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### Bryan

1. 1. Primero dibujamos el diagrama de cuerpo libre: ! 0,28!! !! 0,18!! !! ! 0,10!! ! ! ! 30° 150!! Ahora 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: !! Ahora aplicando las ecuaciones de equilibrio tenemos: (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. 𝑨 =   Chapter 4, Solution 19. e-Body Diagram: 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                    𝑨lever𝟐𝟒𝟒  𝑵   →     diagram of = BCD ∴ TAB = 300 ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 ⎛ 2.4 in. ⎞ (b) From −⎜ ⎟ A − (0.9 in.)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 𝒙 = 0: C y + − 150 xsin 30 ΣFy= −243,74 0.8 ( 300 N ) = 0 = −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 or Then 8 lb ⎞ ( 3 lb ) sin 30° + Bx + ⎛ ⎜ ⎟=0 ⎝ cos30° ⎠ and Then Bx = −10.7376 lb 0: − ( 3 lb ) cos 30° + B y = 0 0: C = 𝐹! = 0:                  𝐷2 − 150 cos 30 = 0 ! 2 2 2 C x + C y C = 380 ) N ( 240 ) = 449.44240 N ∴ = y ( −240 + or Cy = N C 𝑫 ⎛ =   ⎞ 150 cos 30 = 𝟏𝟐𝟗, 𝟗𝟎𝟒  𝑵 2 2 − 240 ⎞ 2 C 1 y 2 = C y =⎛ ) 32240 ) θ = tan −=𝒚⎜ C x⎟ + tan −1 ⎜ ( 380⎟ =+ ( .276° = 449.44 N ⎜C ⎟ ⎝ − 380 ⎠ ⎠ ⎛ Cy ⎞ ⎛ − 240 ⎞ C = 449 N or = 32.276° 32.3° ▹ and θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ ⎜ C!⎟ ! ⎝ − 380 ⎠ ! +   129,904 ! = 𝟑𝟒𝟒, 𝟐𝟎  𝑵 x ⎠ By = 2.5981 lb 𝑃𝑜𝑟  𝑙𝑜  𝑡𝑎𝑛𝑡𝑜:      𝑫 =   𝐷! +   ⎝ 𝐷!   =   −318,74 or 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 ⎝ x 𝑎𝑑𝑒𝑚𝑎𝑠                          𝜽 =   tan!! ! 𝐷! =   tan!! B    𝟑𝟒𝟒  𝑵 𝑜                𝑫 = = 11.05 lb 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° 𝟐𝟐, 𝟐° 𝜽= ▹ = −𝟐𝟐, 𝟏𝟕𝟒°
3. 3. Primero dibujamos el diagrama de cuerpo libre: !! !! 2! + ! os !c !! ! !! ! !! !! !! !! OS: Complete Online Solutions Manual Organization System ! ! pter 4, Solution 19. e-Body Diagram: Ahora aplicando las ecuaciones de equilibrio tenemos: (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 ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 ∴ C x = −380 N or C x = 380 N ΣFy = 0: C y + 0.8 ( 300 N ) = 0 ∴ C = −240 N or C = 240 N
4. 4. Chapter 4, Solution 19. Free-Body Diagram: (a) From free-body diagram of lever BCD 𝑷 ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑻= 𝟏 +   𝐜𝐨𝐬 𝜽              (𝐼) ∴ 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 = C + 0.8 ( 300 N ) ΣFy = 𝒙0:  𝑻y 𝐬𝐢𝐧 𝜽          (𝐼𝐼) = 0 ∴ C y = −240 N C y = 240 N or De (I) 2 2 Chapter 4, Solution 19. en (II) se tiene que: C = Cx + C y = ( 380 )2 + ( 240 )2 = 449.44 N Then Free-Body Diagram: 𝑷 𝐬𝐢𝐧 𝜽 ⎛C ⎞ ⎛ − 240 ⎞ y 𝑪𝒙 = θ = tan −1   ⎜ ⎟ = tan −1            (𝐼𝐼𝐼)= 32.276° ⎜ 𝟏 +⎟ 𝐜𝐨𝐬 𝜽 ⎜ − 380 ⎟   ⎠ ⎝ ⎝ Cx ⎠ (a) From free-body diagram of lever BCD and or C = 449 N ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 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 𝒚 =  𝑷 − 𝑻 𝟏 + 𝐜𝐨𝐬 𝜽              (𝐼𝑉) ∴ C x = −380 N C x = 380 N or ΣFy De (I) en (IV) se tiene que:= 0: C y + 0.8 ( 300 N ) = 0 ∴ 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 𝒚 = 𝟎    ,                𝐶 =   =!32.276° ⎝ − 380 ⎠ 𝐶 ⎝ Cx ⎠ 1 +   cos 𝜃 −1 or C = 449 N 32.3° ▹ 𝑷 𝐬𝐢𝐧 𝜽 𝑪 =            (𝑉) 𝟏 +   𝐜𝐨𝐬 𝜽 𝐶𝑜𝑚𝑜    𝜃 = 60°    𝑠𝑒𝑔𝑢𝑛  𝑒𝑛𝑢𝑛𝑐𝑖𝑎𝑑𝑜 De (I) se tiene que: 𝑻= 𝑃 𝑃 𝑃 𝟐 =   =   =     1 1 +   cos 𝜃 1 +   cos 60 𝟑 1 +   2 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. 𝑷
5. 5. Ahora De (V) se tiene que: 𝑪 =   𝑃 sin 𝜃 𝑃 sin 60 𝑃 0,87 =   =   =     𝟎, 𝟓𝟖 𝑷 1 1 +   cos 𝜃 1 +   cos 60 1 +   2