Static Eqm-unsure about forces to be drawn


by makeAwish
Tags: drawn, eqmunsure, forces, static
makeAwish
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#1
Mar27-09, 08:16 PM
P: 128
1. The problem statement, all variables and given/known data
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. Relevant equations

sum of F = 0
sum of moments = 0

3. 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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LowlyPion
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#2
Mar27-09, 11:52 PM
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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.
makeAwish
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#3
Mar28-09, 03:33 AM
P: 128
Quote Quote by LowlyPion View Post
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?

Doc Al
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#4
Mar28-09, 07:53 AM
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Static Eqm-unsure about forces to be drawn


Quote Quote by makeAwish View Post
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.
makeAwish
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#5
Mar29-09, 08:48 AM
P: 128
Quote Quote by Doc Al View Post
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 cant translate horizontally at these points?
(like the support reactions..) hmm.. is it?
Doc Al
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#6
Mar29-09, 09:56 AM
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If you place a book on a table, what's the horizontal reaction force? Why is that case any different than this one?
makeAwish
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#7
Mar29-09, 10:31 AM
P: 128
Quote Quote by Doc Al View Post
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?
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#8
Mar29-09, 10:42 AM
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Quote Quote by makeAwish View Post
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?)
makeAwish
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#9
Mar29-09, 10:45 AM
P: 128
No force?
Doc Al
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#10
Mar29-09, 10:48 AM
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Quote Quote by makeAwish View Post
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.
makeAwish
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#11
Mar29-09, 10:53 AM
P: 128
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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