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









.