Transcript Document

บทที่ 16
Plane Motion of Rigid Bodies:
Forces and Accelerations
1
Introduction
• In this chapter and in Chapters 17 and 18, we will be
concerned with the kinetics of rigid bodies, i.e., relations
between the forces acting on a rigid body, the shape and mass
of the body, and the motion produced.
• Results of this chapter will be restricted to:
- plane motion of rigid bodies
• Our approach will be to consider rigid bodies as made of
large numbers of particles and to use the results of Chapter
14 for the motion of systems of particles. Specifically,




F  ma
and
M H


G
G
• D’Alembert’s principle is applied to prove that the external

forces acting on a rigid body are equivalent a vector ma
attached to the mass center and a couple of moment I .
2
Equations of Motion for a Rigid Body
• Consider a rigid body acted upon
by several external forces.
• Assume that the body is made of
a large number of particles.
• For the motion of the mass center
G of the body with respect to the
Newtonian frame Oxyz,


F

m
a

• For the motion of the body with
respect to the centroidal frame
Gx’y’z’,


 M G  HG
• System of external forces is
equipollent to the

 system
consisting of ma and H G .
3
Angular Momentum of a Rigid Body in Plane Motion
• Angular momentum of the slab may be
computed by
n  

H G   ri viΔmi 
i 1
n
  
  ri   riΔmi 
i 1



   ri 2 Δmi

 I  I  Mass Moment of Inertia
• Consider a rigid slab in
plane motion.
• After differentiation,



H G  I   I 
• Results are also valid for plane motion of bodies
which are symmetrical with respect to the
reference plane or bodies which have a pricipal
centroidal axis of inertia perpendicular to the
reference plane.
• Results are not valid for asymmetrical bodies or
three-dimensional motion.
4
Plane Motion of a Rigid Body: D’Alembert’s Principle
• Motion of a rigid body in plane motion is
completely defined by the resultant and moment
resultant about G of the external forces.
 Fx  max  Fy  ma y  M G  I
• The external forces and the collective effective
forces of the slab particles are equipollent (reduce
to the same resultant and moment resultant) and
equivalent (have the same effect on the body).
• d’Alembert’s Principle: The external forces
acting on a rigid body are equivalent to the
effective forces of the various particles forming
the body.
• The most general motion of a rigid body that is
symmetrical with respect to the reference plane
can be replaced by the sum of a translation and a
centroidal rotation.
5
Kinetics of Rigid Body : General Plane Motion



 F   Feff ma
max
 F   F 
 F   F 
x
x eff
y
y eff
 ma x
may
ma y
yG/O
o



 MG   MG eff  I 




 MO   MO eff  I  ma x yG / O  ma y x G / O
o
xG/O
6
Sample Problem 16.1
SOLUTION:
• Calculate the acceleration during the
skidding stop by assuming uniform
acceleration.
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces.
At a forward speed of 30 m/s, the truck • Apply the three corresponding scalar
brakes were applied, causing the wheels
equations to solve for the unknown
to stop rotating. It was observed that the
normal wheel forces at the front and rear
truck to skidded to a stop in 200 m.
and the coefficient of friction between
the wheels and road surface.
Determine the magnitude of the normal
reaction and the friction force at each
wheel as the truck skidded to a stop.
7
Sample Problem 16.1
SOLUTION:
• Calculate the acceleration during the skidding stop
by assuming uniform acceleration.
v 2  v02  2a  x  x0 
m
v0  30
s
x  200 m
2
 m
0   30   2a 200 m 
 s
a  2.25
m
s
• Draw a free-body-diagram equation expressing the
equivalence of the external and effective forces.
• Apply the corresponding scalar equations.
 Fy   Fy eff
 Fx   Fx eff
N A  NB  W  0
 FA  FB  ma
 k N A  N B  
  kW  W g a
k 
a 2.25

 0.225
g 10
8
Sample Problem 16.1
• Apply the corresponding scalar equations.
 M A   M A eff
 5 m W  12 m N B  4 m ma
NB 
1
W
5
W

4

12 
g
a
 W
a   5  4 
g
 12 
N B  0.650W
N A  W  N B  0.350W
Nrear  12 N A  12 0.350W 
N rear  0.175W
Frear  k N rear  0.2250.175W 
N front  12 NV  12 0.650W 
Frear  0.04W
N front  0.325W
Ffront  k N front  0.2250.325W 
Ffront  0.07W
9
Sample Problem 16.2
SOLUTION:
• Note that after the wire is cut, all
particles of the plate move along parallel
circular paths of radius 150 mm. The
plate is in curvilinear translation.
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces.
The thin plate of mass 8 kg is held in
place as shown.
• Resolve into scalar component equations
parallel and perpendicular to the path of
the mass center.
Neglecting the mass of the links,
determine immediately after the wire
has been cut (a) the acceleration of the
plate, and (b) the force in each link.
• Solve the component equations and the
moment equation for the unknown
acceleration and link forces.
10
Sample Problem 16.2
SOLUTION:
• Note that after the wire is cut, all particles of the
plate move along parallel circular paths of radius
150 mm. The plate is in curvilinear translation.
• Draw the free-body-diagram equation expressing
the equivalence of the external and effective
forces.
• Resolve the diagram equation into components
parallel and perpendicular to the path of the mass
center.
 Ft   Ft eff
W cos 30  ma
mg cos 30 


a  9.81m/s 2 cos 30
a  8.50 m s2
60o
11
Sample Problem 16.2
• Solve the component equations and the moment
equation for the unknown acceleration and link
forces.
 M G   M G eff
FAE sin 30250 mm   FAE cos 30100 mm 
FDF sin 30250 mm   FDF cos 30100 mm   0
38.4 FAE  211.6 FDF  0
FDF  0.1815 FAE
 Fn   Fn eff
a  8.50 m s2
60o
FAE  FDF  W sin 30  0
FAE  0.1815 FAE  W sin 30  0

FAE  0.6198 kg  9.81m s 2
FDF  0.1815 47.9 N 

FAE  47.9 N T
FDF  8.70 N C
12
Sample Problem 16.3
SOLUTION:
• Determine the direction of rotation by
evaluating the net moment on the
pulley due to the two blocks.
• Relate the acceleration of the blocks to
the angular acceleration of the pulley.
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces on the
complete pulley plus blocks system.
A pulley weighing 12 N and having a
radius of gyration of 8 cm is connected to
• Solve the corresponding moment
two blocks as shown.
equation for the pulley angular
Assuming no axle friction, determine the
acceleration.
angular acceleration of the pulley and the
acceleration of each block.
13
Sample Problem 16.3
SOLUTION:
• Determine the direction of rotation by evaluating the net
moment on the pulley due to the two blocks.
 M G  10 N6 cm5 N10 cm  10 N × cm
rotation is counterclockwise.
W 2
note: I  mk  k
g
12 N
2



8
cm
10 m/s2
2
 76.8 kg cm2
• Relate the acceleration of the blocks to the angular
acceleration of the pulley.
a A  rA
 10 cm 
a B  rB
 6 cm 
14
Sample Problem 16.3
• Draw the free-body-diagram equation expressing the
equivalence of the external and effective forces on the
complete pulley and blocks system.
• Solve the corresponding moment equation for the pulley
angular acceleration.
 M G   M G eff
10 N6 cm  5 N10 cm  Iα  mBa B 6 cm  mAa A 10 cm
106  510  76 .8α  1010 6α6  105 10α10
I  76.8 kg cm 2
a A  10 cm/s2
a B  6 cm/s
2
  0.06rad s 2
Then,
a A  rA 

 10 cm  0.06 rad s 2
a B  rB

 6 cm  0.06 rad s 2


a A  0.6 cm/s2
a B  0.36 cm/s2
15
Sample Problem 16.4
SOLUTION:
• Draw the free-body-diagram equation
expressing the equivalence of the external
and effective forces on the disk.
• Solve the three corresponding scalar
equilibrium equations for the horizontal,
vertical, and angular accelerations of the
disk.
A cord is wrapped around a
homogeneous disk of mass 15 kg.
The cord is pulled upwards with a
force T = 180 N.
• Determine the acceleration of the cord by
evaluating the tangential acceleration of
the point A on the disk.
Determine: (a) the acceleration of the
center of the disk, (b) the angular
acceleration of the disk, and (c) the
acceleration of the cord.
16
Sample Problem 16.4
SOLUTION:
• Draw the free-body-diagram equation expressing the
equivalence of the external and effective forces on the
disk.
• Solve the three scalar equilibrium equations.
 Fx   Fx eff
ax  0
0  ma x
 Fy   Fy eff
T  W  ma y

T  W 180 N - 15 kg  9.81m s 2
ay 

m
15 kg
 M G   M G eff
 Tr  I  
 

a y  2.19 m s 2
12 mr 2 
2T
2180 N 

15 kg 0.5 m 
mr
  48.0 rad s2
17
Sample Problem 16.4
• Determine the acceleration of the cord by evaluating the
tangential acceleration of the point A on the disk.

acord  a A t  a  a A G t

 2.19 m s 2  0.5 m  48 rad s 2

acord  26.2 m s2
ax  0
a y  2.19 m s 2
  48.0 rad s2
18
Constrained Plane Motion
• Most engineering applications involve rigid
bodies which are moving under given
constraints, e.g., cranks, connecting rods, and
non-slipping wheels.
• Constrained plane motion: motions with
definite relations between the components of
acceleration of the mass center and the angular
acceleration of the body.
• Solution of a problem involving constrained
plane motion begins with a kinematic analysis.
• e.g., given q, , and , find P, NA, and NB.
- kinematic analysis yields ax and a y .
- application of d’Alembert’s principle yields
P, NA, and NB.
19
Constrained Motion: Noncentroidal Rotation
• Noncentroidal rotation: motion of a body is
constrained to rotate about a fixed axis that does
not pass through its mass center.
• Kinematic relation between the motion of the mass
center G and the motion of the body about G,
at  r 
an  r  2
• The kinematic relations are used to eliminate
at and an from equations derived from
d’Alembert’s principle or from the method of
dynamic equilibrium.
20
Constrained Plane Motion: Rolling Motion
• For a balanced disk constrained to
roll without sliding,
x  rq  a  r
• Rolling, no sliding:
F  s N
a  r
Rolling, sliding impending:
F  s N
a  r
Rotating and sliding:
F  k N
a, r independent
• For the geometric center of an
unbalanced disk,
aO  r
The acceleration of the mass center,



a G  a O  aG O



 aO  aG O  aG O

t 
n
21
Sample Problem 16.6
SOLUTION:
• Draw the free-body-equation for AOB,
expressing the equivalence of the
external and effective forces.
mE  4 kg
k E  85 mm
mOB  3 kg
The portion AOB of the mechanism is
actuated by gear D and at the instant
shown has a clockwise angular velocity
of 8 rad/s and a counterclockwise
angular acceleration of 40 rad/s2.
Determine: a) tangential force exerted
by gear D, and b) components of the
reaction at shaft O.
• Evaluate the external forces due to the
weights of gear E and arm OB and the
effective forces associated with the
angular velocity and acceleration.
• Solve the three scalar equations
derived from the free-body-equation
for the tangential force at A and the
horizontal and vertical components of
reaction at shaft O.
22
Sample Problem 16.6
SOLUTION:
• Draw the free-body-equation for AOB.
• Evaluate the external forces due to the weights of
gear E and arm OB and the effective forces.


WE  4 kg  9.81m s 2  39.2 N


WOB  3 kg  9.81m s 2  29.4 N

I E  mE k E2  4kg 0.085 m 2 40 rad s 2
 1.156 N  m
mE  4 kg
k E  85 mm
mOB  3 kg


mOB aOB t  mOB r    3 kg 0.200 m  40 rad s 2
  40 rad s2
  8 rad/s

 24.0 N
 
mOB aOB n  mOB r  2  3 kg 0.200 m 8 rad s 2
 38.4 N
I OB 
121 mOB L2   121 3kg0.400 m2 40 rad s2 
 1.600 N  m
23
Sample Problem 16.6
• Solve the three scalar equations derived from the freebody-equation for the tangential force at A and the
horizontal and vertical components of reaction at O.
 MO   MO eff
F 0.120m  I E  mOB aOB t 0.200m  IOB
 1.156 N  m  24.0 N 0.200m  1.600N  m
F  63.0 N
 Fx   Fx eff
WE  39.2 N
WOB  29.4 N
I E  1.156 N  m
Rx  mOB aOB t  24.0 N
 Fy   Fy eff
mOB aOB t  24.0 N
R y  F  WE  WOB  mOB aOB 
mOB aOB n  38.4 N
R y  63.0 N  39.2 N  29.4 N  38.4 N
IOB  1.600 N  m
Rx  24.0 N
Ry  24.0 N
24
Sample Problem 16.8
SOLUTION:
• Draw the free-body-equation for the
sphere, expressing the equivalence of
the external and effective forces.
• With the linear and angular accelerations
related, solve the three scalar equations
derived from the free-body-equation for
the angular acceleration and the normal
A sphere of weight W is released with
and tangential reactions at C.
no initial velocity and rolls without
• Calculate the friction coefficient required
slipping on the incline.
for the indicated tangential reaction at C.
Determine: a) the minimum value
• Calculate the velocity after 10 m of
of the coefficient of friction, b) the
uniformly accelerated motion.
velocity of G after the sphere has
• Assuming no friction, calculate the linear
rolled 10 m and c) the velocity of
acceleration down the incline and the
G if the sphere were to move 10 m
corresponding velocity after 10 m.
down a frictionless incline.
25
Sample Problem 16.8
SOLUTION:
• Draw the free-body-equation for the sphere, expressing
the equivalence of the external and effective forces.
• With the linear and angular accelerations related, solve
the three scalar equations derived from the free-bodyequation for the angular acceleration and the normal
and tangential reactions at C.
 MC   MC eff
a  r
W sin q r  ma r  I
 mr r  52 mr 2 
W
  2W 2
  r  r  
r 
g
 5 g 
a  r 


5 g sin q
7r
5 g sin 30
7

5 10 m/s 2 sin 30

7
a  3.57 m/s 2
26
Sample Problem 16.8
• Solve the three scalar equations derived from the freebody-equation for the angular acceleration and the
normal and tangential reactions at C.
 Fx   Fx eff W sin q  F  ma
W 5 g sin q
g
7
2
F  W sin 30  0.143W
7

 Fy   Fy eff

5 g sin q
7r
a  r  3.57 m/s 2
N  W cosq  0
N  W cos 30  0.866W
• Calculate the friction coefficient required for the
indicated tangential reaction at C.
F  s N
s 
F 0.143W

N 0.866W
 s  0.165
27
Sample Problem 16.8
• Calculate the velocity after 10 m of uniformly
accelerated motion.
v 2  v02  2a  x  x0 


 0  2 3.57 m/s2 10 m 

v  71.4 m/s
• Assuming no friction, calculate the linear acceleration
and the corresponding velocity after 10 m.
5 g sin q

7r
 MG   MG eff
0  I
 Fx   Fx eff
W 
W sin q  ma   a
g

 0

a  10 m/s2 sin 30  5 m/s2
a  r  3.57 m/s 2
v 2  v02  2a  x  x0 


 0  2 5 m/s2 10 m 

v  100 m/s
28
Sample Problem 16.9
SOLUTION:
• Draw the free-body-equation for the
wheel, expressing the equivalence of
the external and effective forces.
A cord is wrapped around the
inner hub of a wheel and pulled
horizontally with a force of 200 N.
The wheel has a mass of 50 kg
and a radius of gyration of 70 mm.
Knowing s = 0.20 and k = 0.15,
determine the acceleration of G and
the angular acceleration of the wheel.
• Assuming rolling without slipping and
therefore, related linear and angular
accelerations, solve the scalar equations
for the acceleration and the normal and
tangential reactions at the ground.
• Compare the required tangential reaction
to the maximum possible friction force.
• If slipping occurs, calculate the kinetic
friction force and then solve the scalar
equations for the linear and angular
accelerations.
29
Sample Problem 16.9
SOLUTION:
• Draw the free-body-equation for the wheel,.
• Assuming rolling without slipping, solve the scalar
equations for the acceleration and ground reactions.
 MC   MC eff
200N 0.040 m   ma 0.100 m   I
8.0 N  m  50 kg 0.100 m 2  0.245 kg  m 2 
I  mk 2  50 kg 0.70 m 2
 0.245 kg  m
2
Assume rolling without slipping,
a  r
 0.100 m 
  10.74 rad s 2


a  0.100 m  10.74 rad s 2  1.074 m s 2
 Fx   Fx eff

F  200 N  ma  50 kg  1.074 m s 2

F  146.3 N
 Fx   Fx eff
N W  0


N  mg  50kg  1.074 m s 2  490.5 N
30
Sample Problem 16.9
• Compare the required tangential reaction to the
maximum possible friction force.
Fmax   s N  0.20490.5 N   98.1 N
F > Fmax , rolling without slipping is impossible.
Without slipping,
F  146.3 N N  490.5 N
• Calculate the friction force with slipping and solve the
scalar equations for linear and angular accelerations.
F  Fk   k N  0.15490.5 N   73.6 N
 Fx   Fx eff
200 N  73.6 N  50 kg a
a  2.53 m s2
 MG   MG eff
73.6 N 0.100 m   200 N 0.0.060 m 


 0.245 kg  m 2 
  18.94 rad s 2
  18.94 rad s2
31
Sample Problem 16.10
SOLUTION:
• Based on the kinematics of the constrained
motion, express the accelerations of A, B,
and G in terms of the angular acceleration.
The extremities of a 4-m rod
weighing 50 N can move freely and
with no friction along two straight
tracks. The rod is released with no
velocity from the position shown.
• Draw the free-body-equation for the rod,
expressing the equivalence of the
external and effective forces.
• Solve the three corresponding scalar
equations for the angular acceleration and
the reactions at A and B.
Determine: a) the angular
acceleration of the rod, and b) the
reactions at A and B.
32
Sample Problem 16.10
SOLUTION:
• Based on the kinematics of the constrained motion,
express the accelerations of A, B, and G in terms of
the angular acceleration.
Express the acceleration of B as



aB  a A  aB A
With aB A  4 , the corresponding vector triangle and
the law of signs yields
a A  5.46
aB  4.90
The acceleration of G is now obtained from
 


a a G  a A  aG A where aG A  2
Resolving into x and y components,
ax  5.46  2 cos 60  4.46
a y  2 sin 60  1.732
33
Sample Problem 16.10
• Draw the free-body-equation for the rod, expressing
the equivalence of the external and effective forces.
• Solve the three corresponding scalar equations for the
angular acceleration and the reactions at A and B.
 M E   M E eff
60o
501.732  22.3 4.46  8.66 1.732  6.67
  0.715 rad s 2
I  121 ml 2 
1 50 N
2


4
m
12 10 m/s2
 6.67 kgm 2
I   6.67
50
4.46   22.3
10
50
ma y   1.732   8.66
10
ma x 
  0.715 rad s 2
 Fx   Fx eff
RB sin 45  22.30.715
RB  22.5 N
 Fy   Fy eff

RB  22.5 N
RA  22.5cos 45  50  8.660.715
RA  40.3 N
45o
34
Suggestive Problem 1

ก้าน BD ยาว 300 mm มีมวล 3 kg ถูกยึดติดกับก้าน AB และปลอก D ที่ไม่คานึงถึงมวล
ถ้าก้าน AB หมุนด้วยความเร็ วรอบคงที่ 300 rpm ในทิศทางทวนเข็มนาฬิกา จงหาแรง
ปฏิกิริยาที่ D เมื่อ q = 90o
35
Suggestive Problem 2
36
Suggestive Problem 3
37
Suggestive Problem 4
38
Suggestive Problem 5
39
Suggestive Problem 6
40
Suggestive Problem 7
41
Suggestive Problem 8
42
Suggestive Problem 9
30.5 cm
30.5 cm
43
Suggestive Problem 10
44
Suggestive Problem 11
45
Suggestive Problem 12
2.3 cm
7.6 cm
46
Suggestive Problem 13
40.6 cm
40.6 cm
47
Suggestive Problem 14
7.6 cm
15.2 cm
20.3 cm
48
Suggestive Problem 15
49
Suggestive Problem 16
50
Suggestive Problem 17
51
Suggestive Problem 18
52
Suggestive Problem 19
53
Suggestive Problem 20
The 4.5 kg uniform rod ABD is attached to the crank BC and is fitted
with a small wheel that can roll without friction along a vertical slot.
Knowing that at the instant shown crank BC rotates with an angular
velocity of 6 rad/s clockwise and an angular acceleration of 15 rad/s2
counterclockwise, determine the reaction at A.
12.7 cm
25.4 cm
25.4 cm
54