Determine x and y coordinates of resultant force through object

Click For Summary
SUMMARY

The discussion focuses on determining the x and y coordinates of the resultant force acting on a concrete slab subjected to multiple vertical loads. The resultant force, calculated as Rz = -212 kN, indicates a net downward force. Moments about the x-axis and y-axis were computed, yielding Mx = -1027.1 kNm and My = 552.4 kNm, respectively. The final coordinates of the resultant force were determined to be x = -2.60566 and y = 4.8448.

PREREQUISITES
  • Understanding of static equilibrium and resultant forces
  • Knowledge of moment calculations in mechanics
  • Familiarity with coordinate systems in structural analysis
  • Proficiency in basic algebra for solving equations
NEXT STEPS
  • Study the principles of static equilibrium in structural engineering
  • Learn about calculating moments in two-dimensional systems
  • Explore the application of coordinate systems in force analysis
  • Review examples of resultant force calculations in engineering problems
USEFUL FOR

Students and professionals in civil engineering, structural analysis, and mechanics who are involved in calculating forces and moments on structures.

Northbysouth
Messages
241
Reaction score
2

Homework Statement


The concrete slab supports the six vertical loads shown. Determine the x- and y-coordinates of the point on the slab through with the resultant of the loading system passes.

Image attached


Homework Equations





The Attempt at a Solution



I started off by finding the resultant force. Seeing as how all the forces in the system act in only the z direction I summed them to get:

Rz = 69Kn - 69kN -42kN - 44kN - 67kN - 59kN
= -212 kN

But I'm not sure how I should proceed from this step. I would appreciate any suggestions.
 

Attachments

  • Statics hwk 10 2.152.png
    Statics hwk 10 2.152.png
    22.3 KB · Views: 2,030
Physics news on Phys.org
Hi Northbysouth! :smile:

Assume that it goes through the position (x,y).

You know it's -212 kN, so what is the moment of that about the x-axis or the y-axis? :wink:
 
I've managed to solve it now. I did the following:

Rz = -212 kN

Mx = (69kN)(2.3) -(67kN)(3.1+2.3) -(44)(8) -59*(2.3+3.1+2.6)
= -1027. kNm

My = 67*2.3 + 44*6.7 + 69*6.7 -69*5.2
= 552.4 kNm

Mz = 0

M = 1166.224 kNm

Rz*y = -1027.1
y = 4.8448

Rzx = 552.4
x = -2.60566

Thanks for your help.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
1
Views
11K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K
Replies
1
Views
3K