Download Systems of Particles

Tenth Edition
CHAPTER
14
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
Systems
Systemsof
ofParticles
Particles
California Polytechnic State University
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Application of Newton’s Laws:
Effective Forces
Linear and Angular Momentum
Motion of Mass Center of System
of Particles
Angular Momentum About Mass
Center
Conservation of Momentum
Sample Problem 14.2
Kinetic Energy
Work-Energy Principle.
Conservation of Energy
Principle of Impulse and Momentum
Sample Problem 14.4
Sample Problem 14.5
Variable Systems of Particles
Steady Stream of Particles
Steady Stream of Particles.
Applications
Streams Gaining or Losing Mass
Sample Problem 14.6
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14 - 2
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Vector Mechanics for Engineers: Dynamics
Engineers often need to analyze the dynamics of systems of
particles – this is the basis for many fluid dynamics
applications, and will also help establish the principles used
in analyzing rigid bodies
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2-3
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Vector Mechanics for Engineers: Dynamics
Introduction
• In the current chapter, you will study the motion of systems
of particles.
• The effective force of a particle is defined as the product of
it mass and acceleration. It will be shown that the system of
external forces acting on a system of particles is equipollent
with the system of effective forces of the system.
• The mass center of a system of particles will be defined
and its motion described.
• Application of the work-energy principle and the
impulse-momentum principle to a system of particles will
be described. Result obtained are also applicable to a
system of rigidly connected particles, i.e., a rigid body.
• Analysis methods will be presented for variable systems
of particles, i.e., systems in which the particles included
in the system change.
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Vector Mechanics for Engineers: Dynamics
Application of Newton’s Laws. Effective Forces
• Newton’s second law for each particle Pi
in a system of n particles,

Fi 
n

j 1
 
ri  Fi 


fij  mi ai
 


ri  fij  ri  mi ai

n
j 1



Fi  external force
fij  internal forces

mi ai  effective force
• The system of external and internal forces on
a particle is equivalent to the effective force
of the particle.
• The system of external and internal forces
acting on the entire system of particles is
equivalent to the system of effective forces.
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Vector Mechanics for Engineers: Dynamics
Application of Newton’s Laws. Effective Forces
• Summing over all the elements,
n 
n n 
n

F

f

m
a
 i   ij  i i
i 1
i 1 j 1
i 1
n n 
n 

 




r

F

r

f

r

m
a
 i i   i ij  i i i 
n
i 1

i 1 j 1

i 1
• Since the internal forces occur in equal
and opposite collinear pairs, the resultant
force and couple due to the internal
forces are zero,


 Fi   mi ai
 





r

F

r

m
a
 i i  i i i
• The system of external forces and the
system of effective forces are
equipollent but not equivalent.
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Vector Mechanics for Engineers: Dynamics
Linear & Angular Momentum
• Linear momentum of the system of
particles,
 n 
L   mi vi

L
i 1
n
i 1
n
n 




H O   ri  mi vi    ri  mi vi 
n


 mi vi   mi ai
i 1
i 1
n
i 1
• Resultant of the external forces is
equal to rate of change of linear
momentum of the system of
particles,
 
FL

• Angular momentum about fixed point O
of system of particles,
n 


H O   ri  mi vi 
i 1


  ri  mi ai 
i 1
• Moment resultant about fixed point O of
the external forces is equal to the rate of
change of angular momentum of the
system of particles,


M H

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O
O
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Vector Mechanics for Engineers: Dynamics
Motion of the Mass Center of a System of Particles
• Mass center G of system of particles is defined

by position vector rG which satisfies
n


mrG   mi ri
i 1
• Differentiating twice,
n


mrG   mi ri
i 1
n



mvG   mi vi  L
i 1



maG  L   F
• The mass center moves as if the entire mass and
all of the external forces were concentrated at
that point.
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Vector Mechanics for Engineers: Dynamics
Angular Momentum About the Mass Center
• The angular momentum of the system of
particles about the mass center,
n 


   ri  mi vi 
HG
i 1
n
n 



 
   ri  mi ai    ri  mi ai  aG 
HG
 

ai  aG  ai
i 1
 n
i 1
i 1


 
  ri  mi ai     mi r    aG
 i 1

i 1
n 
n 


  ri  mi ai    ri  Fi 
• Consider the centroidal frame
of reference Gx’y’z’, which
translates with respect to the
Newtonian frame Oxyz.
• The centroidal frame is not,
in general, a Newtonian
frame.
i 1
n

  MG
• The moment resultant about G of the external
forces is equal to the rate of change of angular
momentum about G of the system of particles.
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Vector Mechanics for Engineers: Dynamics
Angular Momentum About the Mass Center
• Angular momentum about G of particles in
their absolute motion relative to the
Newtonian Oxyz frame of reference.
n 


H G   ri mi vi 
i 1
n
 

vi  vG  vG
• Angular momentum about G of
the particles in their motion
relative to the centroidal Gx’y’z’
frame of reference,
n 


H G   ri mi vi 
i 1



  ri mi vG  vi 
i 1
 n
n 
 

   mi ri  vG   ri mi vi 
 i 1

i 1



   MG
HG  HG
• Angular momentum about G of the particle
momenta can be calculated with respect to
either the Newtonian or centroidal frames of
reference.
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Vector Mechanics for Engineers: Dynamics
Conservation of Momentum
• If no external forces act on the
particles of a system, then the linear
momentum and angular momentum
about the fixed point O are
conserved.




L  F  0
HO   M O  0


L  constant
H O  constant
• Concept of conservation of momentum
also applies to the analysis of the mass
center motion,




HG   M G  0
L  F  0


L  mvG  constant


vG  constant
H G  constant
• In some applications, such as
problems involving central forces,




L  F  0
HO   M O  0


L  constant
H O  constant
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Vector Mechanics for Engineers: Dynamics
Concept Question
Three small identical spheres A, B, and C,
which can slide on a horizontal, frictionless
surface, are attached to three 200-mm-long
strings, which are tied to a ring G. Initially, each
of the spheres rotate clockwise about the ring
with a relative velocity of vrel.
Which of the following is true?
a)
b)
c)
d)
vrel
vrel
x
vrel
The linear momentum of the system is in the positive x direction
The angular momentum of the system is in the positive y direction
The angular momentum of the system about G is zero
The linear momentum of the system is zero
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Vector Mechanics for Engineers: Dynamics
Sample Problem 14.2
SOLUTION:
• Since there are no external forces, the
linear momentum of the system is
conserved.
• Write separate component equations
for the conservation of linear
momentum.
A 10-kg projectile is moving with a
velocity of 30 m/s when it explodes into
2.5 and 7.5-kg fragments. Immediately
after the explosion, the fragments travel
in the directions qA = 45o and qB = 30o.
• Solve the equations simultaneously
for the fragment velocities.
Determine the velocity of each fragment.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 14.2
SOLUTION:
• Write separate component equations for
the conservation of linear momentum.
• Since there are no external forces, the
linear momentum of the system is
conserved.
m A v A + mB vB = mv0
(2.5) vA + (7.5) vB = (10 ) v0
x components:
2.5 vA cos45° + 7.5 vB cos30° = 10 (30 )
y components:
y
2.5 vA sin 45° - 7.5 vB sin30° = 0
x
• Solve the equations simultaneously for the
fragment velocities.
vA = 62.2m s a 45
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vB = 29.3m s c 30
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
vA
v0
vC
vB
In a game of pool, ball A is moving with a
velocity v0 when it strikes balls B and C,
which are at rest and aligned as shown.
Knowing that after the collision the three
balls move in the directions indicated and
that v0 = 4 m/s and vC= 2 m/s, determine
the magnitude of the velocity of (a) ball A,
(b) ball B.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
SOLUTION:
• Since there are no external forces, the
linear momentum of the system is
conserved.
• Write separate component equations
for the conservation of linear
momentum.
• Solve the equations simultaneously
for the pool ball velocities.
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Write separate component equations for the conservation of
linear momentum
x:
y:
m(4)cos 30  mvA sin 7.4  mvB sin 49.3  m(2)cos 45
0.12880vA  0.75813vB  2.0499
(1)
m(4)sin 30  mvA cos 7.4 mvB cos 49.3  m(2)sin 45
0.99167vA 0.65210vB  0.5858
(2)
Two equations, two unknowns - solve
0.65210 (0.12880vA  0.75813vB  2.0499 )
+ 0.75813 ( 0.99167vA 0.65210vB  0.5858 )
0.83581 vA  1.7809
vA  2.13 m/s
Sub into (1) or (2) to get vB
vB  2.34 m/s
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Vector Mechanics for Engineers: Dynamics
Concept Question
In a game of pool, ball A is moving with a
velocity v0 when it strikes balls B and C,
which are at rest and aligned as shown.
After the impact, what is true
about the overall center of mass
of the system of three balls?
vA
v0
vC
vB
a) The overall system CG will move in the same direction as v0
b) The overall system CG will stay at a single, constant point
c) There is not enough information to determine the CG location
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14 - 17
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Vector Mechanics for Engineers: Dynamics
Kinetic Energy
• Kinetic energy of a system of particles,
T
n

1

m
v
2 i i
i 1
n

1
 vi   2  mi vi2
i 1
• Expressing the velocity in terms of the
centroidal reference frame,
T

 

vi  vG  vi

n
1
2
i 1
mi vG  vi  vG  vi
n
n
 1 n
1  m v 2  v 
   mi vi2
m
v


i
G
G
i
i


2
2
 i 1 
i 1
i 1
n
1 mv 2  1
2
m
v

G
i
i
2
2
i 1
• Kinetic energy is equal to kinetic energy of
mass center plus kinetic energy relative to
the centroidal frame.
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14 - 18
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Vector Mechanics for Engineers: Dynamics
Work-Energy Principle. Conservation of Energy
• Principle of work and energy can be applied to each particle Pi ,
T1  U12  T2

where U12 represents the workdone by the internal forces f ij
and the resultant external force Fi acting on Pi .
• Principle of work and energy can be applied to the entire system by
adding the kinetic energies of all particles and considering the work
done by all external and internal forces.


• Although f ij and f ji are equal and opposite, the work of these
forces will not, in general, cancel out.
• If the forces acting on the particles are conservative, the work is
equal to the change in potential energy and
T1  V1  T2  V2
which expresses the principle of conservation of energy for the
system of particles.
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14 - 19
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Vector Mechanics for Engineers: Dynamics
Principle of Impulse and Momentum
 
F  L
t2



  Fdt  L2  L1
t1


 M O  HO
t2



  M O dt  H 2  H1
t1
t2 


L1    Fdt  L2
t1
t2 


H1    M O dt  H 2
t1
• The momenta of the particles at time t1 and the impulse of the forces
from t1 to t2 form a system of vectors equipollent to the system of
momenta of the particles at time t2 .
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14 - 20
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 14.4
SOLUTION:
Ball B, of mass mB,is suspended from a
cord, of length l, attached to cart A, of
mass mA, which can roll freely on a
frictionless horizontal tract. While the
cart is at rest, the ball is given an initial
velocity v0  2 gl .
• With no external horizontal forces, it
follows from the impulse-momentum
principle that the horizontal component
of momentum is conserved. This
relation can be solved for the velocity of
B at its maximum elevation.
• The conservation of energy principle
can be applied to relate the initial
kinetic energy to the maximum potential
energy. The maximum vertical distance
is determined from this relation.
Determine (a) the velocity of B as it
reaches it maximum elevation, and (b)
the maximum vertical distance h
through which B will rise.
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14 - 21
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Vector Mechanics for Engineers: Dynamics
Sample Problem 14.4
SOLUTION:
• With no external horizontal forces, it follows from the
impulse-momentum principle that the horizontal
component of momentum is conserved. This relation can
be solved for the velocity of B at its maximum elevation.
t2 


L1    Fdt  L2
t1
y
x
x component equation:
m Av A,1  mB vB,1  m Av A,2  mB vB,2
Velocities at positions 1 and 2 are
v A,1  0 vB,1  v0
vB,2  v A,2  vB
A, 2
 v A,2 (velocity of B relative to
A is zero at position 2)
mB v0  m A  mB v A, 2
v A,2  vB,2 
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mB
v0
m A  mB
14 - 22
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Vector Mechanics for Engineers: Dynamics
Sample Problem 14.4
• The conservation of energy principle can be applied to relate
the initial kinetic energy to the maximum potential energy.
T1  V1  T2  V2
Position 1 - Potential Energy: V1  mA gl
Kinetic Energy:
T1  12 mBv02
Position 2 - Potential Energy: V2  mA gl  mB gh
Kinetic Energy:
1 m v2
2 B 0
T2  12 mA  mB v 2A,2
 mA gl  12 mA  mB v 2A,2  mA gl  mB gh
2

v02 m A  mB v A, 2 v02 m A  mB  mB

h



v0 
2g
mB
2g 2g
2 g mB  m A  mB 
v02
mB
v02
h

2 g mA  mB 2 g
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2
mA
v02
h
m A  mB 2 g
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Vector Mechanics for Engineers: Dynamics
Sample Problem 14.5
SOLUTION:
• There are four unknowns: vA, vB,x, vB,y,
and vC.
• Solution requires four equations:
conservation principles for linear
momentum (two component equations),
angular momentum, and energy.
Ball A has initial velocity v0 = 3 m/s
parallel to the axis of the table. It hits
ball B and then ball C which are both at
rest. Balls A and C hit the sides of the
table squarely at A’ and C’ and ball B
hits obliquely at B’.
• Write the conservation equations in
terms of the unknown velocities and
solve simultaneously.
Assuming perfectly elastic collisions,
determine velocities vA, vB, and vC with
which the balls hit the sides of the table.
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14 - 24
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Vector Mechanics for Engineers: Dynamics
Sample Problem 14.5
• The conservation of momentum and energy equations,
SOLUTION:



• There are four unknowns: vA,
L1    Fdt  L2
vB,x, vB,y, and vC.
mv0  mvB, x  mvC
0  mv A  mvB, y


vA  vA j
HO ,1 + å ò M O dt = HO ,2



vB  vB , x i  vB , y j
- ( 0.6 m ) mv0 = ( 2.4m) mvA - ( 2.1m) mvB ,y - ( 0.9 m ) mvC


vC  vC i
T1  V1  T2  V2
1 mv 2
0
2


 12 mv 2A  12 m vB2 , x  vB2 , y  12 mvC2
Solving the first three equations in terms of vC,
vA = vB ,y = 3 vC - 6
vB ,x = 3 - vC
Substituting into the energy equation,
2
2
2 (3 vC - 6) + (3 - vC ) + vC2 = 9
20 vC2 - 78 vC + 72 = 0
y
v A = 1.2 m s vC = 2.4 m s
x
(
)
vB = 0.6 i - 1.2 j m s vB = 1.342 m s
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14 - 25
Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Three small identical spheres A, B, and C, which can slide on a horizontal,
frictionless surface, are attached to three 200-mm-long strings, which are tied
to a ring G. Initially, the spheres rotate clockwise about the ring with
a relative velocity of 0.8 m/s and the ring moves along the x-axis with a
velocity v0= (0.4 m/s)i. Suddenly, the ring breaks and the three spheres move
freely in the xy plane with A and B following paths parallel to the y-axis at a
distance a= 346 mm from each other and C following a path parallel to the x
axis. Determine (a) the velocity of each sphere, (b) the distance d.
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14 - 26
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Given: vArel= vBrel = vCrel = 0.8
m/s, v0= (0.4 m/s)i , L= 200
mm, a= 346 mm
Find: vA, vB, vC (after ring
breaks), d
SOLUTION:
• There are four unknowns: vA, vB, vB, d.
Apply the conservation of
linear momentum equation
– find L0 before ring breaks
• Solution requires four equations:
conservation principles for linear
momentum (two component equations),
L0  (3m) v  3m (0.4i )  m (1.2 m/s)i
angular momentum, and energy.
• Write the conservation equations in
terms of the unknown velocities and
solve simultaneously.
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What is Lf (after ring breaks)?
L f  mvA j  mvB j  mvC i
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
xA
Set L0= Lf
m(1.2 m/s)i  mvC i  m(v A  vB ) j
From the y components,
vA  vB
From the x components,
vC  1.200 m/s vC  1.200 m/s
Apply the conservation of angular momentum equation
H0:
( H G )0  3mlvrel  3m (0.2 m) (0.8 m/s)  0.480m
Hf:
( H G ) f  mvA xA  mvB ( xA  a)  mvC d 
Since vA= vB, and
vC = 1.2 m/s, then:
0.480m  0.346mvA  mvC d
0.480  0.346vA  1.200d
d  0.400  0.28833vA
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14 - 28
Tenth
Edition
;
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
;
xA
Need another equationtry work-energy, where
T0 = Tf
T0:
1
1

(3m)v 2  3  mvrel 2 
2
2

3
3
 m v02  vrel 2  [(0.4)2  (0.8)2 ]m  1.200m
2
2
T0 


Substitute in known values:
1 2
v A  v A2  (1.200) 2   1.200

2
v A2  0.480
v A  vB  0.69282 m/s
Tf:
Tf 
1 2 1 2 1 2
mv A  mvB  mvC
2
2
2
Solve for d:
d  0.400  0.28833(0.69282)  0.20024 m
vA  0.693 m/s
v C  1.200 m/s
v B  0.693 m/s
d  0.200 m
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14 - 29
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Variable Systems of Particles
• Kinetics principles established so far were derived for
constant systems of particles, i.e., systems which
neither gain nor lose particles.
• A large number of engineering applications require the
consideration of variable systems of particles, e.g.,
hydraulic turbine, rocket engine, etc.
• For analyses, consider auxiliary systems which consist
of the particles instantaneously within the system plus
the particles that enter or leave the system during a
short time interval. The auxiliary systems, thus
defined, are constant systems of particles.
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14 - 30
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Steady Stream of Particles. Applications
• Fan
• Fluid Stream Diverted by Vane
or Duct
• Jet Engine
• Helicopter
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14 - 31
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Steady Stream of Particles
• System consists of a steady stream of particles
against a vane or through a duct.
• Define auxiliary system which includes
particles which flow in and out over t.
• The auxiliary system is a constant system of
particles over t.
t2 


L1    Fdt  L 2
t1

 mi vi  m vA    F t   mi vi  m vB 
 dm 


F

v

v

B
A
dt
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14 - 32
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Streams Gaining or Losing Mass
• Define auxiliary system to include particles
of mass m within system at time t plus the
particles of mass m which enter the system
over time interval t.
• The auxiliary system is a constant system of
particles.
t2 


L1    Fdt  L 2
t1



mv  m va    F t  m  m v  v 


 





F

t

m

v


m
v

v


m

v

a


dv dm 
F  m  u
dt dt
 dm 

ma   F 
u
dt
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 14.6
SOLUTION:
• Define a system consisting of the mass
of grain on the chute plus the mass that
is added and removed during the time
interval t.
Grain falls onto a chute at the rate of
120 kg/s. It hits the chute with a
velocity of 10 m/s and leaves with a
velocity of 7.5 m/s. The combined
weight of the chute and the grain it
carries is 3000 N with the center of
gravity at G.
• Apply the principles of conservation of
linear and angular momentum for three
equations for the three unknown
reactions.
Determine the reactions at C and B.
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14 - 34
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 14.6
SOLUTION:
• Define a system consisting of the
mass of grain on the chute plus the
mass that is added and removed
during the time interval t.



L1    Fdt  L2
C x t  m vB cos10


 m v A  C y  W  B t  m vB sin 10
H C ,1 + å ò M C dt = H C ,2
-1.5 ( Dm) vA + ( -3.5W + 6 B) Dt
• Apply the principles of conservation
of linear and angular momentum for
= 3( Dm) vB cos10° - 6 ( Dm) vB sin10°
three equations for the three
unknown reactions.
Solve for Cx, Cy, and B with
Dm
= 120 kg/s
Dt
B = 2340 N C = 886 i + 1704 j N
(
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)
14 - 35
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• Calculate the time rate of change of the
mass of the air.
• Determine the thrust generated by the
airstream.
• Use this thrust to determine the
maximum load that the helicopter can
carry.
The helicopter shown can produce a maximum downward air speed
of 25 m/s in a 10-m-diameter slipstream. Knowing that the weight of
the helicopter and its crew is 18 kN and assuming r= 1.21 kg/m3 for
air, determine the maximum load that the helicopter can lift while
hovering in midair.
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14 - 36
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
Given: vB = 25 m/s, W= 18000 N, r= 1.21 kg/m3
Find: Max load during hover
Choose the relationship you will
use to determine the thrust
F
dm
(vB  v A )
dt
Calculate the time rate of change (dm/dt) of the mass of the air.
mass  density  volume  density  area  length
m  r AB (l )  r AB vB (t )
Dm
dm
 r AB vB 
Dt
dt
AB is the area of the slipstream
vB is the velocity in the slipstream.
Well above the blade, vA ≈ 0
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2 - 37
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Use the relationship for dm/dt
to determine the thrust
F
dm 
 AB vB
dt
g
dm
(vB  v A )
dt
F = dm ( vB – vA)
dt
F
()
= (1.21 kg/m 3) p (10 m)2 ( 25 m/s)2
4
= 59, 396 N
WH W
P
Use statics to determine the maximum payload during hover
 Fy  F  WH  WP  0
WP  F – WH  59,396 – 8000  41396 N
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W = 41400 N
2 - 38
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Question
In the previous problem with the
maximum payload attached,
what happens if the helicopter
tilts (or pitches) forward?
a)
b)
c)
d)
The area of displaced air becomes smaller
The volume of displaced air becomes smaller
The helicopter will accelerate upward
The helicopter will accelerate forward
*The helicopter will also accelerate downward
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