B A = 20.5 ft s 2
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