Exercise 5_2: Reduce the system of parallel forces at the center of the system.
ć� (�-�12,20,-�8)�kN F2 =� (�3,-�5,2)�kN F3 =� (�6,-�10,4)�kN ��
F1 =� �� ć� �� ć�
�� �� �� �� �� ��
, ,
�� �� �� �� �� ��
A1(�2,-�5,2)�m A2(�-�1,2,0)�m A3(�1,0,3)�m
Ł� ł� Ł� ł� Ł� ł�
Remark:
The forces are parallel because their components are proportional.
F1 =� -�4 �� F2 ,
F3 =� 2 �� F2
Solution:
1. Calculating the sum vector
S =� F1 +� F2 +� F3 =� (�-� 3,5,-�2)�kN
2. Choosing the unit vector of the system
F2 (�3,-�5,2)�kN 1
e =� =� =� (�3,-�5,2)�
F2 2 10kN 2 10
3. Determining the coefficients ai for each vector in the system.
ai =� Fi o� e
F2 o� F2
a2 =� F2 o� e =� =� F2 =� 2 10kN
F2
=� -�2 10kN
a1 =� F1 o� e =� -�4F2 o� e =� -�4 �� 2 10kN =� -�8 10kN
��ai
a3 =� F3 o� e =� 2F2 o� e =� 2 �� 2 10kN =� 4 10kN
4. The components of the position vectors of the points of application of the forces are the same as the
coordinates of these points
r1 =� OA =� (�2,-�5,2)�m
1
r2 =� OA2 =� (�-�1,2,0)�m
r3 =� OA3 =� (�1,0,3)�m
5. Calculating the position vector of the center of the force system
ri 10(�-� 8(�2,-�5,2)�+� 2(�-�1,2,0)�+� 4(�1,0,3)�)�kNm
(�-�14,44,-�4)�m =� (�7,-�22,2)�m
��ai
*�
OO*� =� r =� =� =�
-� 2
-� 2 10kN
��ai
6. Answer
At the center of the system O*� =� (�7,-�22,2)�m the system is reduced to the resultant force
W =� S =� (�-� 3,5,-�2)�kN .
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