Weight Distributed at Three Points

Click For Summary

Discussion Overview

The discussion revolves around calculating the weight distribution at three points of support for a large rectangular piece of machinery. Participants explore the implications of static equilibrium and the need for additional information to resolve the weight at each point accurately. The context includes a practical application related to installation estimates.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • The total weight of the system is given as 3301 kg, with specific coordinates for the center of mass and points of support.
  • One participant notes that having more than two points of support may lead to a statically indeterminate situation, requiring more information than provided to resolve the reactions.
  • Another participant suggests a method for calculating the weight at each point based on distances from the center of gravity, but acknowledges the need for additional information about the system.
  • A later reply emphasizes the importance of providing more details about the system's construction to facilitate accurate calculations.
  • One participant proposes summing moments and forces to estimate the support forces, suggesting an approximate value of 1100 kg for each leg, but cautions that this is not for design purposes and should be verified.

Areas of Agreement / Disagreement

Participants generally agree that additional information about the system is necessary to accurately determine the weight distribution at the points of support. There is no consensus on the exact method or calculations to be used, and multiple approaches are discussed.

Contextual Notes

The discussion highlights limitations in the provided information, particularly regarding the system's construction and the assumptions made in the calculations. The reliance on static equilibrium principles is noted, but the exact mathematical steps and dependencies remain unresolved.

dmw08001
Messages
3
Reaction score
0

Homework Statement



Total Weight of System = 3301 kg
Length, Width of Sytem = (4670mm, 1931mm)
Calculated Center of Mass = (2261mm, 1065mm)

Find weight at the following points:

Point I (104, 1046)
Point II (3182, 1867)
Point III (3182, 225)



Homework Equations



Center of Gravity equation already used


The Attempt at a Solution



3182-2261=921
921/2261=.4075
40.75% of 3301kg = 1345
Weight at B+C = 1345*2=2690

3301-2690=610kg <--weight at Point I

I really am lost...
 
Physics news on Phys.org
Without knowing more about your 'system', it is hard to follow your calculations.

From simple statics, having more than two points of support introduces the possibility of a statically indeterminate situation, where one needs more information than just the total weight and the locations of the c.g. and the points of support to resolve the reactions.
 
Sorry SteamKing,

This isn't actually a formal homework problem so I understand more info may be needed. What additional information can I provide that will help determine a solution?

My train of thought is as follows:

(Point 2 X Value) - (Center of Gravity X) = Distance from CG

Distance from CG in X Coordinate / CG in X coordinate = Percentage of Weight in X Direction

Weight in X Direction * 2 (because point II and III are along the same plane) = Weight at II + Weight III

Total Weight - Weight II + III = Weight I
 
Last edited:
Well, you need to provide more information about this 'system' and its construction. Don't be bashful about telling what it is (unless it's secret or something.)

(Hint: it's against the rules at PF to post the same or similar threads in multiple forums. All you do is cause confusion.)
 
It is just a big rectangular piece of machinery that we manufacture in china and we need to determine weights at each of the three feet so that we can give the customer and operations guys estimates for installation.

Static solution is all that is needed.
 
You can sum moments in 2 directions, and sum forces in the vertical direction, to solve. We can do the math, but essentially your support forces in each leg are close to equal, about 1100 kg force in each leg. I didn't do the exact math, so this number is approximate, I'd use it for estimating only, and the value needs to be checked! Do not use it for design!
 

Similar threads

  • · Replies 29 ·
Replies
29
Views
10K
Replies
25
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
5K
Replies
7
Views
6K
Replies
2
Views
3K
  • · Replies 13 ·
Replies
13
Views
71K