Zasada prac przygotowanych w ruchu
Zadanie nr : 19
Dane:
G1=4G, G2=0,3G, G3=0,2G, G4=3G
R2
= 3 , r3=1,2r2, R3=1,2r3
r2
i2 x = 2r2 , i3x = 1,2r2 , µ=0,1
Szukane:
przyspieszenia
2 2
(09.1) J2 = m2 2r2 = 4m2r22, J3 = m3 1, 2r2 = 1, 44m3r22
( ) ( )
a1
(09.2) a1 = µ2R2 Ò! µ2 =
R2
µ3 r3 1, 2 r2
(09.3) = = Ò! µ3 = 1, 2µ2 Ò! ´Õ3 = 1, 2´Õ2
µ2 r2 r2
a1
(09.4) µ3 = 1, 2
R2
a4
(09.5) a4 = µ3R3 Ò! µ3 =
R3
a1 a1 R2
(09.6) a4 =1, 2µ21, 2r3 =1,2 1, 22 r2 =1,23 = 0,576a1
R2 R2 3
1
´ s1
(09.7) ´ s1 = R2´Õ2 Ò! ´Õ2 =
R2
R2
(09.8) ´ s4 = R3´Õ3 Ò!1,2r31,2´Õ2 Ò!1,23r2´Õ2 Ò!1,23 ´Õ2 = 0,576R2´Õ2
3
Ogólne równanie dynamiki otrzymuje postać:
(09.9)
m1g sin 60° - µm1g cos 60° - m1a1 ´ s1 + 0 - 4m2r22µ2 ´Õ2 + 0 -1, 44m3r22µ3 ´Õ3 + m4g - m4a4 ´ s4 = 0
( ) ( )
( ) ( )
Podstawiając do równania wcześniejsze zależności:
ëÅ‚ öÅ‚
a1
m1g sin 60° - µm1g cos60° - m1a1 R2´Õ2 + 0 - 4m2r22 ÷Å‚´Õ2 +
( )
ìÅ‚
R2 Å‚Å‚
íÅ‚
(09.10)
ëÅ‚ öÅ‚
a1
+ìÅ‚ 0 -1,728m3r22 ÷Å‚1, 2´Õ2 + m4g - m40,576a1 0,576R2´Õ2 = 0
( )
R2 Å‚Å‚
íÅ‚
Po uproszczeniu:
ëÅ‚ öÅ‚
R2 2 a1
m1g sin 60° - µm1g cos 60° - m1a1 R2´Õ2 + - 4m2 +
( )
ìÅ‚0
3 R2 ÷Å‚´Õ2
íÅ‚ Å‚Å‚
(09.11)
ëÅ‚ öÅ‚
R2 2 a1
+ìÅ‚ 0 -1,728m3 + m4g - m40,576a1 0,576R2´Õ2 = 0
( )
3 R2 ÷Å‚1,2´Õ2
íÅ‚ Å‚Å‚
4
ëÅ‚0
m1g sin 60° - µm1g cos 60° - m1a1 R2´Õ2 + - m2 a1 öÅ‚ R2´Õ2 +
( )
ìÅ‚ ÷Å‚
3
íÅ‚ Å‚Å‚
(09.12) R2´Õ2
a1
+ëÅ‚0 -1,728m3 öÅ‚1, 2R2´Õ2 + m4g - m40,576a1 0,576R2´Õ2 = 0
( )
ìÅ‚ ÷Å‚
3
íÅ‚ Å‚Å‚
4
(09.13) m1g sin 60° - µm1g cos60° - m1a1 - m2 a1 - 0,6912m3a1 + 0,576m4g - 0,331776m4a1 = 0
3
4m2
g m1 sin 60° - µm1 cos 60° + 0,576m4 = a1 ëÅ‚ m1 + + 0,6912m3 + 0,331776m4 öÅ‚
( )
ìÅ‚ ÷Å‚
3
íÅ‚ Å‚Å‚
(09.14) &
0,866µ + 0,003
a1 = g H" 2, 293
0,383392
(09.15) a3 = 0,576a1 H" 1,321
2