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DESCRIPTION
Calculo de velocidades y aceleraciones para el mecanismo caja negra
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CΓ‘lculos para Ι΅=0.
VELOCIDADES
VA = Ο2 β rA
VA = 200 mm s
VB = VA + VB/A
100 πππ
sin 94.54=
ππ΅
sin 75.11
ππ΅ = 96.95 πππ
ππ΅π΄
= 100 πππ 2 + (96.95 ππ
π )2 β 2 100 πππ 96.95 ππ
π β cos 10. 35
π π΅π΄ = 18.02 ππ
π
π4 =ππ΅
ππ΅
π4 = 1.43 ππππ
π3 =π π΅
π΄
π π΅π΄
π3 = 0.277 πππ π
ππ· = π4 β ππ·
ππ· = 1.43 β 115
ππ· = 164.45 πππ
VE = VD + VE/D
164.45 πππ
sin 17.6=
ππΈ
sin 90
ππΈ = 543.87 πππ
543.87 πππ
sin 17.6=
ππΈ/π·
sin 72.4
ππΈ/π· = 518.87 πππ
π5 =ππΈ/π·
ππΈ/π·
π5 = 4.5 ππππ
π6 =ππΈ
ππΈ
π6 = 2.6 πππ π
ACELERACIONES
ππ΄ = ππ΄π+ ππ΄π‘
Como Ο2 es constante, entonces Ξ±2 es 0 y por ende aceleraciΓ³n tangencial de A es cero.
ππ΄ = ππ΄π= π2
2ππ΄ = 22 β 50 = 200 πππ 2
ππ΅π+ ππ΅π‘
= ππ΄π+ ππ΅/π΄π
+ ππ΅/π΄π‘
ππ΅π=π4 π π4 π ππ΅
ππ΅π= 1.43 π π (1.43 πΎ π(66.7 π β 12.18 π)
ππ΅π= β136.39 π + 26.18 π
ππ΅π‘= πΌ4 π ππ΅ = πΌ4π π 66.7 π β 12.18 π
ππ΅π‘= 66.70 β πΌ4π + 12.18 β πΌ4 π
ππ΅/π΄π=π3 π π3 π ππ΅/π΄
ππ΅/π΄π= 0.277 π π 4.62 π β 17.39 π = β1.27 π β 4.81π
ππ΅/π΄π‘= πΌ3 π ππ΅/π΄
ππ΅π΄π‘
= βπΌ3 β 62.81 π + πΌ3 β 16.702 π
π β πΌ4 β 12.18 + πΌ362.81 = β64.88
π β πΌ4 β 66.70 β πΌ3 β 16.702 = β31.69
πΌ4 = β0.0699 ππππ 2
πΌ3 = 1.01 ππππ 2
ππ΅ = 139.01 πππ 2
ππ·π= 42.24 π β 231.23 π
ππ·π‘= 1.46 π + 8.005 π
ππ· = 235.9 ππ π 2
ππΈπ+ ππΈπ‘
= ππ·π+ ππ·π‘
+ ππΈ/π·π+ ππΈ/π·π‘
ππΈπ =π6 π π6 π ππΈ
ππΈπ= β2.6 π π (β2.6 πΎ π(β63.5 π β 200.17 π)
ππΈπ= β429.26 π β 1353.15 π
ππΈπ‘= πΌ6 π ππΈ = πΌ6 π β63.5 π + 200.17 π
ππΈπ‘= β200.17 β πΌ6π β 63.5 β πΌ6 π
ππΈ/π·π =π5 π π5 π ππΈ/π·
ππΈ/π·π= β4.6 π π 84.18 π + 522.19 π = 387.22 π β 2402.047π
ππΈ/π·π‘= πΌ5 π ππΈ/π·
ππΈ/π·π‘= βπΌ5 β 18.37 π + πΌ5 β 113.52 π
π β βπΌ6 β 200.77 β πΌ5113.5 = β273.27
π β βπΌ6 β 63.5 β πΌ5 β 18.37 = β3712.97
πΌ6 = β38.67 ππππ 2
πΌ5 = 70.81 ππππ 2
ππΈ = 8266.85 πππ 2
CΓ‘lculos para Ι΅=45
VELOCIDADES
ππ΄ = π2 π ππ
ππ΄ = β2π π 35.36 π + 35.36 π = β70.72 π + 70.72 π = 100 πππ 2
VB = VA + VB/A
ππ΅ = π4 π ππ΅
ππ΅ = π4π π 65.37 π + 18 π = π4 β 65.3 π β π4 β 18 π
ππ΅π΄
= π3 π ππ΅/π΄
ππ΅π΄
= π3π π 30.02 π + 57.65 π = π3 β 30.02 π β π3 β 57.65 π
π β βπ3 β 57.65 + π4 β 18 = 70.72
π β π3 β 30.02 β π4 β 65.37 = β70.72
π3 = β1.82 πππ π
π4 = β1.91 πππ π
ππ΅ = 129.50 πππ
ππ· = π4 π ππ·
ππ· = β1.92 πΎ π 110.8 π + 30.53 π = β212.73 π + 58.61 π = 220.65 πππ
VE = VD + VE/D
VE = Ο6 X rE
ππΈ = π6 π π 5.46 π + 209.9 π = π6 β 5.46 π β π6 β 209,9 π
ππΈ/π· = π5 π ππΈ/π·
ππΈ/π· = π5 π π 89.59π + 72.09 π = π5 β 85.59 π β π5 β 72.09 π
π β βπ6 β 209.9 + π5 β 72.09 = 55.56
π β π6 β 5.46 β π5 β 89.59 = β201.65
π6 = β1.05 πππ π
π5 = β2.28 πππ π
ππΈ = 220.46 πππ
ACELERACIONES
ππ΄ = ππ΄π+ ππ΄π‘
Como Ο2 es constante, entonces Ξ±2 es 0 y por ende aceleraciΓ³n tangencial de A es cero.
ππ΄π = π2 π(π2 π ππ΄)
ππ΄π = β2π π β70.72 π + 70.72 π = β141.44 π β 141.44 π = 200 ππs2
ππ΅π+ ππ΅π‘
= ππ΄π+ ππ΅/π΄π
+ ππ΅/π΄π‘
ππ΅π=π4 π π4 π ππ΅
ππ΅π= β1.91 π π (β124.85 π + 34.81 π)
ππ΅πβ 238.46 π β 65.66 π
ππ΅π‘= πΌ4 π ππ΅ = πΌ4 π π 65.37 π + 18 π
ππ΅π‘= β18 β πΌ4π + 65.37 β πΌ4 π
ππ΅/π΄π=π3 π π3 π ππ΅/π΄
ππ΅/π΄π= β1.82 π π β54.63 π + 104.9 π = β99.4 π β 190.9π
ππ΅/π΄π‘= πΌ3 π ππ΅/π΄
ππ΅/π΄ π‘ = πΌ3 β 30.02 π β πΌ3 β 57.65 π
π β βπΌ4 β 18 + πΌ3 β 57.65 = β2.28
π β πΌ4 β 65.37 β πΌ3 β 30.02 = β266.68
πΌ4 = β4.76 ππππ 2
πΌ3 = β1.53 ππππ 2
ππ΅ = 400.61 πππ 2
ππ· = ππ·π+ ππ·π‘
ππ·π = π4 π π4 π ππ΅ = β1.92 π π β212.73 π + 58.61 π = β408.44 π β 112.53 π
ππ·π‘= πΌ4 π ππ΅ = β4.76 π π 110.8 π + 30.53 π = β527.408 π β 639, .93 π
ππ· = 691.91 πππ 2
ππΈπ+ ππΈπ‘
= ππ·π+ ππ·π‘
+ ππΈ/π·π+ ππΈ/π·π‘
ππΈπ =π6 π π6 π ππΈ
ππΈπβ 1.05 π π (β5.73 π + 220.39 π)
ππΈπ= β β 6.01 π β 231.4 π
ππΈπ‘= πΌ6 π ππΈ = πΌ6π π 5.46 π + 209.9 π
ππΈπ‘= β209.9 β πΌ6π + 5.46 β πΌ6 π
ππΈ/π·π =π5 π π5 π ππΈ/π·
ππΈ/π·π= β2.28π π β204.2 π + 164.36 π = β465.71 π β 374.74 π
ππΈ/π·π‘= πΌ5 π ππΈ/π·
ππΈ/π·π‘= πΌ5 β 89.59 π β πΌ5 β 72.09 π
π β βπΌ6 β 209.9 + πΌ572.09 = β722.82
π β πΌ6 β 5.46 β πΌ5 β 81.59 = β783.27
πΌ6 = β3.6 ππππ 2
πΌ5 = 0.45 ππππ 2
ππΈ = 6770.24 πππ 2
CΓ‘lculos para Ι΅=105
VELOCIDADES
ππ΄ = π2 π ππ
ππ΄ = β2π π 12.94 π + 98.29 π = β25.88 π + 96.59 π = 100 πππ 2
VB = VA + VB/A
ππ΅ = π4 π ππ΅
ππ΅ = π4π π 48.77 π β 47.11 π = π4 β 48.77 π + π4 β 47.11 π
ππ΅π΄
= π3 π ππ΅/π΄
ππ΅π΄
= π3π π 61.71 π β 20.40 π = π3 β 61.71 π + π3 β 20.40 π
π β βπ3 β 20.9 + π4 β 47.11 = 96.59
π β βπ3 β 61.71 + π4 β 48.77 = β25.88
π3 = 3.13 πππ π
π4 = 3.44 πππ π
ππ΅ = 233.25 πππ
ππ· = π4 π ππ·
ππ· = 3.44πΎ π 82.71 π β 79.9 π = 284.52 π + 274.85 π = 295.544 πππ
VE = VD + VE/D
VE = Ο6 X rE
ππΈ = π6 π π β165.39 π + 129.4π = βπ6 β 165.39 π β π6 β 129.4 π
ππΈ/π· = π5 π ππΈ/π·
ππΈ/π· = π5 π π β102 π β 53.1 π = βπ5 β 102 π + π5 β 53.1 π
π β βπ6 β 129.4 β π5 β 53.1 = 274.85
π β βπ6 β 165.39 + π5 β 102 = 284.85
π6 = β1.96 πππ π
π5 = β0.39 πππ π
ππΈ = 411.59 πππ
ACELERACIONES
ππ΄ = ππ΄π+ ππ΄π‘
Como Ο2 es constante, entonces Ξ±2 es 0 y por ende aceleraciΓ³n tangencial de A es cero.
ππ΄π = π2 π(π2 π ππ΄)
ππ΄π = β2π π β25.88 π + 70.72 π = β51.76 π β 193.18 π = 200 ππs2
ππ΅π+ ππ΅π‘
= ππ΄π+ ππ΅/π΄π
+ ππ΅/π΄π‘
ππ΅π=π4 π π4 π ππ΅
ππ΅π= 3.44 π π (167.76 π + 162.05 π)
ππ΅πβ 577.09 π + 557.45 π
ππ΅π‘= πΌ4 π ππ΅ = πΌ4 π π 48.77 π β 47.11 π
ππ΅π‘= 48.77 β πΌ4π + 47.11 β πΌ4 π
ππ΅/π΄π=π3 π π3 π ππ΅/π΄
ππ΅/π΄π= 3.13 π π 193.1 π + 63.85 π = β604.4 π + 199.85π
ππ΅/π΄π‘= πΌ3 π ππ΅/π΄
ππ΅/π΄ π‘ = πΌ3 β 61.71 π β πΌ3 β 20.4 π
π β βπΌ4 β 18 + πΌ3 β 57.65 = β2.28
π β πΌ4 β 65.37 β πΌ3 β 30.02 = β266.68
πΌ4 = 11.90 ππππ 2
πΌ3 = 31.36 ππππ 2
ππ΅ = 580.36 πππ 2
ππ· = ππ·π+ ππ·π‘
ππ·π = π4 π π4 π ππ΅ = 3.44 π π 284.5 π + 274.85 π = β978.68 π + 945.48 π
ππ·π‘= πΌ4 π ππ΅ = 11.90 π π 82.71 π β 79.9 π = 984.24 π + 950.81 π
ππ· = 1929.92 πππ 2
ππΈπ+ ππΈπ‘
= ππ·π+ ππ·π‘
+ ππΈ/π·π+ ππΈ/π·π‘
ππΈπ =π6 π π6 π ππΈ
ππΈπ= β1.96 π π (324.16 π + 253.62 π)
ππΈπ= 635.35 π β 497.09 π
ππΈπ‘= πΌ6 π ππΈ = πΌ6π π β165.39 π + 129.4 π
ππΈπ‘= β165.39 β πΌ6π β 129.4 β πΌ6 π
ππΈ/π·π =π5 π π5 π ππΈ/π·
ππΈ/π·π= β0.39π π 39.78 π β 20.7 π = 15.51 π + 8.073 π
ππΈ/π·π‘= πΌ5 π ππΈ/π·
ππΈ/π·π‘= βπΌ5 β 102 π + πΌ5 β 53.1 π
π β βπΌ6 β 129.4 β πΌ5 β 53.1 = β147.71
π β βπΌ6 β 165.39 + πΌ5 β 102 = 2068.737
πΌ6 = β4.31 ππππ 2
πΌ5 = 13.28 ππππ 2
ππΈ = 2902.97 πππ 2