Static Eqm-unsure about forces to be drawn

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Homework Help Overview

The discussion revolves around a static equilibrium problem involving a uniform beam supporting two blocks at different positions. The original poster is uncertain about how to represent the forces in their free body diagram, particularly regarding the normal forces and the role of the knife edge supports.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of treating the knife edge at point O as non-existent and question how this affects the moments about point P. There is discussion about the presence of horizontal reaction forces at points P and O, and whether the normal forces between the beam and blocks are internal or external forces.

Discussion Status

Some participants suggest focusing on the moments about point P while disregarding point O, and there is a back-and-forth regarding the nature of forces acting on the beam and blocks. The conversation indicates a productive exploration of the concepts of internal and external forces, though no consensus has been reached on the specifics of the forces involved.

Contextual Notes

Participants are navigating the complexities of static equilibrium and the definitions of forces in the context of the problem setup. There is a lack of explicit consensus on the treatment of forces at the support points and the implications for the overall analysis.

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Homework Statement


A uniform beam of mass mb and length supports blocks with masses m1 and m2 at two positions, as in Fig. P12.3. The beam rests on two knife edges. For what value of x will the beam be balanced at P such that the normal force at O is zero?

2.jpg



Homework Equations



sum of F = 0
sum of moments = 0

The Attempt at a Solution



I'm quite unsure regarding the forces to be drawn in my free body diagram.

Example, the two normal forces by beam on the each masses. Do i draw them out?

And at points P and O, is there a horizontal reaction force?

How do i determine the forces at point P?
Like when i looking at point P, do i think as the way below?

(treat the knife edge at O as not existing) the beam can rotate at P, so no moments.
but the beam cannot translate horiz and vert at P, so there are horiz and vert reaction forces.

or I cannot treat the knife edge at O as not existing? If so, means the beam cannot rotate at P and there will be a moments?



Do you all know what i mean? =x
Can explain to me? Thanks!
 
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The best treatment is to consider that point O doesn't exist.

Then all you care about are the sum of the moments about P, which you can write out by inspection.

As to the forces at P, that's just the Σ m*g if it is in balance.
 
LowlyPion said:
The best treatment is to consider that point O doesn't exist.

Then all you care about are the sum of the moments about P, which you can write out by inspection.

As to the forces at P, that's just the Σ m*g if it is in balance.

okay thanks. So at points P and O, is there a horizontal reaction force?
Then for the two normal forces by beam on the each masses, are they considered as internal forces?
 
makeAwish said:
So at points P and O, is there a horizontal reaction force?
Why would there be?

Then for the two normal forces by beam on the each masses, are they considered as internal forces?
If you consider the beam + blocks as a single system, then the normal force between them would be an internal force. But you could also treat the blocks separately, then the normal force would be an external force. Either way is fine.
 
Doc Al said:
Why would there be?


If you consider the beam + blocks as a single system, then the normal force between them would be an internal force. But you could also treat the blocks separately, then the normal force would be an external force. Either way is fine.

Yup. Think i understand the internal forces :) Thanks a lot!

There are horizontal forces cos the beam can't translate horizontally at these points?
(like the support reactions..) hmm.. is it?
 
If you place a book on a table, what's the horizontal reaction force? Why is that case any different than this one?
 
Doc Al said:
If you place a book on a table, what's the horizontal reaction force? Why is that case any different than this one?

If i push the book, there will be a frictional force, provided surface not smooth. Correct?
 
makeAwish said:
If i push the book, there will be a frictional force, provided surface not smooth. Correct?
Sure, if you push it horizontally. (But what if you don't?)
 
No force?
 
  • #10
makeAwish said:
No force?
Right. If you just place a book on a (horizontal) table, the reaction force of the table on the book is strictly vertical--no horizontal component.
 
  • #11
Means it is not like those support reactions? Like if we cannot rotate abt that point, the there is a moment; if we cannot translate vertically or horizontally, then there is a reaction vertically or horizontally?
 

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