Eligheor

Mediante el método de los nodos: Dibujando el diagrama de cuerpo libre:

Aplicando las ecuaciones de equilibrio para calcular las reacciones en B y C obtenemos que:

𝐹! = 0 , 𝑅!! = 0

𝐹! = 0 , 𝑅!! + 𝑅!! − 945 = 0

𝑅!! + 𝑅!! = 945 (𝑎)

!!!

945!!"! !!

!! !!

!!!

! !

!!!

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 4, Solution 19.

Free-Body Diagram:

(a) From free-body diagram of lever BCD

( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB

T∴ =(b) From free-body diagram of lever BCD

( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

CO

SM

OS

: C

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te O

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on

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izat

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tics

and

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erd

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eer,

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iot

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ure

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l ©

20

07

Th

e M

cGra

w-H

ill

Co

mp

anie

s.

Ch

ap

ter

4, S

olu

tion

19.

Fre

e-B

od

y D

iag

ram

:

(a)

Fro

m f

ree-

bo

dy

dia

gra

m o

f le

ver

BCD

()

()

0:

50

mm

200

N75 m

m0

CAB

MT

Σ=

−=

300

AB

T∴

=(b

) F

rom

fre

e-b

od

y d

iag

ram

of

lev

er BCD

()

0:

200

N0.6

300

N0

xx

FC

Σ=

++

=

3

80

N

o

r3

80

Nx

xC

∴=

−=

C

()

0:

0.8

30

0 N

0y

yF

CΣ

=+

=

N

240

or

N 240

=−

=∴

yy

CC

Th

en

()

()

22

22

38

02

40

449

.44

Nx

yC

CC

=+

=+

=

and

°

=⎟ ⎠⎞

⎜ ⎝⎛

−−=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

=−

−276

.32

380

240

tan

tan

11

xy

CCθ

or

4

49

N=

C32.3

°▹

𝑀! = 0:

−12 945 + 12+ 3,75 𝑅!! = 0

𝑅!! = 12 94512+ 3,75 = 720 𝑙𝑏 (𝑏)

Ahora de (b) en (a) tenemos:

𝑅!! + 𝑅!! = 945

𝑅!! + 720 = 945

𝑅!! = 945− 720 = 225 𝑙𝑏 Aplicando el método de los nodos para obtener las fuerzas internas en los elementos BA, AC y BC, y así poder conocer si están en tracción o compresión En el nodo B tenemos:

tan𝛼 =

𝐶𝑂𝐶𝐴 =

912 =

34

𝛼 = tan!!34 = 36,87°




Free-Body Diagram:


( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB


( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

!! !!"

!!"

225!!"!

36,87°

Aplicando las ecuaciones de equilibrio obtenemos que:

𝐹! = 0 , − 𝐹!" + 𝐹!" cos 36,87 = 0 (𝑎)

𝐹! = 0 , 225 + 𝐹!" sin 36,87 = 0

𝐹!" sin 36,87 = −225

𝐹!" =−225

sin 36,87 = −375 𝑙𝑏,

𝐹!" = 375 𝑙𝑏 𝑏 , 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐵𝐴 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛

Ahora de (b) en (a) tenemos:

−𝐹!" + 𝐹!" cos 36,87 = 0

−𝐹!" + 375 cos 36,87 = 0

𝐹!" = 375 cos 36,87 = 300 𝑙𝑏

𝐹!! = 300 𝑙𝑏 , 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐵𝐶 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛 En el nodo C tenemos que:

tan𝛽 =

𝐶𝑂𝐶𝐴 =

93,75 =

125

𝛽 = tan!!125 = 67,38°




Free-Body Diagram:


( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB


( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

CO

SM

OS

: C

om

ple

te O

nli

ne

So

luti

on

s M

anu

al O

rgan

izat

ion

Sy

stem

V

ecto

r M

echa

nic

s fo

r E

ng

inee

rs:

Sta

tics

and

Dyn

am

ics,

8/e

, F

erd

inan

d P

. B

eer,

E.

Ru

ssel

l Jo

hn

sto

n,

Jr.,

Ell

iot

R.

Eis

enb

erg

, W

illi

am E

. C

lau

sen

, D

avid

Maz

ure

k,

Ph

illi

p J

. C

orn

wel

l ©

20

07

Th

e M

cGra

w-H

ill

Co

mp

anie

s.

Ch

ap

ter

4, S

olu

tion

19.

Fre

e-B

od

y D

iag

ram

:

(a)

Fro

m f

ree-

bo

dy

dia

gra

m o

f le

ver

BCD

()

()

0:

50

mm

200

N75 m

m0

CAB

MT

Σ=

−=

300

AB

T∴

=(b

) F

rom

fre

e-b

od

y d

iag

ram

of

lev

er BCD

()

0:

200

N0.6

300

N0

xx

FC

Σ=

++

=

3

80

N

o

r3

80

Nx

xC

∴=

−=

C

()

0:

0.8

30

0 N

0y

yF

CΣ

=+

=

N

240

or

N 240

=−

=∴

yy

CC

Th

en

()

()

22

22

38

02

40

449

.44

Nx

yC

CC

=+

=+

=

and

°

=⎟ ⎠⎞

⎜ ⎝⎛

−−=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

=−

−276

.32

380

240

tan

tan

11

xy

CCθ

or

4

49

N=

C32.3

°▹

!!

!!"

!!"

720!!"!

67,38°

Aplicando las ecuaciones de equilibrio tenemos que:

𝐹! = 0 , 𝐹!" − 𝐹!" cos 67,38 = 0

𝐹!" cos 67,38 = 𝐹!"

𝐹!" = 𝐹!"

cos 67,38 = 300

cos 67,38 = 780 𝑙𝑏

𝐹!" = 780 𝑙𝑏 , 𝑻 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐶𝐴 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛




Free-Body Diagram:


( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB


( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

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