Mass and Force distribution between two legs

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Discussion Overview

The discussion revolves around the distribution of forces acting on a mass mounted on two fixed supports, exploring how to calculate the forces F1 and F2 at these supports based on the given angles and the total force acting on the mass. The conversation includes attempts to derive these forces mathematically and questions about the validity of different approaches.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the total force acting on the mass as F = 98N and proposes that F1 should be greater than F2 due to their positions relative to a point on the ground.
  • Another participant challenges the method of calculating F1 and F2 by stating that forces cannot be split in the manner suggested, emphasizing the need to consider all forces acting on the mass together.
  • A participant suggests an alternative approach by treating the configuration as a straight line and attempts to derive F1 and F2 using equilibrium equations, claiming to prove F1 > F2.
  • There is a discussion about whether it is necessary to consider the x-components of the forces, with some participants arguing that the bar is rigid and thus the total mass acts on the two points, while others question the assumptions about rigidity.
  • One participant expresses confusion about the rigidity of the bar and the implications for the calculations, indicating a need for clarification on these concepts.
  • A later reply points out that the new diagram presented corresponds to a different situation than the original, suggesting that the original diagram should be used for calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method for calculating the forces F1 and F2. There are competing views on whether to consider x-components and the implications of rigidity in the system.

Contextual Notes

Participants express uncertainty regarding the assumptions about rigidity and the necessity of considering different components of the forces. The discussion reflects varying interpretations of the problem setup and the mathematical approach required.

akhter900
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...o..M
.... /|\
.Theta 2 /..|.\.Theta 1
.../...v..\
...F2 /...F...\.F1
.../.....\
.../... \
-----------|---------ground
...|...p...|
...v..f2.....v..f1

Hi every one,
On the above diagram, M=10g is mounted on two fixed rods from the ground. So, the force acting on the point 'o' is F=10*9.8 = 98N. This force is going to distribute through the support as F1 and F2 and force on the ground for that two support will be f1 and f2. Here f1 is closer to p than f2. So, f1 should be grater than f2, ie. f1 > f2, is not it? Theta 1 = 30 degree and theta 2 = 45 degree.

To find out F1 and F2 I did something like,
F1 = F/Cos(30) = 113.16N
F2 = F/Cos(45) = 138.59N
Is it correct for force vector? I need to proof that F1 > F2.

I could not find out the distributed force F1 and F2 again f1 and f2 also. Is there anybody to give me some suggestions on this. I will be so grateful to you.
 

Attachments

  • ForceDistribution.jpg
    ForceDistribution.jpg
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Last edited:
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welcome to pf!

hi akhter900! welcome to pf! :smile:

(have a theta: θ and a degree: ° and try using the X2 tag just above the Reply box :wink:)
akhter900 said:
To find out F1 and F2 I did something like,
F1 = F/Cos(30) = 113.16N
F2 = F/Cos(45) = 138.59N

no, you can't do that, you can't split up the forces on the same body …

F1 and F2 are both acting on M, so they must both be in any equation for M

start again, doing components of all forces on M in the x and y directions (separately) :smile:​
 
Dear Tiny-tim,
Thanks a lot for the quick reply. What if I do the following way...

If I consider the above configuration as a straight line where M acting on its center of mass then it is possible to find out F1 and F2, isn't it?

...x2.....x1
...M2......M, F...M1
...O---------------o----------O
---|---------------------------|----Ground
...v..F1. ......v...F2

So, F = 98N. x1 = 2 and x2 = 4.
F = F1 + F2------------(1)
=> F1 = F-F2
F1*x1 = F2*x2---------(2)
=> (98-F2)2 = (F2*4)
=> (4*F2 + 2*F2) = 2*98
=> F2 = (2*98)/6 = 32.66N
Similarly, F1 = 65.33N

So, F1 > F2 Proved. and the friction for F1 is grater than friction for F2.

Is the procedure correct. Need your valuable suggestion.
 
hi akhter900! :smile:
akhter900 said:
If I consider the above configuration as a straight line where M acting on its center of mass then it is possible to find out F1 and F2, isn't it?

...x2.....x1
...M2......M, F...M1
...O---------------o----------O
---|---------------------------|----Ground
...v..F1. ......v...F2

hmm :confused: … you seem to be laying the whole thing out on the ground …

there might be some merit in that, but i don't see it yet :redface:
So, F = 98N. x1 = 2 and x2 = 4.
F = F1 + F2------------(1)
=> F1 = F-F2
F1*x1 = F2*x2---------(2)

no … (1) should be the equation for the y-components, and (2) should be the equation for the x-components

try again :smile:​
 
Do I need to consider x component here? The bar is considered as rigid. The total mass is acting on the two points F1 and F2. based on that, F1 > F2. its proved.

If you have any other idea or technique, please tell me. I will be so much grateful to you.
 
akhter900 said:
The bar is considered as rigid.

what bar is considered as rigid? :confused:

i see nothing about rigidity in the question or in the diagram …

unless the whole joint at M is rigid, yes you must consider the x components
 
According to the figure...(just uploaded)

If I consider the bar AB is rigid then the force F applied on the point O will divide and act on the two points A and B. Am I right?

The x component for X1 or X2 makes 0 degree with the horizontal line, so Cos0 will be 1. So the x component will be only X1 and X2. Is it correct?

F1 and F2 can be found by the mass distribution law mentioned earlier. F1 is grater than F2, so the friction on the point B will be more than the other point A.

Is it okay? Your suggestion please...
 

Attachments

  • ForceDistribution_2.jpg
    ForceDistribution_2.jpg
    3.4 KB · Views: 449
Would you please give me some links or documents about this type of problem? I want to study on...
 
hi akhter900! :smile:
akhter900 said:
If I consider the bar AB is rigid then the force F applied on the point O will divide and act on the two points A and B. Am I right?

This is a completely different situation (and incidentally your F1 and F2 in the new diagram correspond to f1 and f2 in your original diagram, not to F1 and F2).

You must use the original diagram.

I do not understand why you will not try using the component method, as suggested. :redface:
 

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