TMS
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TMS
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ENGINEERING MECHANICS
DYNAMICS
6
6
Mulyadi Bur
LDS
Unand
Structural Dynamics Laboratory
ANDALAS UNIVERSITY
TMS
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6
/1/. Beer, F.P.; Johnston, E.R.
Mechanics for Engineer: Dynamics, 5th Ed., McGraw-Hill, New York, 2008
/2/. Meriam, J.L.; Kraige, L.G.
Engineering Mechanics: Dynamics, 6th. Ed., John Wiley, 2008.
/3/. Hibbeler, R.C.
Engineering Mechanics: Dynamics, 12 Ed., Prentice Hall, New Jersey,
2010.
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Contents
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Introduction
• Previously, problems dealing with the motion of particles
r were
r
solved through the fundamental equation of motion, F = ma.
Current chapter introduces two additional methods of analysis.
Introduction
Work of a Force
Principle of Work & Energy
Applications of the Principle of Work & Energy
Power and Efficiency
Sample Problem 13.1
Sample Problem 13.2
Sample Problem 13.3
6
• Method of work and energy: directly relates force, mass,
velocity and displacement.
• Method of impulse and momentum: directly relates force,
mass, velocity, and time.
LDS
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LDS
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seperti tercantum berikut ini:
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Unand
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Work of a Force
Work of a Force
r
• Differential vector dr is the particle
displacement.
• Work of a force during a finite displacement,
A2 r
r
U1→2 = ∫ F • dr
A1
• Work of the force is
r r
dU = F ⋅ dr
= F ds cos α
= Fx dx + Fy dy + Fz dz
6
6
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(
• Work is represented by the area under the
curve of Ft plotted against s.
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• Magnitude of the force exerted by a spring is
proportional to deflection,
F = kx
k = spring constant (N/m or lb/in.)
U1→2 = ( F cos α ) ∆x
• Work of the force of gravity,
• Work of the force exerted by spring,
dU = Fx dx + Fy dy + Fz dz
dU = − F dx = −kx dx
6
= −W dy
x2
y2
U1→2 = − ∫ kx dx = 12 kx12 − 12 kx22
U1→2 = − ∫ W dy
x1
y1
= −W ( y 2 − y1 ) = −W ∆y
• Work of the force exerted by spring is positive
when x2 < x1, i.e., when the spring is returning to
its undeformed position.
• Work of the weight is equal to product of
weight W and vertical displacement ∆y.
Unand
• Work of the force exerted by the spring is equal to
negative of area under curve of F plotted against x,
• Work of the weight is positive when ∆y < 0,
i.e., when the weight moves down.
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Work of a Force
• Work of a constant force in rectilinear motion,
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)
A1
Work of a Force
6
s1
= ∫ Fx dx + Fy dy + Fz dz
1ft ⋅ lb = 1.356 J
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s1
A2
• Dimensions of work are length × force.
Units are
1 J ( joule ) = (1 N )(1 m )
s2
= ∫ ( F cos α )ds = ∫ Ft ds
• Work is a scalar quantity, i.e., it has
magnitude and sign but not direction.
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s2
U1→2 = − 12 ( F1 + F2 ) ∆x
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Work of a Force
Forces which do not do work (ds = 0 or cos α = 0):
Work of a gravitational force (assume particle M
occupies fixed position O while particle m follows
path shown),
dU = − Fdr = −G
6
r2
U1→2 = − ∫ G
Mm
r1
r
2
Mm
r2
• reaction at frictionless pin supporting rotating body,
dr
dr = G
Work of a Force
6
Mm
Mm
−G
r2
r1
• reaction at frictionless surface when body in contact
moves along surface,
• reaction at a roller moving along its track, and
• weight of a body when its center of gravity moves
horizontally.
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Particle Kinetic Energy: Principle of Work &
Energy
• Consider
r a particle of mass m acted upon by
dv
force F
Ft = mat = m
dt
dv
dv ds
=m
= mv
ds
ds dt
F t ds = mv dv
• Integrating from A1 to A2 ,
6
s2
v2
s1
v1
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Applications of the Principle of Work and Energy
• Wish to determine velocity of pendulum bob
at A2. Consider work & kinetic energy.
r
• Force P acts normal to path and does no
work.
T1 + U1→2 = T2
6
0 + Wl =
2
T = 12 mv 2 = kinetic energy
r
• The work of the force F is equal to the change in
kinetic energy of the particle.
• Units of work and kinetic energy are the same:
2
 m
m
T = 12 mv 2 = kg  =  kg 2 m = N ⋅ m = J
s
 s 
• Velocity found without determining
expression for acceleration and integrating.
U1→2 = T2 − T1
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1W 2
v2
2 g
v2 = 2 gl
∫ Ft ds = m ∫ v dv = 12 mv2 − 12 mv1
2
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• All quantities are scalars and can be added
directly.
• Forces which do no work are eliminated from
the problem.
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Applications of the Principle of Work and Energy
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• Power = rate at which work is done.
r r
dU F • dr
=
=
dt
dt
r r
= F •v
• Principle of work and energy cannot
be applied to directly determine the
acceleration of the pendulum bob.
• Calculating the tension in the cord
requires supplementing the method
of work and energy with an
application of Newton’s second law.
6
6
W v22
g l
W 2 gl
P =W +
= 3W
g l
P −W =
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or 1 hp = 550
ft ⋅ lb
= 746 W
s
• η = efficiency
output work
=
input work
power output
=
power input
∑ Fn = m an
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• Dimensions of power are work/time or force*velocity.
Units for power are
J
m
1 W (watt) = 1 = 1 N ⋅
s
s
• As the bob passes through A2 ,
v2 = 2 gl
Power and Efficiency
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Sample Problem 13.1
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Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
SOLUTION:
 mi  5280 ft  h 
v1 =  60 

 = 88 ft s
h  mi  3600 s 

• Evaluate the change in
kinetic energy.
6
• Determine the distance
required for the work to equal
the kinetic energy change.
6
T1 = 12 mv12 = 12 (4000 32.2 )(88)2 = 481000 ft ⋅ lb
v2 = 0
T2 = 0
• Determine the distance required for the work to
equal the kinetic energy change.
An automobile weighing 4000 lb is
driven down a 5o incline at a speed of
60 mi/h when the brakes are applied
causing a constant total breaking force
of 1500 lb.
U1→2 = (− 1500 lb )x + (4000 lb )(sin 5°)x
= −(1151 lb )x
T1 + U1→2 = T2
Determine the distance traveled by the
automobile as it comes to a stop.
481000 ft ⋅ lb − (1151 lb )x = 0
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x = 418 ft
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Sample Problem 13.2
Sample Problem 13.2
SOLUTION:
• Apply the principle of work and energy separately
to blocks A and B.
SOLUTION:
• Apply the principle of work
and energy separately to
blocks A and B.
6
Two blocks are joined by an inextensible
cable as shown. If the system is released
from rest, determine the velocity of block
A after it has moved 2 m. Assume that the
coefficient of friction between block A and
the plane is µk = 0.25 and that the pulley is
weightless and frictionless.
• When the two relations are
combined, the work of the
cable forces cancel. Solve
for the velocity.
FA = µ k N A = µ k W A = 0.25(1962 N ) = 490 N
6
T1 + U1→2 = T2 :
0 + FC (2 m ) − FA (2 m ) = 12 m A v 2
FC (2 m ) − (490 N )(2 m ) = 12 (200 kg )v 2
(
T1 + U1→2 = T2 :
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− Fc (2 m ) + (2940 N )(2 m ) = 12 (300 kg )v 2
Unand
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Sample Problem 13.2
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Sample Problem 13.3
SOLUTION:
• When the two relations are combined, the work
of the cable forces cancel. Solve for the
velocity.
FC (2 m ) − (490 N )(2 m ) = 12 (200 kg )v 2
6
6
− Fc (2 m ) + (2940 N )(2 m ) = 12 (300 kg )v 2
(2940 N )(2 m ) − (490 N )(2 m ) = 12 (200 kg + 300 kg )v 2
4900 J = 12 (500 kg )v 2
v = 4.43 m s
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)
0 − Fc (2 m ) + WB (2 m ) = 12 mB v 2
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)
WB = (300 kg ) 9.81 m s 2 = 2940 N
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(
W A = (200 kg ) 9.81 m s 2 = 1962 N
• Apply the principle of work
and energy between the initial
position and the point at
which the spring is fully
compressed and the velocity is
zero. The only unknown in
the relation is the friction
coefficient.
A spring is used to stop a 60 kg package
which is sliding on a horizontal surface.
The spring has a constant k = 20 kN/m
and is held by cables so that it is initially
compressed 120 mm. The package has
a velocity of 2.5 m/s in the position shown
and the maximum deflection of the spring • Apply the principle of work
and energy for the rebound of
is 40 mm.
Determine (a) the coefficient of kinetic
friction between the package and surface
and (b) the velocity of the package as it
passes again through the position shown.
the package. The only
unknown in the relation is the
velocity at the final position.
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Sample Problem 13.3
Sample Problem 13.3
SOLUTION:
• Apply principle of work and energy between initial
position and the point at which spring is fully
compressed.
T1 = 12 mv12 = 12 (60 kg )(2.5 m s )2 = 187.5 J
(U1→2 ) f = − µ kW x
6
(
= − µ k (60 kg ) 9.81m s
2
• Apply the principle of work and energy for the
rebound of the package.
T2 = 0
T2 = 0
)(0.640 m) = −(377 J )µk
T 3= 12 mv32 = 12 (60kg )v32
U 2→3 = (U 2→3 ) f + (U 2→3 )e = −(377 J )µ k + 112 J
= +36.5 J
6
T2 + U 2→3 = T3 :
Pmin = kx0 = (20 kN m )(0.120 m ) = 2400 N
0 + 36.5 J = 12 (60 kg )v32
Pmax = k ( x0 + ∆x ) = (20 kN m )(0.160 m ) = 3200 N
v3 = 1.103 m s
(U1→2 )e = − 12 (Pmin + Pmax )∆x
= − 12 (2400 N + 3200 N )(0.040 m ) = −112.0 J
U1→2 = (U1→2 ) f + (U1→2 )e = −(377 J )µ k − 112 J
T1 + U1→ 2 = T2 :
187.5 J - (377 J )µ k − 112 J = 0
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µ k = 0.20
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Unand
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