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V 0, 288
-1
  BA   0, 72 s ,
2
l AB 0, 4
V 0, 672
-1
 3  BO 1   2, 688s ,
l O 1 B 0, 25
V 0, 096
  DC   0, 32 s .
-1
4
l CD 0, 3
Angular velocities  ,  ,  have clockwise direction.
2
3
4
Drawing of accelerations polygon. First of all determine
acceleration of the point A
2
2
a  a   2 l  OA  7  0, 12  5, 88 m/s .
n
A
A
1
Acceleration of the point A is parallel to the link OA and is
directed to centre of rotation – to the point O.
Choose free position of the pole of accelerations polygon
(see fig. 1.3). After that, draw straight line through the pole and
parallel to the line segment OA. Put the vector p a a on this line
with length 100 mm, than scale factor of accelerations polygon
will look like:
a 5, 88
2
  A   0, 059 m/s ∙mm.
а
p a a 100
Determine acceleration of the point B. Solve the system of
vector equations
a  a  a n  a  ,
 B A BA BA
 n 
 a  a O 1  a BO 1  a BO 1 .
 B
n
In first equation, vector of acceleration a is parallel to link
BA
AB and is directed from point B till A. Magnitude of this
acceleration will be
2
2
a n BA   2 l  AB  0, 72  0, 4  0, 207 m/s .
2
Length of this vector on the accelerations polygon
a n 0, 207
an  BА   3, 05mm.
2
 a 0, 059
In the second equation, the vector of acceleration a n is
BO 1
parallel to the link О 1B and is directed from the point B till point
O 1. Magnitude of this acceleration will be
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