give answer in 2 step with explanation at the end of each step and final answer at the end:Design of Machinery, 5th Edition by Robert Norton 11-3 Table P11-1 shows kinematic and geometric data for several slider-crank mechanisms of the type and orientation shown in Figure P11-1 (p. 426). The point locations are defined as described in the text. TABLA P11-1 Datos para el problema 11-3 (Véase figura P11-1, p. 426 para la nomenclatura) Forte 1 Longitudes en puigadas. Gnguios en grados. masa en blobs, velocidad angular en rad/s Fla Eslabén2 Eslabon3 Descentrado 6, ®, m, my my a a 12 [] 5 10 20 0002 0020 0060 b 3 10 1 0 15 5 005 0100 0200 c 5 15 -1 20 20 15 0010 0020 0030 d 6 2 1 75 -10 -10 0006 0150 005 e 2 8 0 135 25 25 0001 0004 0014 f 10 35 2 120 5 20 0150 0300 00% 9 7 2% 2 45 30 15 0080 0200 0.100 Parte 2 Acelerackn anguiar en rad/s, momentos de Inercia en biob-pulg2, par de forsién en Ib-pulg -— 1, Ra 5 Ra, 6 Fp, orp, Rp, Rp = 80 Ov NG Gh TI 00 Da ®0 a 010 02 2 0 5 0 0 0 [] 0 20 b 020 04 1 2 4 30 0 5 4 30 -35 c 005 01 3 40 9 50 32 270 0 0 -65 d 012 03 3 1200 2 © 15 180 2 60 -12 e 03 08 05 30 3 75 6 60 2 75 40 ¢ 024 06 6 45 15 135 25 270 0 0 -75 g 045 09 4 45 0 2% 9 120 5 25 -9%0 Pore 3 Fuerzas en Ib, aceleraciones lineales en puigadas/s? Fila 0 a 9g, ag, 9g, ag; ag, ag, 0° mw aw mw ap mg ag a 166.40 240 20396 21369 37108 20084 3717 180 b 177.13 3433 2506 231.27 58043 20005 M97 180 c 19517-13476 120084 37.85 208804 4343 929.12 0 d 199.86 2074 30150 23071 51174 7452 2397 180 e 169.82 N32 31275 17.9 97679 5813 849.76 0 f 169.03 329 19209 23.66 30250 -2093 301.92 0 g 18678 17220 360050 9095 805235 13466 4909.27 180 £5 ’e) A 7 => * ’ c) ’ |.
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:
Design of Machinery, 5th Edition by Robert Norton
11-3 Table P11-1 shows kinematic and geometric data for several slider-crank
mechanisms of the type and orientation shown in Figure P11-1 (p. 426). The point
locations are defined as described in the text.
TABLA P11-1 Datos para el problema 11-3 (Véase figura P11-1, p. 426 para la
nomenclatura)
Forte 1 Longitudes en puigadas. Gnguios en grados. masa en blobs, velocidad angular en rad/s
Fla Eslabén2 Eslabon3 Descentrado 6, ®, m, my my
a a 12 [] 5 10 20 0002 0020 0060
b 3 10 1 0 15 5 005 0100 0200
c 5 15 -1 20 20 15 0010 0020 0030
d 6 2 1 75 -10 -10 0006 0150 005
e 2 8 0 135 25 25 0001 0004 0014
f 10 35 2 120 5 20 0150 0300 00%
9 7 2% 2 45 30 15 0080 0200 0.100
Parte 2 Acelerackn anguiar en rad/s, momentos de Inercia en biob-pulg2, par de forsién en Ib-pulg
-— 1, Ra 5 Ra, 6 Fp, orp, Rp, Rp
= 80 Ov NG Gh TI 00 Da ®0
a 010 02 2 0 5 0 0 0 [] 0 20
b 020 04 1 2 4 30 0 5 4 30 -35
c 005 01 3 40 9 50 32 270 0 0 -65
d 012 03 3 1200 2 © 15 180 2 60 -12
e 03 08 05 30 3 75 6 60 2 75 40
¢ 024 06 6 45 15 135 25 270 0 0 -75
g 045 09 4 45 0 2% 9 120 5 25 -9%0
Pore 3 Fuerzas en Ib, aceleraciones lineales en puigadas/s?
Fila 0 a 9g, ag, 9g, ag; ag, ag,
0° mw aw mw ap mg ag
a 166.40 240 20396 21369 37108 20084 3717 180
b 177.13 3433 2506 231.27 58043 20005 M97 180
c 19517-13476 120084 37.85 208804 4343 929.12 0
d 199.86 2074 30150 23071 51174 7452 2397 180
e 169.82 N32 31275 17.9 97679 5813 849.76 0
f 169.03 329 19209 23.66 30250 -2093 301.92 0
g 18678 17220 360050 9095 805235 13466 4909.27 180
£5
’e)
A
7 => *
’ c)
’ |.
Asked by: Ratan
Created at: 2025-06-19 22:43:46
Final Answer:
Question:
Solving in step by step in all parts.
Asked by: Ratan
Created at: 2025-06-19 22:45:38
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