Ejercicios Resueltos Beer Jhonson

Universidad Nacional de San Cristobal deHuamanga

Facultad de Ingeniera Minas, Geologa y Civil

Escuela de Formacion Profesional de Ingeniera Civil

CURSO

DINAMICA (IC-244)

SOLUCION DE PROBLEMAS -

CINETICA DE UNA PARTICULA Y CUERPO

RIGIDOBeer - Jhonston

DOCENTE:

Ing. CASTRO PEREZ Cristian

ALUMNOS:

AYALA BIZARRO Rocky G.

CONTRERAS VENTURA Samir

VARGAS NAUPA Hilmar

ZARATE LAZO Dick F.

Ayacucho, Julio de 2013

Indice GeneralIndice General

Captulo 1 Problemas de Dinamica Pagina 1

1.1 Leyes de Newton Cuerpo Rgido 2Ejericio Nro 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Trabajo y Energa en Cuerpo Rgido 5Ejericio Nro 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Problema de Computadora . . . . . . . . . . . . . . . . . . . . . . . . . 7

Ejericio Nro 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Ejericio Nro 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Ejericio Nro 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Ejericio Nro 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Ejericio Nro 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Ejericio Nro 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Ejericio Nro 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Ejericio Nro 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Ejericio Nro 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Ejericio Nro 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Ejericio Nro 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Ejericio Nro 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Ejericio Nro 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

1ING. CIVILFG

F

r

x

y

z

x0

y0

z0

0

0

+y

+xN

f R

T

Mg

+

T0

mg

v

y

x

60

30

z

A

B

120 ft

vB

vA

C H A P T E R

691

Kinetics of Particles:

Newtons Second Law

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UNSCH

Newtons Second Law

N I N T H ED I T I O N

VECTOR MECHANICS

FOR ENGINEERS

Statics and Dynamics

Ferdinand P. Beer

Late of Lehigh University

E. Russell Johnston, Jr.

University of Connecticut

David F. Mazurek

U.S. Coast Guard Academy

Phillip J. Cornwell

Rose-Hulman Institute of Technology

Elliot R. Eisenberg

The Pennsylvania State University

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Ferdinand P. Beer

Late of Lehigh University

E.

Russell

Johnston, Jr.

University of Connecticut

DINMICA:

DINAMICA

FOR ENGINEERS

v

v0

y

x

z

Problemas de Dinamica

Dinamica IC-244 Problemas de Dinamica

1.1 Leyes de Newton Cuerpo Rgido

Ejercicio 1.1 El movimiento de una barra uniforme AB de 5kg de masa y longitud

L=750mm se gua por medio de dos ruedas pequenas de masa despreciable

que ruedan sobre la superficie mostrada. Si la barra se suelta desde

el reposo cuando q = 20, determine inmediatamente despues de laaceleracion a) la aceleracion angular de la barra y b) la reaccion en

A.(Problema Nro 122)

A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1

Problems 1069 16.120 The 4-lb uniform rod AB is attached to collars of negligible mass which may slide without friction along the xed rods shown. Rod AB

is at rest in the position u 5 25 when a horizontal force P is applied

to collar A, causing it to start moving to the left with an acceleration

of 12 ft/s2. Determine (a) the force P, (b) the reaction at B.

25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124

16.121 The 4-lb uniform rod AB is attached to collars of negligible mass

which may slide without friction along the xed rods shown. If rod

AB is released from rest in the position u 5 25, determine imme-

diately after release (a) the angular acceleration of the rod, (b) the

reaction at B.

16.122 The motion of the uniform rod AB of mass 5 kg and length L 5

750 mm is guided by small wheels of negligible mass that roll on

the surface shown. If the rod is released from rest when u 5 20,

determine immediately after release (a) the angular acceleration of

the rod, (b) the reaction at A.

16.123 End A of the 8-kg uniform rod AB is attached to a collar that can

slide without friction on a vertical rod. End B of the rod is attached

to a vertical cable BC. If the rod is released from rest in the posi-

tion shown, determine immediately after release (a) the angular

acceleration of the rod, (b) the reaction at A.

L = 750 mm

30 = q

A

B

C

Fig. P16.123

16.124 The 4-kg uniform rod ABD is attached to the crank BC and is t-

ted 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.


Solucion:

Datos

m = 5kg

L = 750mm

q = 20

LA ACELERACION RELATIVA SERA

aB = aA + aA/B

[aB ] = [aA] + [L.]aA/B = L.

En las siguientes imagenes se vera de donde salen la aB, aA y aA/B.

Ingeniera Civil2

Ing. Civil


DIAGRAMA DE ACELERACIONES

A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1





25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124





reaction at B.











L = 750 mm

30 = q

A

B

C

Fig. P16.123







60

750

PROBLEMS

1063

16.75 Show that the couple IA of Fig. 16.15 can be eliminated by attach-

ing the vectors mat and ma n at a point P called the center of per-

cussion, located on line OG at a distance GP 5 k2/r from the mass

center of the body.

16.76 A uniform slender rod of length L 5 36 in. and weight W 5 4 lb hangs

freely from a hinge at A. If a force P of magnitude 1.5 lb is applied

at B horizontally to the left (h 5 L), determine (a) the angular

acceleration of the rod, (b) the components of the reaction at A.

G

a

r

mat

ma n

P

O

Fig. P16.75

C

G

B

A

P

L

2

L

2

r

Fig. P16.78

A

B

h

L

P

Fig. P16.76

A

A'

l

x

w

Fig. P16.80

16.77 In Prob. 16.76, determine (a) the distance h for which the hori-

zontal component of the reaction at A is zero, (b) the correspond-

ing angular acceleration of the rod.

16.78 A uniform slender rod of length L 5 900 mm and mass m 5 4 kg

is suspended from a hinge at C. A horizontal force P of magnitude

75 N is applied at end B. Knowing that r 5 225 mm, determine

(a) the angular acceleration of the rod, (b) the components of the

reaction at C.

16.79 In Prob. 16.78, determine (a) the distance r for which the hori-

zontal component of the reaction at C is zero, (b) the correspond-


16.80 A uniform slender rod of length l and mass m rotates about a verti-

cal axis AA9 with a constant angular velocity V . Determine the

tension in the rod at a distance x from the axis of rotation.

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aA

20

aB

GA

B

750

20

aA/B

GA

B

20

Punto Fijo

Fig. P16.120 and P16.121

Aplicando Ley de Senos

A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1

Problems 106916.120 The 4-lb uniform rod AB is attached to collars of negligible masswhich may slide without friction along the xed rods shown. Rod AB




25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124



AB is released from rest in the position u5 25, determine imme-


reaction at B.











L = 750 mm

30 = q

A

B

C

Fig. P16.123







60

750

PROBLEMS

1063


ing the vectors mat and man at a point P called the center of per-


center of the body.

16.76 A uniform slender rod of length L 5 36 in. and weight W5 4 lb hangs




G

a

r

mat

ma n

P

O

Fig. P16.75

C

G

B

A

P

L

2

L

2

r

Fig. P16.78

A

B

h

L

P

Fig. P16.76

A

A'

l

x

w

Fig. P16.80








reaction at C.








aA

20

aB

GA

B

750

20

aA/B

GA

B

20

Punto Fijo

Fig. P16.120 and P16.121

aA/B

aA

aB

60

50

70

aBsin70 =

aAsin50 =

L.sin60

aA = 1.085LaB = 0.884L

Descomponiendo las ecuaciones de la aceleracion en coordenadas xy

ax = 1.085Lcos60 + 0.5L sin20

ax = 0.542L+ 0.171Lax = 0.7135L (1.1)

ay = 1.085L sin60 0.5Lcos20ay = 0.938L+ 0.469L

ay = 0.469L (1.2)

DIAGRAMAS DE CUERPO LIBRE

En el triangulo ABE

] ABE = 70] BAE = 50

Ingeniera Civil3

Ing. Civil


A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1





25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124





reaction at B.











L = 750 mm

30 = q

A

B

C

Fig. P16.123







60

750

PROBLEMS

1063




center of the body.





G

a

r

mat

ma n

P

O

Fig. P16.75

C

G

B

A

P

L

2

L

2

r

Fig. P16.78

A

B

h

L

P

Fig. P16.76

A

A'

l

x

w

Fig. P16.80








reaction at C.








aA

20

aB

GA

B

750

20

aA/B

GA

B

20

Punto Fijo

Fig. P16.120 and P16.121

aA/B

aA

aB

60

50

70

750

W=mg

A

B

1031stated, without proof (Sec. 3.3), that the conditions of equilibrium or

motion of a rigid body remain unchanged if a force F acting at a given

point of the rigid body is replaced by a force F9 of the same magni-

tude and same direction, but acting at a different point, provided

that the two forces have the same line of action. But since F and F9

have the same moment about any given point, it is clear that they

form two equipollent systems of external forces. Thus, we may now

prove, as a result of what we established in the preceding section,

that F and F9 have the same effect on the rigid body (Fig. 3.3).

The principle of transmissibility can therefore be removed from

the list of axioms required for the study of the mechanics of rigid

bodies. These axioms are reduced to the parallelogram law of addi-

tion of vectors and to Newtons laws of motion.

16.6 SOLUTION OF PROBLEMS INVOLVING

THE MOTION OF A RIGID BODY

We saw in Sec. 16.4 that when a rigid body is in plane motion, there

exists a fundamental relation between the forces F1, F2, F3, . . . , acting

on the body, the acceleration a of its mass center, and the angular

acceleration A of the body. This relation, which is represented in Fig.

16.7 in the form of a free-body-diagram equation, can be used to deter-

mine the acceleration a and the angular acceleration A produced by a

given system of forces acting on a rigid body or, conversely, to deter-

mine the forces which produce a given motion of the rigid body.

The three algebraic equations (16.6) can be used to solve prob-

lems of plane motion. However, our experience in statics suggests

that the solution of many problems involving rigid bodies could be

simplied by an appropriate choice of the point about which the

moments of the forces are computed. It is therefore preferable to

remember the relation existing between the forces and the accelera-

tions in the pictorial form shown in Fig. 16.7 and to derive from this

fundamental relation the component or moment equations which t

best the solution of the problem under consideration.

The fundamental relation shown in Fig. 16.7 can be presented

in an alternative form if we add to the external forces an inertia vec-

tor 2m a of sense opposite to that of a, attached at G, and an inertia

couple 2 IA of moment equal in magnitude to Ia and of sense oppo-

site to that of A (Fig. 16.10). The system obtained is equivalent to

zero, and the rigid body is said to be in dynamic equilibrium.

Whether the principle of equivalence of external and effective

forces is directly applied, as in Fig. 16.7, or whether the concept of dy-

namic equilibrium is introduced, as in Fig. 16.10, the use of free-body-

diagram equations showing vectorially the relationship existing between

the forces applied on the rigid body and the resulting linear and angular

accelerations presents considerable advantages over the blind application

of formulas (16.6). These advantages can be summarized as follows:

1. The use of a pictorial representation provides a much clearer under-

standing of the effect of the forces on the motion of the body.

We recall that the last of Eqs. (16.6) is valid only in the case of the plane motion of

a rigid body symmetrical with respect to the reference plane. In all other cases, the

methods of Chap. 18 should be used.

Fig. 16.10

F1

F2

= 0F3

F4

am

aa

aI G

16.6 Solution of Problems Involving the

Motion of a Rigid Body

Fig. 3.3 (repeated)

F

=F'


60

20

100 mm

0.469

0.17

60

20

0.469

0.17

Iamax

may

0.714

A

B

EE

Ley de senos

BEsin50 =

Lsin60

BE = 0.885L

Suma de Momentos

ME =

(ME)

mg(0.4698L) = I+max(0.714) +may(0.469)

0.469mgL= 112mL2 +m(0.714)(0.714) +m(0.469)(0.469)

0.469mgL=mL2(0.813)a (1.3)

Calculando

= 0.5778 gL= 0.57789.810.75

= 7.56rad/s (1.4)

Ingeniera Civil4

Ing. Civil


CALCULANDO LA REACCION EN A

Fx =

(Fx)

Asin60 =maxAsin60 =ma0.714Asin60 = (5)(0.714)(0.75)(7.557)

A= 23.3rad/s (1.5)

1.2 Trabajo y Energa en Cuerpo Rgido

Ejercicio 1.2 El disco A con peso de 10lb y radio r=6in, se encuentra en reposo cuando

esta en contacto con la banda BC, la cual se mueve hacia la derecha con

velocidad constante V= 40ft/s. Si k = 0.20 entre el disco y la bandadetermine el numero de revoluciones ejecutadas por el disco antes de

alcanzar una velocidad angular constante.(Problema Nro 17.8)

A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1





25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124





reaction at B.











L = 750 mm

30 = q

A

B

C

Fig. P16.123







60

750

PROBLEMS

1063




center of the body.





G

a

r

mat

ma n

P

O

Fig. P16.75

C

G

B

A

P

L

2

L

2

r

Fig. P16.78

A

B

h

L

P

Fig. P16.76

A

A'

l

x

w

Fig. P16.80








reaction at C.








aA

20

aB

GA

B

750

20

aA/B

GA

B

20

Punto Fijo

Fig. P16.120 and P16.121

aA/B

aA

aB

60

50

70

750

W=mg

A

B



















































Fig. 16.10

F1

F2

= 0F3

F4

am

aa

aI G



Fig. 3.3 (repeated)

F

=F'


60

20

100 mm

0.469

0.17

60

20

0.469

0.17

Iamax

may

0.714

A

B

EE

1096 Plane Motion of Rigid Bodies: Energy and Momentum Methods

17.7 Disk A is of constant thickness and is at rest when it is placed in

contact with belt BC , which moves with a constant velocity v .

Denoting by mk the coefcient of kinetic friction between the disk

and the belt, derive an expression for the number of revolutions

executed by the disk before it attains a constant angular velocity.

17.8 Disk A, of weight 10 lb and radius r 5 6 in., is at rest when it is

placed in contact with belt BC, which moves to the right with a

constant speed v 5 40 ft/s. Knowing that m 5k 0.20 between the

disk and the belt, determine the number of revolutions executed

by the disk before it attains a constant angular velocity.

17.9 Each of the gears A and B has a mass of 2.4 kg and a radius of gyra-

tion of 60 mm, while gear C has a mass of 12 kg and a radius of

gyration of 150 mm. A couple M of constant magnitude 10 N ? m is

applied to gear C. Determine (a) the number of revolutions of gear

C required for its angular velocity to increase from 100 to 450 rpm,

(b) the corresponding tangential force acting on gear A.

17.10 Solve Prob. 17.9, assuming that the 10-N ? m couple is applied to

gear B.

17.11 The double pulley shown weighs 30 lb and has a centroidal radius

of gyration of 6.5 in. Cylinder A and block B are attached to cords

that are wrapped on the pulleys as shown. The coefcient of

kinetic friction between block B and the surface is 0.25. Knowing

that the system is released from rest in the position shown, deter-

mine (a) the velocity of cylinder A as it strikes the ground, (b) the

total distance that block B moves before coming to rest.

B

rA

C

v

Fig. P17.7 and P17.8

A B

80 mm 80 mm

200 mmC

M

Fig. P17.9

A

C

3 ft

25 lb

B

20 lb

10 in.

6 in.

Fig. P17.11

P

10 in.

15 in.

A

B

C

D

6 in.

8 in.

Fig. P17.12

17.12 The 8-in.-radius brake drum is attached to a larger ywheel that

is not shown. The total mass moment of inertia of the ywheel and

drum is 14 lb ? ft ? s2 and the coefcient of kinetic friction between

the drum and the brake shoe is 0.35. Knowing that the initial

angular velocity of the ywheel is 360 rpm counterclockwise,

determine the vertical force P that must be applied to the pedal

C if the system is to stop in 100 revolutions.


B

rA

C

v

rA

C

W=mg

N=mg

F=usmgSolucion:

Datos

o = 0

r = 6in

V = 40ft/s

s = 0.20

Ingeniera Civil5

Ing. Civil


La unica fuerza que hace el trabajo es F . el momento en A es M = rF :

U12 =MU12 = rFU12 = r(smg)

La velocidad angular se vuelve constante cuando o2 =V

r:

consideramos a T como el trabajo realizado

A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1





25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124





reaction at B.











L = 750 mm

30 = q

A

B

C

Fig. P16.123







60

750

PROBLEMS

1063




center of the body.





G

a

r

mat

ma n

P

O

Fig. P16.75

C

G

B

A

P

L

2

L

2

r

Fig. P16.78

A

B

h

L

P

Fig. P16.76

A

A'

l

x

w

Fig. P16.80








reaction at C.








aA

20

aB

GA

B

750

20

aA/B

GA

B

20

Punto Fijo

Fig. P16.120 and P16.121

aA/B

aA

aB

60

50

70

750

W=mg

A

B



















































Fig. 16.10

F1

F2

= 0F3

F4

am

aa

aI G



Fig. 3.3 (repeated)

F

=F'


60

20

100 mm

0.469

0.17

60

20

0.469

0.17

Iamax

may

0.714

A

B

EE



















gear B.








B

rA

C

v


A B

80 mm 80 mm

200 mmC

M

Fig. P17.9

A

C

3 ft

25 lb

B

20 lb

10 in.

6 in.

Fig. P17.11

P

10 in.

15 in.

A

B

C

D

6 in.

8 in.

Fig. P17.12









B

rA

C

v

rA

C

W=mg

N=mg

F=usmg

T1 = 0

T2 =12(12mr

2)(

Vr

)2T2 =

mV 2

4

T1+U12 = T2 (2.6)

0+r(smg) =mV 2

4 (2.7)

=mV 2

4r(smg)rad (2.8)

=V 2

8pir(sg)rev (2.9)

PARA r=6in , s= 0.20, V=40ft/s

= 402

8pir(0.6)(32.2)

= 19.77rev (2.10)

Ingeniera Civil6

Ing. Civil


Ejercicio 1.3 Cada una de las dos barras ligeras e identicas que se muestran tiene

una longitud L=30in. Si el sistema se suelta desde el reposo cuando

las barras estan horizontales, utilice software para calcular y graficas

la velocidad angular de la barra BC y la velocidad del punto AB para

valores de desde 0 hasta 90(Problema de computadora Nro 17.C5)

A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1





25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124





reaction at B.











L = 750 mm

30 = q

A

B

C

Fig. P16.123







60

750

PROBLEMS

1063




center of the body.





G

a

r

mat

ma n

P

O

Fig. P16.75

C

G

B

A

P

L

2

L

2

r

Fig. P16.78

A

B

h

L

P

Fig. P16.76

A

A'

l

x

w

Fig. P16.80








reaction at C.








aA

20

aB

GA

B

750

20

aA/B

GA

B

20

Punto Fijo

Fig. P16.120 and P16.121

aA/B

aA

aB

60

50

70

750

W=mg

A

B



















































Fig. 16.10

F1

F2

= 0F3

F4

am

aa

aI G



Fig. 3.3 (repeated)

F

=F'


60

20

100 mm

0.469

0.17

60

20

0.469

0.17

Iamax

may

0.714

A

B

EE



















gear B.








B

rA

C

v


A B

80 mm 80 mm

200 mmC

M

Fig. P17.9

A

C

3 ft

25 lb

B

20 lb

10 in.

6 in.

Fig. P17.11

P

10 in.

15 in.

A

B

C

D

6 in.

8 in.

Fig. P17.12









B

rA

C

v

rA

C

W=mg

N=mg

F=usmg

1143Computer Problems 17.C4 Collar C has a mass of 2.5 kg and can slide without friction on rod AB. A spring of constant 750 N/m and an unstretched length r0 5

500 mm is attached as shown to the collar and to the hub B. The total mass

moment of inertia of the rod, hub, and spring is known to be 0.3 kg2

? m

about B. Initially the collar is held at a distance of 500 mm from the axis

of rotation by a small pin protruding from the rod. The pin is suddenly

removed as the assembly is rotating in a horizontal plane with an angular

velocity V 0 of 10 rad/s. Denoting by r the distance of the collar from the

axis of rotation, use computational software to calculate and plot the angular

velocity of the assembly and the velocity of the collar relative to the rod for

values of r from 500 to 700 mm. Determine the maximum value of r in the

ensuing motion.

17.C5 Each of the two identical slender bars shown has a length L 5 30 in.

Knowing that the system is released from rest when the bars are horizontal,

use computational software to calculate and plot the angular velocity of rod

AB and the velocity of point D for values of u from 0 to 90.

A

B

r0

w0

C

Fig. P17.C4

A

B

Dq

LL

Fig. P17.C5


A

B

Dq

LL

q

B

q

L

A

B

q

LB

q

L

L L L LLLLL/2 LLLL/2 LLLL/2 LLLL/2

WAB=w

W=mg

W=mg W=mg

AB

D

W=mg

LLLL

h=L/2(sen(q))

qA

B

D

q

LLLL

h=L/2(sen(q))

qA

B

D

vAB=wL/2

WBD=w

WBD=w

C

vB=wL/2

vAB=wL/2

vBD=wL/2

vD=(CD)w

EED=L/2

CENTRO INSTANTNEO

DE ROTACIN

Solucion:

Datos

L = 30in

Las barras parten desde el reposo cuando = 0 primera imagen de la izquierda,hallamos la Vd para angulos desde = 0 hasta = 90 usando incrementos de = 10

A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1





25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124





reaction at B.











L = 750 mm

30 = q

A

B

C

Fig. P16.123







60

750

PROBLEMS

1063




center of the body.





G

a

r

mat

ma n

P

O

Fig. P16.75

C

G

B

A

P

L

2

L

2

r

Fig. P16.78

A

B

h

L

P

Fig. P16.76

A

A'

l

x

w

Fig. P16.80








reaction at C.








aA

20

aB

GA

B

750

20

aA/B

GA

B

20

Punto Fijo

Fig. P16.120 and P16.121

aA/B

aA

aB

60

50

70

750

W=mg

A

B



















































Fig. 16.10

F1

F2

= 0F3

F4

am

aa

aI G



Fig. 3.3 (repeated)

F

=F'


60

20

100 mm

0.469

0.17

60

20

0.469

0.17

Iamax

may

0.714

A

B

EE



















gear B.








B

rA

C

v


A B

80 mm 80 mm

200 mmC

M

Fig. P17.9

A

C

3 ft

25 lb

B

20 lb

10 in.

6 in.

Fig. P17.11

P

10 in.

15 in.

A

B

C

D

6 in.

8 in.

Fig. P17.12









B

rA

C

v

rA

C

W=mg

N=mg

F=usmg

1143Computer Problems 17.C4 Collar C has a mass of 2.5 kg and can slide without friction on rod AB. A spring of constant 750 N/m and an unstretched length r0 5

500 mm is attached as shown to the collar and to the hub B. The total mass

moment of inertia of the rod, hub, and spring is known to be 0.3 kg2

? m

about B. Initially the collar is held at a distance of 500 mm from the axis

of rotation by a small pin protruding from the rod. The pin is suddenly

removed as the assembly is rotating in a horizontal plane with an angular

velocity V 0 of 10 rad/s. Denoting by r the distance of the collar from the

axis of rotation, use computational software to calculate and plot the angular

velocity of the assembly and the velocity of the collar relative to the rod for

values of r from 500 to 700 mm. Determine the maximum value of r in the

ensuing motion.

17.C5 Each of the two identical slender bars shown has a length L 5 30 in.

Knowing that the system is released from rest when the bars are horizontal,

use computational software to calculate and plot the angular velocity of rod

AB and the velocity of point D for values of u from 0 to 90.

A

B

r0

w0

C

Fig. P17.C4

A

B

Dq

LL

Fig. P17.C5


A

B

Dq

LL

q

B

q

L

A

B

q

LB

q

L

L L L LLLLL/2 LLLL/2 LLLL/2 LLLL/2

WAB=w

W=mg

W=mg W=mg

AB

D

W=mg

LLLL

h=L/2(sen(q))

qA

B

D

q

LLLL

h=L/2(sen(q))

qA

B

D

vAB=wL/2

WBD=w

WBD=w

C

vB=wL/2

vAB=wL/2

vBD=wL/2

vD=(CD)w

EED=L/2

CENTRO INSTANTNEO

DE ROTACIN

Dimensiones:

AB = BC = L

Ingeniera Civil7

Ing. Civil


A B

(v ) = 96 km/hB 0(v ) = 0A 0

x

d

B

(v ) = 62 mi /hB 0d

(v ) = 96 km/hB 1(v ) = 5aA 1 A

90m

v = vA B

t = 5+t1

t = 5s

A

(v ) = 0A 0

t = 0t = 5s

B

(v ) = 96 km/hB 0

t = 0

desacelera

90m

x

1.0

0.5

0

0.5

1.0

0.2 0.40.6

y/y /x11

x

y

xz

2y

2A

2x

2A

2z

2B = 1





25 in.

q 70

A

B

Fig. P16.120 and P16.121

60 q

L

A

B

200 mm

200 mm

100 mm

A

B

D

C

Fig. P16.124





reaction at B.











L = 750 mm

30 = q

A

B

C

Fig. P16.123







60

750

PROBLEMS

1063




center of the body.





G

a

r

mat

ma n

P

O

Fig. P16.75

C

G

B

A

P

L

2

L

2

r

Fig. P16.78

A

B

h

L

P

Fig. P16.76

A

A'

l

x

w

Fig. P16.80








reaction at C.








aA

20

aB

GA

B

750

20

aA/B

GA

B

20

Punto Fijo

Fig. P16.120 and P16.121

aA/B

aA

aB

60

50

70

750

W=mg

A

B




































Ejercicios Resueltos Beer Jhonson

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Transcript of Ejercicios Resueltos Beer Jhonson