Mediante	
  el	
  método	
  de	
  los	
  nodos:	
  
	
  
Dibujando	
  el	
  diagrama	
  de	
  cuerpo	
  libre:	
  
...
Chapter 4, Solution 19.
Free-Body Diagram:

	
  

(a) From free-body diagram of lever BCD

ΣM C = 0: TAB ( 50 mm ) − 200 N...
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟=3
⎜
Cx ⎟
⎝ − 380 ⎠
⎠
⎝

( 380 )2 + ( 24

or

ΣFy = 0: C y + 0.8 ( 300 N ) = 0

or

ΣFx = 0: ...
Free-Body Diagram:
(a) From free-body diagram of lever BCD

ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0

Aplicando	
  la...
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Eligheor

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Eligheor

  1. 1.     Mediante  el  método  de  los  nodos:     Dibujando  el  diagrama  de  cuerpo  libre:     or C = 449 N ! ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ ( 380 )2 + ( 240 )2 = 449.44 N ! or ΣFy = 0: C y + 0.8 ( 300 N ) = 0 C y = 240 N C x = 380 N !!! or Chapter 4, Solution 19. ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 !!! !! COSMOS: Complete Online Solutions Manual Organization System Free-Body Diagram: 945!!"! 32.3° ▹ ∴ TAB = 300 !! !! !!! and θ = tan −1 ⎜ ⎜ C = ΣFy = 0: C y + 0.8 ( 300 N ) = 0 𝐹! = 0      , 𝑅! +   𝑅! − 945   =  0     ∴ C y!= −240 ! N or C y = 240 N 2 𝑅!! =   C!!+= y = ( 380 )2 + ( 240 )2 = 449.44 N  945          (𝑎)   C + 𝑅x C 2 and Then 2 2 Cx + C y = ∴ C y = −240 N ∴ C x = −380 N ΣFx = 0: 0      , N + Cx + = 0  300 N ) = 0 𝐹! = 200 𝑅!! 0.6 ( ∴ C x = −380 N or C x = 380 N Then   ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ θ = tan −1 ⎜ ⎜ or C = 449 N ram: tion 19. nline Solutions Manual Organization System   (b) From free-body diagram of lever BCD (a) From free-body diagram of lever BCD   (a) From free-body diagram of lever BCD   Aplicando  las  ecuaciones  de  equilibrio  para  calcular  las  (reacciones  en  B  y  C   ΣM C = 0: TAB ( 50 mm ) − 200 N 75 mm ) = 0 obtenemos  que:   ∴ TAB = 300   (b) From free-body diagram of lever BCD   32.3° ▹
  2. 2. Chapter 4, Solution 19. Free-Body Diagram:   (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0:         (b) From free-body diagram of lever BCD ΣFx 3,75 N C + −12 945 + 12 += 0: 200𝑅!! + =x  0  0.6 ( 300 N ) = 0 ∴ C x = −380 N   𝑅!! =     Ahora  de  (b)    en    (a)  tenemos:   Then     ∴ TAB = 300 and C x = 380 N or 12 945 = 0: C + 0.8 300 N = 0 ΣFy = 720  𝑙𝑏        (𝑏)   ( ) y 12 + 3,75 ∴ C y = −240 N C = 2 2 Cx + C y = C y = 240 N or ( 380 )2 + ( 240 )2 = 449.44 N 𝑅!! +   𝑅!! = ⎛  945   Cy ⎞ −1 − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ θ = tan ⎜ ⎜ 𝑅!! +  720 =  945     or C = 449 N 32.3° ▹ 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:     !!" 36,87° !! !!" 225!!"!   𝐶𝑂 9 3 tan 𝛼 = = =     𝐶𝐴 12 4 𝛼 =   tan!! 3 = 36,87°   4   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.  
  3. 3. ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟=3 ⎜ Cx ⎟ ⎝ − 380 ⎠ ⎠ ⎝ ( 380 )2 + ( 24 or ΣFy = 0: C y + 0.8 ( 300 N ) = 0 or ΣFx = 0: 200 N + Cx + 0.6 ( 300 (b) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( (a) From free-body diagram of lever BCD ∴ C x = −380 N ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 2 2 Cx + C y = ∴ C y = −240 N (a) From free-body diagram of lever BCD -Body Diagram: Aplicando  las  ecuaciones  de  equilibrio  obtenemos  que:   (b) From free-body diagram of lever BCD   θ = tan −1 ⎜ ⎜ C = ΣFy = 0: C y + 0.8 ( 300 N ) = 0 𝐹! = 0      , 225   +   𝐹!" sin 36,87 = 0   ∴ C y = −240 N C = C y = 240 N or 2 2 C x + C y𝐹!" sin 36,87( 240 −225   = ( 380 ) + = ) = 449.44 N 2 and   Then ΣFx = 0: 0      ,         − Cx +   0.6 ( 𝐹 cos = 0 𝐹! = 200 N + 𝐹!" +   300 N ) 36,87 = 0            (𝑎)   !" ∴ C x = −380 N or C x = 380 N Then   2 ⎛ Cy ⎞ −1 ⎛ − 240 ⎟ = tan−225 ⎞ = 32.276° ⎟ ⎜ ⎟= !" ⎝ − 380 =   −375  𝑙𝑏,     ⎝ C x ⎠ sin 36,87 ⎠ θ = tan −1 ⎜ 𝐹 ⎜ and   or C = 449 N Free-Body Diagram: Chapter 4, Solution 19. COSMOS: Complete Online Solutions Manual Organization System   ∴ TAB = 300 32.3° ▹ 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:     !!" 67,38° !! !!" 720!!"! 𝐶𝑂 9 12 tan 𝛽 =   = =     𝐶𝐴 5 Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., 3,75 Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell   The McGraw-Hill Companies. 12 𝛽 =   tan!! = 67,38°   5      
  4. 4. Free-Body Diagram: (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 Aplicando  las  ecuaciones  de  equilibrio  tenemos  que:   (b) From free-body diagram of lever BCD   ΣFx = 0: 0      ,        𝐹 C  x−   0.6 ( 300 N ) = 0 = 0   𝐹! = 200 N + + 𝐹!" cos 67,38 !" ∴ C x = −380 N or C x = 380 N   ΣFy = 0:𝐹!" cos 0.8 ( 300 N ) = 𝐹!"   C y + 67,38 =   0   or C y = 240 N 300 = 780  𝑙𝑏   cos2 + ( 240 2 2 2 C = cos 67,38 ( 380 ) 67,38 ) = 449.44 N Cx + C y = 𝐹!" =             ∴ TAB = 300 Then ∴ 𝐹C y = −240 N !" =   ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ 𝐹!" = 780  𝑙𝑏  ,                    𝑻      𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐶𝐴  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛   ⎝ − 380 ⎠ ⎝ Cx ⎠ and θ = tan −1 ⎜ ⎜ ector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., lliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. or C = 449 N 32.3° ▹

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