DYNAMICS Ferdinand P. Beer Plane Motion of Rigid Bodies: E. - - PowerPoint PPT Presentation
Seventh Edition VECTOR MECHANICS FOR ENGINEERS: CHAPTER DYNAMICS Ferdinand P. Beer Plane Motion of Rigid Bodies: E. Russell Johnston, Jr. Forces and Accelerations Lecture Notes: J. Walt Oler Texas Tech University
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Seventh Edition Ferdinand P. Beer
Lecture Notes:
Texas Tech University CHAPTER
Forces and Accelerations
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 2
Contents
Introduction Equations of Motion of a Rigid Body Angular Momentum of a Rigid Body in Plane Motion Plane Motion of a Rigid Body: d’Alembert’s Principle Axioms of the Mechanics of Rigid Bodies Problems Involving the Motion of a Rigid Body Sample Problem 16.1 Sample Problem 16.2 Sample Problem 16.3 Sample Problem 16.4 Sample Problem 16.5 Constrained Plane Motion Constrained Plane Motion: Noncentroidal Rotation Constrained Plane Motion: Rolling Motion Sample Problem 16.6 Sample Problem 16.8 Sample Problem 16.9 Sample Problem 16.10
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 3
Introduction
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.
numbers of particles and to use the results of Chapter 14 for the motion of systems of particles. Specifically,
G G
H M a m F
=
symmetrical with respect to the reference plane.
forces acting on a rigid body are equivalent a vector attached to the mass center and a couple of moment a m . α I
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 4
Equations of Motion for a Rigid Body
several external forces.
large number of particles.
Newtonian frame Oxyz, a m F
respect to the centroidal frame Gx’y’z’,
G G
H M
equipollent to the system consisting
. and
G
H a m
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 5
Angular Momentum of a Rigid Body in Plane Motion
plane motion.
by
( ) ( ) [ ]
( )
ω ω ω
m r m r r m v r H
i i n i i i i n i i i i G
= ′ = ′ × × ′ = ′ × ′ =
=
1 1
α ω
I H G = =
which are symmetrical with respect to the reference plane.
three-dimensional motion.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 6
Plane Motion of a Rigid Body: D’Alembert’s Principle
α I M a m F a m F
G y y x x
= = =
defined by the resultant and moment resultant about G of the external forces.
same resultant and moment resultant) and equivalent (have the same effect on the body).
symmetrical with respect to the reference plane can be replaced by the sum of a translation and a centroidal rotation.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 7
Axioms of the Mechanics of Rigid Bodies
act at different points on a rigid body but but have the same magnitude, direction, and line of action. F F
and
point and are therefore, equipollent external forces.
whereas it was previously stated as an axiom.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 8
Problems Involving the Motion of a Rigid Body
acting on a rigid body in plane motion and the acceleration of its mass center and the angular acceleration of the body is illustrated in a free- body-diagram equation.
equilibrium may be applied to solve problems
problems involving plane motion of connected rigid bodies by drawing a free-body-diagram equation for each body and solving the corresponding equations of motion simultaneously.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 9
Sample Problem 16.1
At a forward speed of 10 m/s, the truck brakes were applied, causing the wheels to stop rotating. It was observed that the truck to skidded to a stop in 6 m. Determine the magnitude of the normal reaction and the friction force at each wheel as the truck skidded to a stop. SOLUTION:
skidding stop by assuming uniform acceleration.
equations to solve for the unknown normal wheel forces at the front and rear and the coefficient of friction between the wheels and road surface.
expressing the equivalence of the external and effective forces.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 10
Sample Problem 16.1
m 6 s / m 10 = = x v
SOLUTION:
assuming uniform acceleration.
( )
( ) ( )
m 6 2 s / m 10 2
2 2 2
a x x a v v + = − + =
2
s / m 33 . 8 − = a
equivalence of the external and effective forces.
= − + W N N
B A
( )
eff y y
F F
( ) ( )
849 . 81 . 9 33 . 8 = = = − = − = + − − = − − g a a g W W N N a m F F
k k B A k B A
µ µ µ
( )
eff x x
F F
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 11
Sample Problem 16.1
kN 37 . 7 = − =
B A
N W N kN 37 . 7
2 1
= =
A rear
N N kN 69 . 3 =
rear
N
( )
kN 16 . 17
2 1 2 1
= =
V front
N N
kN 58 . 8 =
front
N
( )( )
kN 69 . 3 849 . = =
rear k rear
N F µ
kN 13 . 3 =
rear
F
( )( )
kN 58 . 8 849 . = =
front k front
N F µ
kN 29 . 7 =
front
F
( ) ( ) ( )
kN 16 . 17 2 . 1 5 . 1 6 . 3 2 . 1 5 6 . 3 1 m 2 . 1 m 6 . 3 m 5 . 1 =
=
= = + −
B B B
N g a W a g W W N a m N W
( )
eff A A
M M
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 12
Sample Problem 16.2
The thin plate of mass 8 kg is held in place as shown. 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. SOLUTION:
paths of radius 150 mm. The plate is in curvilinear translation.
expressing the equivalence of the external and effective forces.
parallel and perpendicular to the path of the mass center.
moment equation for the unknown acceleration and link forces.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 13
Sample Problem 16.2
SOLUTION:
plate move along parallel circular paths of radius 150 mm. The plate is in curvilinear translation.
equivalence of the external and effective forces.
parallel and perpendicular to the path of the mass center.
( )
eff t t
F F = ° = ° 30 cos 30 cos mg a m W
( )
° = 30 cos m/s 81 . 9
2
a
2
s m 50 . 8 = a 60o
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 14
Sample Problem 16.2
2
s m 50 . 8 = a 60o
equation for the unknown acceleration and link forces.
( )eff
G G
M M
( )( ) ( )( ) ( )( ) ( )( )
mm 100 30 cos mm 250 30 sin mm 100 30 cos mm 250 30 sin = ° + ° ° − °
DF DF AE AE
F F F F
AE DF DF AE
F F F F 1815 . 6 . 211 4 . 38 − = = +
( )
eff n n
F F
( )(
)
2
s m 81 . 9 kg 8 619 . 30 sin 1815 . 30 sin = = ° − − = ° − +
AE AE AE DF AE
F W F F W F F T FAE N 9 . 47 =
( )
N 9 . 47 1815 . − =
DF
F C FDF N 70 . 8 =
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 15
Sample Problem 16.3
A pulley weighing 6 kg and having a radius of gyration of 200 mm is connected to two blocks as shown. Assuming no axle friction, determine the angular acceleration of the pulley and the acceleration of each block. SOLUTION:
evaluating the net moment on the pulley due to the two blocks.
the angular acceleration of the pulley.
expressing the equivalence of the external and effective forces on the complete pulley plus blocks system.
for the pulley angular acceleration.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 16
Sample Problem 16.3
2 2 2 2 2
m kg 24 . ft 12 8 s ft 32.2 kg 6 ⋅ =
= = k g W k m I
note: SOLUTION:
moment on the pulley due to the two blocks.
( )( ) ( )( )
lb in 10 mm 250 kg 5 . 2 mm 150 kg 5 ⋅ = − =
M
rotation is counterclockwise.
acceleration of the pulley.
m) 15 . ( m) 25 . ( = = = = α α α
b B A A
r a r a
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 17
Sample Problem 16.3
equivalence of the external and effective forces on the complete pulley and blocks system.
( ) ( )α
α m 15 . m 25 . m kg 24 .
2
= = ⋅ =
B A
a a I
( )
eff G G
M M
( )( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( )( ) ( )( )( )
25 . 25 . 5 . 2 15 . 15 . 5 24 . 25 . 5 . 24 15 . 1 . 49 m 0.25 m 0.15 m 0.25 N 5 . 24 m 0.15 N 1 . 49 − + = − − + = − α α α
A A B B
a m a m I
angular acceleration.
2
s rad 374 . 2 = α
( )
) s rad 2.374 ( m 0.15
2
= = α
B B
r a
2
s m 37 . =
B
a
( )
) s rad 2.374 ( m 0.25
2
= = α
A A
r a
2
s m 61 . =
A
a
Then,
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 18
Sample Problem 16.4
A cord is wrapped around a homogeneous disk of mass 15 kg. The cord is pulled upwards with a force T = 180 N. Determine: (a) the acceleration of the center of the disk, (b) the angular acceleration of the disk, and (c) the acceleration of the cord. SOLUTION:
expressing the equivalence of the external and effective forces on the disk.
equilibrium equations for the horizontal, vertical, and angular accelerations of the disk.
evaluating the tangential acceleration of the point A on the disk.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 19
Sample Problem 16.4
SOLUTION:
equivalence of the external and effective forces on the disk.
( )
eff y y
F F
( )
kg 15 ) s m 81 . 9 ( kg 15
180
2
= − = = − m W T a a m W T
y y
2
s m 19 . 2 =
y
a
( )
eff G G
M M
( )
( ) ( )( )
m 5 . kg 15 N 180 2 2
2 2 1
− = − = = = − mr T mr I Tr α α α
2
s rad . 48 = α
( )
eff x x
F F
x
a m = =
x
a
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 20
Sample Problem 16.4
2
s m 19 . 2 =
y
a
2
s rad . 48 = α =
x
a
tangential acceleration of the point A on the disk. ) s rad 48 ( m) 5 . ( s m 19 . 2 ) ( ) (
2 2 +
= + = =
t G A t A cord
a a a a
s m 2 . 26 =
cord
a
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 21
Sample Problem 16.5
A uniform sphere of mass m and radius r is projected along a rough horizontal surface with a linear velocity v0. The coefficient of kinetic friction between the sphere and the surface is µk. Determine: (a) the time t1 at which the sphere will start rolling without sliding, and (b) the linear and angular velocities of the sphere at time t1. SOLUTION:
expressing the equivalence of the external and effective forces on the sphere.
equilibrium equations for the normal reaction from the surface and the linear and angular accelerations of the sphere.
uniformly accelerated motion to determine the time at which the tangential velocity of the sphere at the surface is zero, i.e., when the sphere stops sliding.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 22
Sample Problem 16.5
SOLUTION:
equivalence of external and effective forces on the sphere.
( )
eff y y
F F = −W N mg W N = =
( )
eff x x
F F = − = − mg a m F
k
µ g a
k
µ − =
( )
( )α
µ α
2 3 2 mr
r mg I Fr
k
= = r g
k
µ α 2 5 =
( )
eff G G
M M NOTE: As long as the sphere both rotates and slides, its linear and angular motions are uniformly accelerated.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 23
Sample Problem 16.5
g a
k
µ − = r g
k
µ α 2 5 =
to determine the time at which the tangential velocity of the sphere at the surface is zero, i.e., when the sphere stops sliding.
( )t
g v t a v v
k
µ − + = + = t r g t
k
= + = µ α ω ω 2 5
1 1
2 5 t r g r gt v
k k
− µ µ g v t
k
µ
1
7 2 =
g v r g t r g
k k k
µ µ µ ω
1 1
7 2 2 5 2 5 r v0
1
7 5 = ω
= r v r r v
1 1
7 5 ω
7 5 1
v v =
At the instant t1 when the sphere stops sliding,
1 1
ω r v =
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 24
Constrained Plane Motion
bodies which are moving under given constraints, e.g., cranks, connecting rods, and non-slipping wheels.
relations between the components of acceleration
the body.
motion begins with a kinematic analysis.
NA, and NB.
. and
y x
a a
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 25
Constrained Motion: Noncentroidal Rotation
constrained to rotate about a fixed axis that does not pass through its mass center.
center G and the motion of the body about G,
2
ω α r a r a
n t
= =
from equations derived from d’Alembert’s principle or from the method of dynamic equilibrium.
n t
a a and
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 26
Constrained Plane Motion: Rolling Motion
without sliding, α θ r a r x = → =
N F
s
µ ≤ α r a = Rolling, sliding impending: N F
s
µ = α r a = Rotating and sliding: N F
k
µ = α r a, independent
unbalanced disk, α r aO = The acceleration of the mass center,
( ) ( )n
O G t O G O O G O G
a a a a a a
+ = + =
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 27
Sample Problem 16.6
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.
kg 3 mm 85 kg 4 = = =
OB E E
m k m
SOLUTION:
expressing the equivalence of the external and effective forces.
weights of gear E and arm OB and the effective forces associated with the angular velocity and acceleration.
from the free-body-equation for the tangential force at A and the horizontal and vertical components of reaction at shaft O.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 28
Sample Problem 16.6
rad/s 8 = ω
2
s rad 40 = α kg 3 mm 85 kg 4 = = =
OB E E
m k m SOLUTION:
and arm OB and the effective forces.
( ) ( )
N 4 . 29 ) s m 81 . 9 ( kg 3 N 2 . 39 ) s m 81 . 9 ( kg 4
2 2
= = = =
OB E
W W
( )( )
m N 156 . 1 ) s rad 40 ( m 085 . kg 4
2 2 2
⋅ = = = α α
E E E
k m I
( ) ( ) ( )( )
N . 24 ) s rad 40 ( m 200 . kg 3
2
= = = α r m a m
OB t OB OB
( )
( ) (
)( )( )
N 4 . 38 s rad 8 m 200 . kg 3
2 2
= = = ω r m a m
OB n OB OB
( )
( )( )
m N 600 . 1 ) s rad 40 ( m .400 kg 3
2 2 12 1 2 12 1
⋅ = = = α α L m I
OB OB
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 29
Sample Problem 16.6
N 4 . 29 N 2 . 39 = =
OB E
W W m N 156 . 1 ⋅ = α
E
I
( )
N . 24 =
t OB OB a
m
( )
N 4 . 38 =
n OB OB a
m m N 600 . 1 ⋅ = α
OB
I
body-equation for the tangential force at A and the horizontal and vertical components of reaction at O.
( )eff
O O
M M
( ) ( ) ( ) ( )( )
m N 600 . 1 m 200 . N . 24 m N 156 . 1 m 200 . m 120 . ⋅ + + ⋅ = + + = α α
OB t OB OB E
I a m I F N . 63 = F
( )eff
x x
F F
( )
N . 24 = =
t OB OB x
a m R N . 24 =
x
R
eff y y
F F ) ( =
)
N 4 . 38 N 4 . 29 N 2 . 39 N . 63 = − − − = − − −
y OB OB OB E y
R a m W W F R N . 24 =
y
R
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 30
Sample Problem 16.8
A sphere of weight W is released with no initial velocity and rolls without slipping
Determine: a) the minimum value of the coefficient of friction, b) the velocity of G after the sphere has rolled 3 m and c) the velocity of G if the sphere were to move 3 m down a frictionless incline. SOLUTION:
expressing the equivalence of the external and effective forces.
related, solve the three scalar equations derived from the free-body-equation for the angular acceleration and the normal and tangential reactions at C.
accelerated motion.
acceleration down the incline and the corresponding velocity after 3 m.
the indicated tangential reaction at C.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 31
Sample Problem 16.8
SOLUTION:
equivalence of the external and effective forces. α r a =
three scalar equations derived from the free-body-equation for the angular acceleration and the normal and tangential reactions at C.
( )
eff C C
M M
( ) ( ) ( )
( )
α α α α α θ
+ = + =
2 2 5 2
5 2 sin r g W r r g W mr r mr I r a m r W r g 7 sin 5 θ α =
( )
7 30 sin s m 81 . 9 5 7 30 sin 5
2
° = ° = = g r a α
2
s m 5 . 3 = a
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 32
Sample Problem 16.8
equation for the angular acceleration and the normal and tangential reactions at C. r g 7 sin 5 θ α =
2
s m 5 . 3 = = α r a ( )
eff x x
F F
143 . 30 sin 7 2 7 sin 5 sin mg mg F g g mg a m F mg = ° = = = − θ θ
y y
F F ) (
mg mg N mg N 866 . 30 cos cos = ° = = − θ
tangential reaction at C.
mg mg N F N F
s s
866 . 143 . = = = µ µ
165 . =
s
µ
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 33
Sample Problem 16.8
r g 7 sin 5 θ α =
2
s m 5 . 3 = = α r a
motion.
( )
( )(
)
m 3 s m 5 . 3 2 2
2 2 2
+ = − + = x x a v v
s m 6 . 4 = v
)
eff G G
M M = = α α I
the corresponding velocity after 3 m.
( )
eff x x
F F
( )
2 2
s m 91 . 4 30 sin s m 81 . 9 30 sin = ° = = ° a a m mg
( )
( )(
)
m 3 s m 91 . 4 2 2
2 2 2
+ = − + = x x a v v
s m 43 . 5 = v
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 34
Sample Problem 16.9
A cord is wrapped around the inner hub
a force of 200 N. The wheel has a mass
determine the acceleration of G and the angular acceleration of the wheel. SOLUTION:
expressing the equivalence of the external and effective forces.
therefore, related linear and angular accelerations, solve the scalar equations for the acceleration and the normal and tangential reactions at the ground.
the maximum possible friction force.
friction force and then solve the scalar equations for the linear and angular accelerations.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 35
Sample Problem 16.9
SOLUTION:
Assume rolling without slipping,
( )α
α m 100 . = = r a
( )( )
2 2 2
m kg 245 . m 70 . kg 50 ⋅ = = = k m I
equations for the acceleration and ground reactions.
( )( ) ( ) ( )( ) ( )
2 2 2 2 2
s m 074 . 1 ) s rad 74 . 10 ( m 100 . s rad 74 . 10 ) m kg 245 . ( m 100 . kg 50 m N . 8 m 100 . m 040 . N 200 = = + = ⋅ + = ⋅ + = a I a m α α α α
( )
eff C C
M M
( )
eff x x
F F
( )
N 5 . 490 ) s m 074 . 1 ( kg 50
2
+ = = = = − mg N W N
( )
eff x x
F F N 3 . 146 ) s m 074 . 1 ( kg) 50 ( N 200
2
− = = = + F a m F
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 36
Sample Problem 16.9
N 3 . 146 − = F N 5 . 490 = N Without slipping,
possible friction force.
( )
N 1 . 98 N 5 . 490 20 .
max
= = = N F
s
µ F > Fmax , rolling without slipping is impossible.
scalar equations for linear and angular accelerations.
( )
N 6 . 73 N 5 . 490 15 . = = = = N F F
k k
µ
( )
eff G G
M M
( )( ) ( )( )
2 2
s rad 94 . 18 ) m kg 245 . ( m 060 . . N 200 m 100 . N 6 . 73 − = ⋅ = − α α
2
s rad 94 . 18 = α
( )
eff x x
F F
( )a
kg 50 N 6 . 73 N 200 = −
2
s m 53 . 2 = a
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 37
Sample Problem 16.10
The extremities of a 1.2 m rod weighing 25 kg can move freely and with no friction along two straight
velocity from the position shown. Determine: a) the angular acceleration
B. SOLUTION:
motion, express the accelerations of A, B, and G in terms of the angular acceleration.
expressing the equivalence of the external and effective forces.
equations for the angular acceleration and the reactions at A and B.
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 38
Sample Problem 16.10
SOLUTION:
express the accelerations of A, B, and G in terms of the angular acceleration. Express the acceleration of B as
A B A B
a a a
= With the corresponding vector triangle and the law of signs yields , 4α =
A B
a α α 90 . 4 46 . 5 = =
B A
a a The acceleration of G is now obtained from
A G A G
a a a a
= = α 2 where =
A G
a Resolving into x and y components, α α α α α 732 . 1 60 sin 2 46 . 4 60 cos 2 46 . 5 − = ° − = = ° − =
y x
a a
Vector Mechanics for Engineers: Dynamics
Seventh Edition 16 - 39
Sample Problem 16.10
equivalence of the external and effective forces.
( ) ( ) ( )
α α α α α α . 13 520 . 12 25 5 . 33 34 . 1 12 25 3 m . kg 3 m 2 . 1 12 kg 25 12 1
2 2 2 12 1
− = − = = = = = = =
y x
a m a m I ml I
angular acceleration and the reactions at A and B.
( )( ) ( )( ) ( )( )
2
s rad 33 . 2 3 520 . . 13 34 . 1 5 . 33 520 . 25 + = + + = α α α α
( )
eff E E
M M
2
s rad 33 . 2 = α
( )
eff x x
F F
( )( )
N 110 1 . 78 33 . 2 5 . 33 45 sin = = = °
B B
R R N 110 =
B
R
( )
eff y y
F F
( ) ( )( )
33 . 2 . 13 245 45 cos N 110 − = − ° +
A
R N 137 =
A
R