Project “The development of the didactic potential of Cracow University of Technology in the range of modern
construction” is co-financed by the European Union within the confines of the European Social Fund
and realized under surveillance of Ministry of Science and Higher Education.
Exercise 5_2: Reduce the system of parallel forces at the center of the system.
m
A
kN
F
2
,
5
,
2
8
,
20
,
12
1
1
,
m
A
kN
F
0
,
2
,
1
2
,
5
,
3
2
2
,
m
A
kN
F
3
,
0
,
1
4
,
10
,
6
3
3
Remark:
The forces are parallel because their components are proportional.
2
1
4 F
F
,
2
3
2 F
F
Solution:
1. Calculating the sum vector
kN
F
F
F
S
2
,
5
,
3
3
2
1
2. Choosing the unit vector of the system
2
,
5
,
3
10
2
1
10
2
2
,
5
,
3
2
2
kN
kN
F
F
e
3. Determining the coefficients
for each vector in the system.
i
a
e
F
a
i
i
kN
kN
e
F
e
F
a
kN
kN
e
F
e
F
a
kN
F
F
F
F
e
F
a
10
4
10
2
2
2
10
8
10
2
4
4
10
2
2
3
3
2
1
1
2
2
2
2
2
2
kN
a
i
10
2
4. The components of the position vectors of the points of application of the forces are the same as the
coordinates of these points
m
OA
r
m
OA
r
m
OA
r
3
,
0
,
1
0
,
2
,
1
2
,
5
,
2
3
3
2
2
1
1
5. Calculating the position vector of the center of the force system
m
m
kN
kNm
a
r
a
r
OO
i
i
i
2
,
22
,
7
2
4
,
44
,
14
10
2
3
,
0
,
1
4
0
,
2
,
1
2
2
,
5
,
2
8
10
6. Answer
At the center of the system
the system is reduced to the resultant force
m
O
2
,
22
,
7
kN
S
W
2
,
5
,
3
.