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Eligheor

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Eligheor

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Eligheor

  1. 1. Dibujamos el diagrama de cuerpo libre: ! 0,28!! !! 0,18!! !! ! 0,10!! ! ! ! 30° 150!! Llevamos las medidas de mm a metros: ne Solutions Manual Organization System 280  𝑚𝑚 = 0,28  𝑚 180 = 0,18  𝑚 100 = 0,10  𝑚 on 19. m: !! Aplicando las ecuaciones de equilibrio obtenemos: (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0:           − 𝐴 0,18 + 150 sin 30 0,10 +   150 cos 30 0,28 = 0 ∴ TAB = 300 (b) From free-body diagram of lever BCD ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 ∴ C x = −380 N or C x = 380 N
  2. 2. 21. OS: Complete Online Solutions Manual Organization System COSMOS: Complete Online Solutions Manual Organization System pter 4, Solution 19. Chapter 4, Solution 19. e-Body Diagram: 𝐴 =   150 sin 30 0,10 +   150 cos 30 0,28 = 𝟐𝟒𝟑, 𝟕𝟒  𝑵 0,18 (a) From free-body diagram of lever BCD Free-Body Diagram: ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 (a) From free-body        𝑜                    𝐴lever BCD →     diagram of = 244  𝑁   ∴ TAB = 300 ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 ⎛ 2.4 in. ⎞ −⎜ in. From ⎟ A − (0.9 (b))Fsp = 0 free-body diagram of lever BCD Βx = 0 : ⎝ cosα ⎠ ∴ T = 300 ΣFx = 0: 0:                    243,74 +300 N ) = 0 30 +   𝐷 = 0 AB 𝐹! = 200 N + Cx + 0.6 ( 150 sin ! 8 (b) From free-body diagram of lever BCD Fsp = lb = kx = k (1.2 in.) ∴ C x = −380 N or C x = 380 N cos 30° ΣFx = 0: 200 N + C + 0.6 ( 300 N ) = 0 𝐷 F 0: C y + − 300 x) = 0 Σ! y==−243,74 0.8 ( 150N sin 30 = −𝟑𝟏𝟖, 𝟕𝟒  𝑵 ∴ C x = −380=N or k = 7.69800 lb/in. k 7.70 lb/in. ▹ C x = 380 N ∴ C y = −240 N or C y = 240 N ΣFy = 0: C y + 0.8 ( 300 N ) = 0 0: or 0: or = 𝐹! = 0:                  𝐷2 − 150 cos 30 = 0 ! 2 2 2 C = C x + C y C = 380 ) N ( 240 ) = 449.44240 N Then ⎛ 8 lb ⎞ ∴ = y ( −240 + or Cy = N ( 3 lb ) sin 30° + Bx + ⎜ ⎟=0 ⎝ cos30° ⎠ C 𝐷!⎛ =   ⎞ 1502 cos 240 = 𝟏𝟐𝟗, 𝟗𝟎𝟒  𝑵 − 30 ⎞2 C 1 y 2 = C y =⎛ ) 32240 )2 and Then θ = tan −=⎜ C x⎟ + tan −1 ⎜ ( 380⎟ =+ ( .276° = 449.44 N ⎜C ⎟ Bx = −10.7376 lb ⎝ − 380 ⎠ ⎝ x⎠ ⎛ Cy ⎞ ⎛ − 240 ⎞ C = 449 N − ( 3 lb ) cos 30° + B y = 0 or = 32.276° 32.3° ▹ and θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ ⎜C ⎟ ! ! ⎝ x ⎠ ⎝ − 380 ⎠   129,904 ! = 𝟑𝟒𝟒, 𝟐𝟎  𝑵 !+ ∴      𝐷 =   𝐷! +   𝐷!   =   −318,74 By = 2.5981 lb or C = 449 N 32.3° ▹ 2 2 −10.7376 ) + ( 2.5981) = 11.0475 lb, and ( 𝐷 129,904 2.5981 = tan −1 = 13.6020° 10.7376 𝑦                            𝜃 =   tan!! ! 𝐷! =   tan!! 𝑜                𝐷 B = 11.05 lb =    344  𝑁 s: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Clausen, David Mazurek, Phillip J. Cornwell mpanies. Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., . Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 7 The McGraw-Hill Companies. 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. −318,74 13.60° ▹ 𝜃 = 22,2° = −𝟐𝟐, 𝟏𝟕𝟒°
  3. 3. Dibujamos el diagrama de cuerpo libre: !! !! 2! + ! os !c !! ! !! ! !! !! !! !! OS: Complete Online Solutions Manual Organization System ! ! pter 4, Solution 19. e-Body Diagram: Aplicando las ecuaciones de equilibrio obtenemos: (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0:                      𝑇 2𝑎 + 𝑎 cos 𝜃 −  𝑇𝑎 + 𝑃𝑎 = 0 ∴ TAB = 300 (b) From free-body diagram of lever BCD 𝑷 𝑇=              (𝑎) ΣFx = 0: 200 N + Cx   + 0.6 ( 300 N ) = 0 𝟏 + 𝐜𝐨𝐬 𝜽 ∴ C x = −380 N or C x = 380 N ΣFy = 0: C y + 0.8 ( 300 N ) = 0 ∴ C y = −240 N or C y = 240 N
  4. 4. Free-Body Diagram: (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 ∴ TAB = 300 (b) From free-body diagram of lever BCD ΣFx = 0: 0:                    𝐶 x− 0.6 ( 300 = )0= 0 𝐹! = 200 N + C + 𝑇 sin 𝜃 N ! ∴ C x = −380 N or C x = 380 N COSMOS: Complete Online Solutions Manual Organization System 𝐶 =  𝑻 + 0.8 ( 300 N ΣFy =!0: C y 𝐬𝐢𝐧 𝜽          (𝑏)) = 0 ∴ C y = −240 N C y = 240 N or De la 2 (b) 2 Chapter 4, Solution 19. ecuación (a) en la ecuación+ C y se tiene2que: )2 = 449.44 N C = Cx = ( 380 ) + ( 240 Then Free-Body Diagram: ⎛ C𝑷 ⎞ 𝐬𝐢𝐧 𝜽 −1 ⎛ − 240 ⎞ y            (𝑐) = 32.276° ⎟ = tan ⎜ ⎟ 𝟏 ⎟ C x+   𝐜𝐨𝐬 𝜽⎝ − 380 ⎠ ⎠ ⎝ (a) From free-body diagram of lever BCD or C = 449 N ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0   tan −1 ⎜ θ = 𝐶! = ⎜ and 𝐹! = 0:                    𝐶! + 𝑇 + 𝑇 cos 𝜃 − 𝑃 = 0 32.3° ▹ ∴ TAB = 300 (b) From free-body diagram of lever BCD ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 𝐶! =  𝑷 − 𝑻 𝟏 + 𝐜𝐨𝐬 𝜽              (𝑑) ∴ C x = −380 N C x = 380 N or Σ ecuación (d) ( 300 N ) = 0 De la ecuación (a) en laFy = 0: C y + 0.8se tiene que: ∴ C y = −240 N C y = 240 N or Then 2 2 2 C = 𝐶C x=  𝑃y− 𝑃 ( 380 )cos (𝜃 = 0= 449.44 N + C 2 = 1 + + 240 ) ! and ⎛ Cy ⎞ ⎛ − 240 ⎞ θ = tan ⎜ ⎟ = tan −1 ⎜ ⎟ ⎜ 𝐶! = 0    ,                𝐶 =  ⎟ =!32.276° ⎝ − 380 ⎠ 𝐶 ⎝ Cx ⎠ 1 +   cos 𝜃 −1 or C = 449 N 32.3° ▹ 𝑷 𝐬𝐢𝐧 𝜽 𝐶 =            (𝑒) 𝟏 +   𝐜𝐨𝐬 𝜽 𝑃𝑎𝑟𝑎    𝜃 = 60°    𝑎  𝑡𝑟𝑎𝑣𝑒𝑠  𝑑𝑒𝑙  𝑒𝑛𝑢𝑛𝑐𝑖𝑎𝑑𝑜 De la ecuación (a) se tiene que: 𝑇= 𝑃 𝑃 𝑃 𝟐 =   =   =     1 1 +   cos 𝜃 1 +   cos 60 𝟑 1 +   2 Vector Mechanics for Engineers:ecuación (e) se tiene que: E. Russell Johnston, Jr., De la Statics and Dynamics, 8/e, Ferdinand P. Beer, Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. 𝐶 =   𝑷 𝑃 sin 𝜃 𝑃 sin 60 𝑃 0,87 =   =   =     𝟎, 𝟓𝟖 𝑷 1 1 +   cos 𝜃 1 +   cos 60 1 +   2

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