Equilibrium of a stiff plate on inclined planes

[Delta2]

Virtual work.

As I had suspected, something tells me that @aang hasn't been taught about virtual work . But I might be wrong.

 

[haruspex]

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?

 

[Delta2]

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.

 

[haruspex]

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.

 

[aang]

H wrt THETA=L/2*COSθ
IS IT CORRECT.
ARE WE NEAR THE SOLUTION

 

[haruspex]

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'.

 

[aang]

H wrt THETA=L/2*COSθ+SINβ* x'

 

[haruspex]

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.

 

[wrobel]

I have not seen the thread carefully perhaps somebody has already proposed to consider the potential energy's critical points

 

[aang]

LCOSθ=x'*SINα-y'*SINβ
-LSINθ=x'*COSα-y'*COSβ
HOW MUCH FURTHER.

 

[haruspex]

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?