background image

 

 

 

 

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. 

1

Exercise 5_1: For a given system of parallel forces determine the center of the system, and reduce the 
system at that point. 
 

 

 
Course of action: 
 
1. Choosing the unit vector of the system. 
 





 

2

2

,

0

,

2

2

2

2

2

,

0

,

2

m

m

BE

BE

e

 

2. Determining the coefficients 

 for each vector in the system. 

i

a

 







e

F

F

e

F

F

a

i

i

i

 

if

    

 

if

   

 

 
 

kN

F

a

kN

F

a

kN

F

a

kN

F

a

2

2

2

2

4

2

3

4

4

3

3

2

2

1

1

                    

kN

a

2

2

i

 

 
 3. The sum vector of the system can be computed as 
 

kN

kN

e

a

S

i

2

,

0

,

2

2

2

,

0

,

2

2

2

2





 

 

4. Determining the position vectors of the points of application of the forces. 
 

m

OD

r

m

OC

r

m

OB

r

m

OA

r

2

,

6

,

0

2

,

4

,

0

0

,

1

,

2

2

,

0

,

0

4

3

2

1

 

background image

 

 

 

 

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. 

2

 
5. Calculating the position vector of the center of the force system. 
 

 

 

 

m

m

kN

kNm

a

r

a

r

OO

i

i

i

6

,

6

,

4

2

12

,

12

,

8

2

2

2

,

6

,

0

2

2

,

4

,

0

1

0

,

1

,

2

4

2

,

0

,

0

3

2

 

6. Answer 
At the center of the parallel system 

  the system is reduced to the resultant force 

m

O

6

,

6

,

4

kN

S

W

2

,

0

,

2