Seventh Edition
CHAPTER
16
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr.
Plane Motion of Rigid Bodies: Forces and Accelerations
Lecture Notes: J. Walt Oler Texas Tech University
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Vector Mechanics for Engineers: Dynamics D ynamics Contents Introduction
Sample Problem 16.3
Equations of Motion of a Rigid Body
Sample Problem 16.4
Angular Momentum of a Rigid Body in Plane Motion
Sample Problem 16.5
Plane Motion of a Rigid Body: d’Alembert d’Ale mbert’s ’s Princ Principle iple
Constrained Plane Motion: Noncentroidal Rotation
Axioms of the Mechanics of Rigid Bodies
Constrained Plane Motion: Rolling Motion
Problems Involving the Motion of a Rigid Body
Sample Problem 16.6
Sample Problem 16.1 Sample Problem 16.2
Constrained Plane Motion
Sample Problem 16.8 Sample Problem 16.9 Sample Problem 16.10
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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, and - rigid bodies consisting of plane slabs or bodies which are symmetrical with respect to the reference plane. • 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, r
r
∑ F = ma
r
and
r
& ∑ M G = H G
• D’Alembert’s principle is applied to prove that the external r forces acting on a rigid body are equivalent a vector ma attached to the mass center and a couple of moment I α . 16 - 3
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Vector Mechanics for Engineers: Dynamics 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 , r
r
∑ F = ma
• For the motion of the body with respect to the centroidal frame Gx’y’z’ , r r & ∑ M G = H G • System of external forces is equipollent to the system r r & consisting of ma and H G .
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Angular Momentum of a Rigid Body in Plane Motion • Angular momentum of the slab may be computed by r
n
r
r
H G = ∑ (r i′× vi′∆mi ) i =1 n
r
r
r
= ∑ [r i′× (ω × r i′)∆mi ] i =1
= ω ∑ (r i′ 2 ∆mi ) r = ω r
• After differentiation, r
r
r
& & H G = I ω = I α
• Consider a rigid slab in plane motion.
• Results are also valid for plane motion of bodies which are symmetrical with respect to the reference plane. • Results are not valid for asymmetrical bodies or three-dimensional motion.
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16 - 5
Vector Mechanics for Engineers: Dynamics 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.
∑ F x = ma x ∑ F y = ma y ∑ M G = I α • 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.
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Problems Involving the Motion of a Rigid Body • The fundamental relation between the forces 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. • The techniques for solving problems of static equilibrium may be applied to solve problems of plane motion by utilizing - d’Alembert’s principle, or - principle of dynamic equilibrium • These techniques may also be applied to 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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16 - 7
Vector Mechanics for Engineers: Dynamics 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 ft/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 20 ft. 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.
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Sample Problem 16.1 SOLUTION: • Calculate the acceleration during the skidding stop by assuming uniform acceleration. v 2 = v02 + 2a ( x − x0 ) v0 = 30
ft s
2
x = 20 ft
ft 0 = 30 + 2a (20 ft ) s
a = −22.5
ft s
• Draw a free-body-diagram equation expressing the equivalence of the external and effective forces. • Apply the corresponding scalar equations.
∑ F y = ∑
F y
N + N B − W = 0
eff
∑ F x = ∑ (F x )eff
− F A − F B = −ma − µ k ( N A + N B ) = − µ k W = −(W g )a µ k =
a g
=
32.2
= 0.699 16 - 9
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22.5
Vector Mechanics for Engineers: Dynamics Sample Problem 16.1 • Apply the corresponding scalar equations.
∑ M A = ∑ (M A )eff − (5 ft )W + (12 ft ) N B = (4 ft)ma N B =
1 W W a 5W + 4 a = 5 + 4 12 g 12 g
N B = 0.650W N A = W − N B = 0.350W N rear = 12 N A = F rear =
1 2
k N rear
N front = 12 N V = F front =
N rear = 0.175W
(0.350W )
= (0.690 )(0.175W )
1 2
(0.650W )
k N front =
F rear = 0.122W
N front = 0.325W
(0.690 )(0.325W ) F front = 0.0.227W
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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.
16 - 11
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Vector Mechanics for Engineers: Dynamics 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.
∑ F t = ∑ (F t )eff W cos 30° = ma mg cos 30° =
)
a = 9.81m/s2 cos 30° a = 8.50 m s 2
60o
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Sample Problem 16.2 • Solve the component equations and the moment equation for the unknown acceleration and link forces.
∑ M G = (∑ M G )eff ( F AE sin 30°)(250 mm) − ( F AE cos 30°)(100 mm) ( F DF sin 30°)(250 mm) + ( F DF cos 30°)(100 mm) = 0 38.4 F AE + 211.6 F DF = 0 F DF = −0.1815 F AE
∑ F n = ∑ (F n )eff a = 8.50 m s 2
60o
F AE + F DF − W sin 30° = 0 F AE − 0.1815 F AE − W sin 30° = 0
(
F AE = 0.619(8 kg ) 9.81m s 2 F DF = −0.1815 (47.9 N )
F AE = 47.9 N T F DF = 8.70 N
C
16 - 13
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)
Vector Mechanics for Engineers: Dynamics 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 lb and having a radius of gyration of 8 in. 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.
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Sample Problem 16.3 SOLUTION: • Determine the direction of rotation by evaluating the net moment on the pulley due to the two blocks.
∑
G
= (10 lb)(6 in ) − (5 lb)(10 in) = 10 in ⋅ lb
rotation is counterclockwise. note: I = mk 2 =
W g
k 2
8 ft = 32.2 ft s 2 12 12 lb
2
= 0.1656 lb ⋅ ft ⋅ s 2 • Relate the acceleration of the blocks to the angular acceleration of the pulley. a A = r Aα ft )α = (10 12
a B = r Bα
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6 ft )α = (12 16 - 15
Vector Mechanics for Engineers: Dynamics 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 6 (10 lb)(126 ft ) − (5 lb)(10 ft ) = I α + m B a B (12 ft ) − m Aa A (10 ft ) 12 12
) = (0.1656)α + (3210.2 )(126 α )(126 ) − (325.2 )(10 )(10 ) (10)(126 ) − (5)(10 12 12 12 α = 2.374 rad s 2 I = 0.1656 lb ⋅ ft ⋅ s 2 a A = (10 α )ft s 2 12 6 a B = (12 α )ft s 2
Then, a A = r Aα 2 ) ( ) ft 2.374 rad s = (10 12
a A = 1.978 ft s 2
a B = r Bα 6 ft )(2.374 rad s 2 ) = (12
a B = 1.187 ft s 2
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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 - 17
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Vector Mechanics for Engineers: Dynamics 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.
∑ F x = ∑ (F x )eff a x = 0
0 = ma x
∑ F y = ∑
F y
eff
T − W = ma y a y =
(
T − W 180 N - (15 kg ) 9.81m s 2 m
=
15 kg a y = 2.19 m s 2
∑ M G = ∑ (M G )eff − Tr = I α = α = −
2T mr
)
1 mr 2 2
=−
)α
2(180 N )
(15 kg )(0.5 m)
α = 48.0 rad s 2
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Sample Problem 16.4 • Determine the acceleration of the cord by evaluating the tangential acceleration of the point A on the disk. r
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 s 2
a x = 0
a y = 2.19 m s 2
α = 48.0 rad s 2
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16 - 19
Vector Mechanics for Engineers: Dynamics Sample Problem 16.5 SOLUTION: • Draw the free-body-diagram equation expressing the equivalence of the external and effective forces on the sphere.
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 µ k . Determine: (a) the time t 1 at which the sphere will start rolling without sliding, and (b) the linear and angular velocities of the sphere at time t 1.
• Solve the three corresponding scalar equilibrium equations for the normal reaction from the surface and the linear and angular accelerations of the sphere. • Apply the kinematic relations for 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.
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Sample Problem 16.5 SOLUTION: • Draw the free-body-diagram equation expressing the equivalence of external and effective forces on the sphere. • Solve the three scalar equilibrium equations.
∑ F y = ∑
F y
eff
N = W = mg
N − W = 0 ∑ F x = ∑ (F x )eff
− F = ma − µ k mg =
a =−
k g
∑ M G = ∑ (M G )eff Fr = I α
(µ k mg )r = (23 mr 2 )α
α =
5
k g
2 r NOTE: As long as the sphere both rotates and slides, its linear and angular motions are uniformly accelerated. 16 - 21
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Vector Mechanics for Engineers: Dynamics Sample Problem 16.5 • Apply the kinematic relations for 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. v =v 0 + a t =v 0 + (−
)
k g t
5 k g t 2 r
ω = ω 0 + α t = 0 +
At the instant t 1 when the sphere stops sliding, a = − k g α =
5
k
2 r
v1 = r
1
5 k g t 1 2 r
v0 − µ k gt 1 = r
2 v0 5 µ k g t 5 µ k g 1 = 2 r 2 r 7 µ k g
ω 1 =
5 v0 7 r
v1 = r ω 1 = r
t 1 =
2 v0 7 µ k g
ω 1 =
5 v0 7 r
v1 = 75 v0
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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 θ, ω, and α , find P , N A, and N B. - kinematic analysis yields a x and a y . - application of d’Alembert’s principle yields P , N A, and N B.
16 - 23
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Vector Mechanics for Engineers: Dynamics 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 α
a n = r ω 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.
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Constrained Plane Motion: Rolling Motion • For a balanced disk constrained to roll without sliding, x = r θ → a = r α • Rolling, no sliding: F ≤ s N a = r α Rolling, sliding impending: F = s N a = r α Rotating and sliding: a, r α independent F = k N • For the geometric center of an unbalanced disk, aO = r α The acceleration of the mass center, r
r
r
aG = aO + aG O r
r
r
= aO + (aG O ) + (aG O ) t
16 - 25
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n
Vector Mechanics for Engineers: Dynamics Sample Problem 16.6 SOLUTION: • Draw the free-body-equation for AOB, expressing the equivalence of the external and effective forces. m E = 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/s 2. 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.
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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. 2 W E = (4 kg ) 9.81m s = 39.2 N
(
)
W OB = (3 kg ) 9.81m s 2 = 29.4 N 2
2 I E α = m E k E α = (4kg )(0.085 m) 40 rad s 2
= 1.156 N ⋅ m mOB (aOB )t = mOB (r α ) = (3 kg)(0.200 m) 40 rad s 2
m E = 4 kg
α = 40 rad s 2
k E = 85 mm
ω = 8 rad/s
= 24.0 N mOB (aOB )n = mOB r ω 2 = (3 kg)(0.200 m)(8 rad s)
mOB = 3 kg
2
= 38.4 N I OBα =
1 m L2 12 OB
)α = 121 (3kg)(0.400 m)2 40 rad s2 )
= 1.600 N ⋅ m 16 - 27
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Vector Mechanics for Engineers: Dynamics Sample Problem 16.6 • 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 O.
∑
O
= (∑
)
O eff
F (0.120m ) = I E α + mOB (aOB )t (0.200m) + I OBα
= 1.156 N ⋅ m + (24.0 N )(0.200m) + 1.600 N ⋅ m F = 63.0 N
∑ F x = (∑ F x )eff W E = 39.2 N W OB = 29.4 N I E α = 1.156 N ⋅ m
R x = mOB (aOB )t = 24.0 N
∑ F y = ∑ F y
R x = 24.0 N
eff
mOB (aOB )t = 24.0 N
R y − F − W E − W OB = mOB (aOB )
mOB (aOB )n = 38.4 N
R y − 63.0 N − 39.2 N − 29.4 N = 38.4 N
I OBα = 1.600 N ⋅ m
R y = 24.0 N
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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 of • Calculate the velocity after 10 ft of the coefficient of friction, b) the uniformly accelerated motion. velocity of G after the sphere has rolled 10 ft and c) the velocity of G if • Assuming no friction, calculate the linear acceleration down the incline and the the sphere were to move 10 ft down a corresponding velocity after 10 ft. frictionless incline. 16 - 29
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Vector Mechanics for Engineers: Dynamics 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.
∑
C =
∑(
)
C eff
(W sin θ )r = (ma )r + I α = (mr α )r +
a = r α
a = r α =
=
(52 mr 2 )α
W 2 W 2 r α = r α r + g 5 g 5 g sin 30°
α =
7
(
)
5 32.2 ft s 2 sin 30° 7
a = 11.50 ft s 2
5 sin θ 7 r
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Sample Problem 16.8 • Solve the three scalar equations derived from the free body-equation for the angular acceleration and the normal and tangential reactions at C.
∑ F x = ∑ (F x )eff
W sin θ − F = ma
=
W 5 g sin θ g
7
2 F = W sin 30° = 0.143W 7
∑ F y = ∑ α =
F y
eff
N − W cosθ = 0 N = W cos 30° = 0.866W
5 g sin θ 7r
a = r α = 11.50 ft s 2
• Calculate the friction coefficient required for the indicated tangential reaction at C . F = µ s =
s N
F N
=
0.143W s
0.866W
= 0.165
16 - 31
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Vector Mechanics for Engineers: Dynamics Sample Problem 16.8 • Calculate the velocity after 10 ft of uniformly accelerated motion. v 2 = v02 + 2a ( x − x0 )
(
)
= 0 + 2 11.50 ft s 2 (10 ft )
r
v = 15.17 ft s
• Assuming no friction, calculate the linear acceleration and the corresponding velocity after 10 ft.
∑ α =
5 g sin θ 7r
G
= ∑(
)
0 = I α
G eff
α = 0
W a g
∑ F x = ∑ (F x )eff
W sin θ = ma =
(
)
a = 32.2 ft s 2 sin 30° = 16.1ft s 2
a = r α = 11.50 ft s 2 v 2 = v02 + 2a ( x − x0 )
(
)
= 0 + 2 16.1ft s 2 (10 ft )
r
v = 17.94 ft s
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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 µ s = 0.20 and µ 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. 16 - 33
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Vector Mechanics for Engineers: Dynamics 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.
∑
C =
∑(
)
C eff
(200 N )(0.040 m) = ma (0.100 m) + I α
(
2
)
8.0 N ⋅ m = (50 kg )(0.100 m ) α + 0.245 kg ⋅ m 2 α 2
I = mk 2 = (50 kg )(0.70 m )
= 0.245 kg ⋅ m 2 Assume rolling without slipping, a = r α
= (0.100 m )α
α = +10.74 rad s 2
(
)
a = (0.100 m ) 10.74 rad s 2 = 1.074 m s 2
∑ F x = ∑ (F x )eff F + 200 N = ma = (50 kg ) 1.074 m s 2 F = −146.3 N
∑ F x = ∑ (F x )eff N − W = 0
(
)
N = mg = (50kg ) 1.074 m s 2 = +490.5 N
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Sample Problem 16.9 • Compare the required tangential reaction to the maximum possible friction force. F max =
s N =
0.20(490.5 N ) = 98.1 N
F > F max , rolling without slipping is impossible. • Calculate the friction force with slipping and solve the scalar equations for linear and angular accelerations.
Without slipping, F = −146.3 N
N = 490.5 N
F = F k =
k N =
0.15(490.5 N ) = 73.6 N
∑ F x = ∑ (F x )eff 200 N − 73.6 N = (50 kg )a
∑
G
= ∑(
a = 2.53 m s 2
)
G eff
(73.6 N )(0.100 m) − (200 N)(0.0.060 m)
(
)
= 0.245 kg ⋅ m 2 α α = −18.94 rad s 2 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
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α = 18.94 rad s 2 16 - 35
Vector Mechanics for Engineers: Dynamics 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-ft rod weighing 50 lb can move freely and with no friction along two straight tracks. The rod is released with no velocity from the position shown. Determine: a) the angular acceleration of the rod, and b) the reactions at A and B.
• 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.
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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 r r r a B = a A + a B A With a B A = 4α , the corresponding vector triangle and the law of signs yields a A = 5.46α
aB = 4.90α
The acceleration of G is now obtained from r
r
r
r
a = a G = a A + aG A where aG A = 2α Resolving into x and y components, a x = 5.46α − 2α cos 60° = 4.46α a y = −2α sin 60° = −1.732α 16 - 37
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Vector Mechanics for Engineers: Dynamics 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.
∑
E =
∑(
)
E eff
(50)(1.732) = (6.93α )(4.46) + ( 2.69α )(1.732) + 2.07α α = +2.30 rad s 2 α = 2.30 rad s 2 1 I = 12 ml 2 =
1
50 lb
12 32.2 ft s
2 ( ) 4 ft 2
= 2.07 lb ⋅ ft ⋅ s2 I α = 2.07α ma x =
50
(4.46α ) = 6.93α
32.2 50
ma y = −
32.2
(1.732α ) = −2.69α
∑ F x = ∑ (F x )eff R B sin 45° = (6.93)(2.30) R B = 22.5 lb
∑ F y = ∑
F y
r
R B = 22.5 lb eff
R A + (22.5) cos 45° − 50 = −(2.69)(2.30) R A = 27.9 lb
45o
