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2
2
a n   2 l  O B  2, 688  0, 25  1, 806 m/s .
3
BO
1
1
Length of this vector on the accelerations polygon
a n 1, 806
BO
p a n   1  0, 059  30, 61mm.
3
a
We must solve these equations graphically. First of all, on
the accelerations polygon, draw the vector аn 2 from the point a.

Further, draw line of action for a through the point n 2. This line
BA
is perpendicular to the link AB. Next, draw vector р n 3 from the
pole р  and, then, draw line of action for a  through the point n 3
BO
1
and perpendicular to the link O 1B. These two lines cross in the
point b. Determine magnitudes of accelerations by the measuring
of mentioned line segments on the accelerations polygon
2
a   n 2 b  a  68 0, 059  4, 012 m/s ,
BA
2
a  1  n 3 b a  58 0, 059  3, 422 m/s ,
BO
2
a  p a b a  65 0, 059  3, 835 m/s .
B
Determine acceleration of the point C by theorem of
similarity. Write down next proportions:
p a c  O 1 C  p c  p  b O 1 C  65 70  91mm.
p a b O 1 B a a O 1 B 50
Let’s draw vector p a c 91 mm on the accelerations polygon.
This vector is continuing of the vector p c. Then acceleration of
b
a
the point C looks like:
a  p a c   a  91 0, 059  5, 369 m/s .
2
C
Determine acceleration of the point D by next vector
equation

a  a  a DC  a DC .
n
C
D
n
Vector of acceleration a DC is directed along the link CD
from the point D till point C. Magnitude of the acceleration
2
a n DC   2 l  CD  0, 32  0, 3  0, 03 m/s .
2
4
Vector’s length on the accelerations polygon
a n 0, 03
cn  DC   0, 52 mm.
4
 0, 059
a
Let’s solve the vector equation graphically. Draw vector cn 4
from the point c on the accelerations polygon. Further, draw line of
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