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Velocity of the point B is determined by the next two vector
equations
V V A V BA ,
 B

V V O 1 V BO 1 .
 B
We must solve these equations graphically. Let’s draw line
of velocity V ВА action through the point a, which is perpendicular
to the link AB. V O 1  0, because the point O 1 is fixed. This point is
placed in the pole of velocities polygon. Next one draw line of
V BO action through the pole р v. This line should perpendicular to
1
the link О 1В. These both lines will cross and will give wanted the
point b.
Determine velocities
V BA  ab  V  24 0, 012  0, 288m/s,
V B V BO 1  p  v  56 0, 012  0, 672m/s.
b
v
Velocity of the point C is determined by theorem of
similarity. Look at these proportions:
O 1 C  p v c , wherefrom p c  p  b O 1 C  56 70  78 mm.
O 1 B p v b v v O 1 B 50
Draw vector p  c  78 mm on the velocities polygon from
the pole in same direction as the vector p (or continue the
b

vector p v b on distance p v c  p v b  22mm). The point C is gotten
on the velocities polygon as result. Magnitude of this velocity
looks like:
V  p  c   v  78 0, 012  0, 936m/s.
C
We need to solve the vector equation for determination of
the point D velocity
V D  V C V DC .
Draw straight line through the point c perpendicular to the
link CD – direction of velocity V DC. Further, draw vertical line
from the pole p – it is direction of the slider’s trajectory of

motion. We got the point d on intersection.
Determine velocities
V DC  cd    8 0, 012  0, 096m/s,
v
V  p v d   v  76 0, 012  0, 912 m/s.
D
Let’s determine angular velocities for links,   7s -1
1
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