Dibujamos el diagrama de cuerpo libre:

Llevamos las medidas de mm a metros:

280  𝑚𝑚 = 0,28  𝑚 180 = 0,18  𝑚 100 = 0,10  𝑚

Aplicando las ecuaciones de equilibrio obtenemos:

𝑀! = 0:          − 𝐴 0,18 + 150 sin 30 0,10 +   150 cos 30 0,28 = 0

0,28!!

0,18!!

0,10!!

30°

!

!

150!!

!!

!!

!

!

!

!!

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°▹

𝐴 =  150 sin 30 0,10 +   150 cos 30 0,28

0,18 = 𝟐𝟒𝟑,𝟕𝟒  𝑵

       𝑜                    𝐴 = 244  𝑁   →    

𝐹! = 0:                    243,74+ 150 sin 30+  𝐷! = 0

𝐷! = −243,74− 150 sin 30 = −𝟑𝟏𝟖,𝟕𝟒  𝑵

𝐹! = 0:                  𝐷! − 150 cos 30 = 0

𝐷! =   150 cos 30 = 𝟏𝟐𝟗,𝟗𝟎𝟒  𝑵

∴      𝐷 =   𝐷!! +  𝐷!!  =   −318,74 ! +   129,904 ! = 𝟑𝟒𝟒,𝟐𝟎  𝑵

𝑦                            𝜃 =   tan!!𝐷!𝐷!

=   tan!!129,904−318,74 = −𝟐𝟐,𝟏𝟕𝟒°

𝑜                𝐷 =    344  𝑁 𝜃 = 22,2°

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°▹

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°▹

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 21.

Free-Body Diagram:

(a)

( ) 0in.9.0cos

in.2.4:0 =−⎟

⎠

⎞⎜⎝

⎛−=Σ spΒx FAΜα

or ( )8lb 1.2 in.

cos30sp

F kx k= = =°

Solving for k:

7.69800 lb/in.k = 7.70 lb/in.k = ▹

(b)

( ) 8 lb0: 3 lb sin30 0

cos30x x

F B⎛ ⎞Σ = ° + + =⎜ ⎟°⎝ ⎠

or 10.7376 lbx

B = −

( )0: 3 lb cos30 0y y

F BΣ = − ° + =

or 2.5981 lby

B =

( ) ( )2 210.7376 2.5981 11.0475 lb,B = − + = and

1 2.5981tan 13.6020

10.7376θ −= = °

Therefore: 11.05 lb=B 13.60° ▹

Dibujamos el diagrama de cuerpo libre:

Aplicando las ecuaciones de equilibrio obtenemos:

𝑀! = 0:                      𝑇 2𝑎 + 𝑎 cos𝜃 −  𝑇𝑎 + 𝑃𝑎 = 0

𝑇 =𝑷

𝟏+  𝐜𝐨𝐬𝜽              (𝑎)

!! !

2!+ !

cos!

!

! !

!!

!!

!!

!!

!! !!

!!

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°▹

𝐹! = 0:                    𝐶! − 𝑇 sin𝜃 = 0

𝐶! =  𝑻 𝐬𝐢𝐧𝜽          (𝑏) De la ecuación (a) en la ecuación (b) se tiene que:

𝐶! =  𝑷 𝐬𝐢𝐧𝜽𝟏+  𝐜𝐨𝐬𝜽            (𝑐)

𝐹! = 0:                    𝐶! + 𝑇 + 𝑇 cos𝜃 − 𝑃 = 0

𝐶! =  𝑷− 𝑻 𝟏+ 𝐜𝐨𝐬𝜽              (𝑑) De la ecuación (a) en la ecuación (d) se tiene que:

𝐶! =  𝑃 −𝑃 1+ cos𝜃1+  cos𝜃 = 0

𝐶! = 0    ,                𝐶 =  𝐶!

𝐶 =  𝑷 𝐬𝐢𝐧𝜽𝟏+  𝐜𝐨𝐬𝜽          (𝑒)

𝑃𝑎𝑟𝑎    𝜃 = 60°    𝑎  𝑡𝑟𝑎𝑣𝑒𝑠  𝑑𝑒𝑙  𝑒𝑛𝑢𝑛𝑐𝑖𝑎𝑑𝑜 De la ecuación (a) se tiene que:

𝑇 =𝑃

1+  cos𝜃 =  𝑃

1+  cos 60 =  𝑃

1+  12=  

𝟐𝟑   𝑷

De la ecuación (e) se tiene que:

𝐶 =  𝑃 sin𝜃1+  cos𝜃 =  

𝑃 sin 601+  cos 60 =  

𝑃 0,87

1+  12=     𝟎,𝟓𝟖 𝑷

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°▹

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°▹