Equilibrium of a stiff plate on inclined planes

AI Thread Summary
The discussion revolves around the equilibrium of a stiff rectangular plate on inclined planes, where participants express confusion about how to start solving the problem. Key points include the need to visualize the scenario with a diagram, particularly focusing on the angles α and β, and the relationship between the center of gravity (G) and the point O. The conversation emphasizes the importance of geometry and the conditions for equilibrium, such as the height of G and the angles involved. Participants also explore the principle of virtual work to determine equilibrium conditions, indicating a complex interplay of forces and geometry in the problem. The discussion highlights the challenges of applying theoretical concepts to practical scenarios in physics.
aang
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Homework Statement
A thin stiff uniform rectangular plate with width L (L =AB) is lying on two inclined surface as shown in Fig. 3-1. The angle between the horizontal surface and the left inclined surface is α, and that between the horizontal surface and the right inclined surface is β. It is assumed that the plate length is sufficiently larger than L and both edges of the plate are smooth enough to neglect friction.

(1) In case of α+β = 90°, the static balance shown in Fig.3-1 is
Relevant Equations
if three forces act on a body in equilibrium,
The sum of vectors is zero
The Vectors cross or intersect at same point and they are in the same plane.
1625804175704.png
I HAVE NO IDEA HOW TO START.ONLY THINGS I KNOW ARE WHAT I RERAD ON THIS THRED.
https://www.physicsforums.com/threads/equilibrium-of-a-stiff-plate-on-inclined-planes.947601/I can't continue from there. There are also questions where α=β, α+β=45 and where α=45,β=60.How to make use of the fact that G will follow the arc of a circle centered at O, and that at equilibrium it is vertically above O.
 
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aang said:
Homework Statement:: A thin stiff uniform rectangular plate with width L (L =AB) is lying on two inclined surface as shown in Fig. 3-1. The angle between the horizontal surface and the left inclined surface is α, and that between the horizontal surface and the right inclined surface is β. It is assumed that the plate length is sufficiently larger than L and both edges of the plate are smooth enough to neglect friction.

(1) In case of α+β = 90°, the static balance shown in Fig.3-1 is
Relevant Equations:: if three forces act on a body in equilibrium,
The sum of vectors is zero
The Vectors cross or intersect at same point and they are in the same plane.

View attachment 285680I HAVE NO IDEA HOW TO START.ONLY THINGS I KNOW ARE WHAT I RERAD ON THIS THRED.
https://www.physicsforums.com/threads/equilibrium-of-a-stiff-plate-on-inclined-planes.947601/I can't continue from there. There are also questions where α=β, α+β=45 and where α=45,β=60.How to make use of the fact that G will follow the arc of a circle centered at O, and that at equilibrium it is vertically above O.
A good start would be to draw a new diagram making use of what you gleaned from the other thread.
 
isn't the diagram a rectangle which has L as diagonal.
WHAT I DONOT KNOW IS How to make use of the fact that G will follow the arc of a circle centered at O AND HOW TO GET α AND β ONTO THE DIAGRAM(INTO THE GAME)
 
IN THE CASE OF α=45,β=60 Those don't EVEN add to 90.(DON'T THEY HAVE TO)
BUT THE QUESTION SAYS THAT STATIC BALANCE IS POSSIBLE.
HOW IS THAT.
 
help.
 
aang said:
isn't the diagram a rectangle which has L as diagonal.
WHAT I DONOT KNOW IS How to make use of the fact that G will follow the arc of a circle centered at O AND HOW TO GET α AND β ONTO THE DIAGRAM(INTO THE GAME)
A diagram showing
aang said:
that at equilibrium G is vertically above O.
Once you have that, all you need is geometry.
aang said:
IN THE CASE OF α=45,β=60 Those don't EVEN add to 90.(DON'T THEY HAVE TO)
BUT THE QUESTION SAYS THAT STATIC BALANCE IS POSSIBLE.
HOW IS THAT.
The angles adding to 90 just makes the solution simpler. In the general case, equilibrium is still possible but I don't see any nice geometric solution.
If the distances of A and B from O are x and y, and the plate makes angle theta to the horizontal, what two equations can you write to relate them?
In terms of x, y and the given angles, what is the height of G?
 
DO YOU MEAN BETWEEN X,Y,THETA
IN THE GENERAL CASE OR WHEN The angles adding to 90.
 
I TRIED AND GOT X2+Y2= L2 WHEN The angles adding to 90.
NA cosα+NB cosβ=W
And I took a moment at point A,B
NA*X = W*L/2*cosθ
NB*Y = W*L/2*cosθ
I really can't find ANYTHING ELSE.
 
Last edited:
CAN YOU GIVE ME a hint or your first equation please.
 
  • #10
aang said:
I TRIED AND GOT X2+Y2= L2 WHEN The angles adding to 90.
NA cosα+NB cosβ=W
And I took a moment at point A,B
NA*X = W*L/2*cosθ
NB*Y = W*L/2*cosθ
I really can't find ANYTHING ELSE.
As I wrote, you only need geometry for this case, so don't worry about forces and moments.
Have you drawn a diagram showing the angle as 90 and with G vertically above O?
What is the height of G above O? If it is not immediately obvious, complete the rectangle from the points AOB and note its diagonals.
What is the angle GOB? So what is angle GBO?
Write angle GBO in terms of beta and theta.

For the general case, consider the horizontal distance from A to B. Write it in terms of L and theta, then again in terms of x, y, alpha and beta. That gives you an equation.
Do something similar for the vertical direction.
A third equation expresses the height of G in terms of the heights of A and B.
 
  • #11
the height of G above O=L/2
GOB=90-β
GBO=90-β+θ
LCOSθ=XCOSα+YCOSβ
LSINθ=XSINα+YSINβ
A third equation expresses the height of G in terms of the heights of A and B.(I DONOT KNOW HOW TO GET)
 
  • #12
aang said:
the height of G above O=L/2
GOB=90-β
GBO=90-β+θ
LCOSθ=XCOSα+YCOSβ
LSINθ=XSINα+YSINβ
A third equation expresses the height of G in terms of the heights of A and B.(I DONOT KNOW HOW TO GET)
You are mixing up the two sets of advice I gave.

The first is just for the 90 degree case. Having found the height of G above O is L/2:
haruspex said:
What is the angle GOB? So what is angle GBO?
Write angle GBO in terms of beta and theta.

When you've finished that case, proceed to the general one. Yes
LCOSθ=XCOSα+YCOSβ
But not:
LSINθ=XSINα+YSINβ
For G in terms of the heights of A and B, remember that G is halfway between them.
 
  • #13
L/2=X/2SINα+YSINβ
GBO=β+θ=GOB(I HOPE)
 
  • #14
HOW TO MAKE USE OF THESE.
 
  • #15
aang said:
GBO=β+θ=GOB
Yes, but what is the direct relationship between GOB and β?
aang said:
L/2=X/2SINα+YSINβ
No. You seem to be guessing.
Let's say B is lower than A.
What is the height of A above B in terms of:
1. x, y, α, β
2. L, θ
?
 
  • #16
direct relationship between GOB and β MEANING.
i really do not know HOW TO GET WHAT YOU ARE ASKING
 
  • #17
LSINθ=XSINα-YSINβ
 
  • #18
@haruspex just one question, are you trying to do this via the principle of virtual work?
 
  • #19
aang said:
direct relationship between GOB and β MEANING.
i really do not know HOW TO GET WHAT YOU ARE ASKING
Have you drawn a diagram showing AOB as a right angle and G vertically above O?
What angle does OG make to the horizontal?
What angle does OB make to the horizontal?
So what angle does OB make to OG?
(And please stop SHOUTING.)
 
  • #20
What angle does OG make to the horizontal?90
What angle does OB make to the horizontal?β
So what angle does OB make to OG?90-β
 
  • #21
aang said:
What angle does OG make to the horizontal?90
What angle does OB make to the horizontal?β
So what angle does OB make to OG?90-β
Sorry, just realized you already got that in post 11.
aang said:
GOB=90-β
Combine that with your equation in post 13:
aang said:
β+θ=GOB
 
  • #22
2β+θ=90
 
  • #23
aang said:
2β+θ=90
So now you have the value of theta for the right angle case.

On to the general case?
 
  • #24
On the general case WHAT TO DO?
 
  • #25
aang said:
On the general case WHAT TO DO?
See posts 12 and 15.
It's late here, so no more from me for some hours.
 
  • #26
LSINθ=XSINα-YSINβ
LCOSθ=XCOSα+YCOSβ
HOW TO USE THESE
 
  • #27
aang said:
LSINθ=XSINα-YSINβ
LCOSθ=XCOSα+YCOSβ
HOW TO USE THESE
So far, that's just geometry. Now you need to use the equilibrium condition.
That is, for a small displacement of the plate the mass centre does not ascend or descend. I.e. it is at its highest or lowest point.
So next you need the height of G in terms of the heights of A and B. It won't be L/2 in general. Use the fact that G is halfway between A and B, and you know the heights of those.
 
  • #28
OG=L/2*SINθ+YSINβ
 
  • #29
aang said:
OG=L/2*SINθ+YSINβ
Yes, that is a correct expression for the height of G, but it is not the same as OG. OG need not be vertical in the general case.
Call the height H, so we have H=L/2*SINθ+YSINβ.
What mathematical procedure should you use to find its maximum or minimum value?
 
  • #30
differentiation
 
  • #31
aang said:
differentiation
Right.
 
  • #32
H=L/2*SINθ+YSINβ.WANT ME TO differentiate.
but H IS NOT ON THE RIGHT SIDE.
WITH RESPECT TO WHAT SHOULD I differentiate.
 
  • #33
aang said:
H=L/2*SINθ+YSINβ.WANT ME TO differentiate.
but H IS NOT ON THE RIGHT SIDE.
WITH RESPECT TO WHAT SHOULD I differentiate.
You can differentiate H wrt anything that varies as the plate shifts position. I suggest its angle, theta, as the most convenient.
 
  • #34
haruspex said:
Now you need to use the equilibrium condition.
That is, for a small displacement of the plate the mass centre does not ascend or descend. I.e. it is at its highest or lowest point.
From where do you get this equilibrium condition? its not ##\sum F=0 or \sum M=0##...
 
  • #35
Delta2 said:
From where do you get this equilibrium condition? its not ##\sum F=0 or \sum M=0##...
Virtual work.
 
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  • #36
haruspex said:
Virtual work.
As I had suspected, something tells me that @aang hasn't been taught about virtual work . But I might be wrong.
 
  • #37
Delta2 said:
As I had suspected, something tells me that @aang hasn't been taught about virtual work . But I might be wrong.
For this question at least, it doesn’t seem to me it is something that needs to have been taught. Isn't it evident that if even the slightest displacement of the plate (in some direction) lowers its mass centre then it is not at equilibrium?
 
  • #38
haruspex said:
For this question at least, it doesn’t seem to me it is something that needs to have been taught. Isn't it evident that if even the slightest displacement of the plate (in some direction) lowers its mass centre then it is not at equilibrium?
No that doesn't seem so obvious to me but that's my subjective view, most people might find it obvious, I can't tell.
 
  • #39
Delta2 said:
No that doesn't seem so obvious to me but that's my subjective view, most people might find it obvious, I can't tell.
Think of it this way... suppose the least displacement to the right lowers the mass centre and so, the derivative being smooth, the least displacement to the left raises the mass centre. So displacing to the left does work, and it can only do that by opposing a force. So there must be a net force on the object to the right.
That is the essence of the principle of virtual work.
 
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  • #40
H wrt THETA=L/2*COSθ
IS IT CORRECT.
ARE WE NEAR THE SOLUTION
 
  • #41
aang said:
H wrt THETA=L/2*COSθ
IS IT CORRECT.
ARE WE NEAR THE SOLUTION
No, that is incorrect, and there is still a way to go.
As theta varies, x and y change. So these are functions of theta, and when you differentiate an expression containing them you get terms like ##\frac{dx}{d\theta}##. For simplicity, write these as x', y'.
 
  • #42
H wrt THETA=L/2*COSθ+SINβ* x'
 
  • #43
aang said:
H wrt THETA=L/2*COSθ+SINβ* x'
Right. And since we want the extremum of H that derivative is zero.
So this has introduced another unknown, x'. What you need to do next is to obtain other equations with x' and maybe y'. You can get those by differentiating the other equations you have.
 
  • #44
I have not seen the thread carefully perhaps somebody has already proposed to consider the potential energy's critical points
 
  • #45
LCOSθ=x'*SINα-y'*SINβ
-LSINθ=x'*COSα-y'*COSβ
HOW MUCH FURTHER.
 
  • #46
aang said:
LCOSθ=x'*SINα-y'*SINβ
-LSINθ=x'*COSα-y'*COSβ
HOW MUCH FURTHER.
You have a sign error in the second equation.

It seems to me that this problem is beyond your level. I am having to lead you every step. Repeatedly asking how much further doesn't fill me with confidence either.
Was this given to you as homework or is it something you found for yourself somewhere?
 
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