Equilibrium of a stiff plate on inclined planes

[thevinciz]

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
(a) possible only provided α≠β
(b) always possible
(c) always impossible

Homework Equations

The Attempt at a Solution

From the picture, at point A and B,
Calculating in horizontal axis : NA(normal force at a)*sinα = NBsinβ
vertically direction : NAcosα+NBcosβ=mg
And I took a moment at point B, so NA*L = mg*Lcosα.
And I tried to substitue all variables into α and β only, but I got quadratic equation that looked like wrong : (cosα)^2+cosα(sinα)/sinβ-2=0 because I have solved with x=-b+-b^2... but it couldn't be solved.

Which way should I do to solve this problem?

 

[haruspex]

moment at point B, so NA*L = mg*Lcosα.

What angle does N_A make to the plate?
How far along the plate is its mass centre from B?
What angle does the weight of the plate make to the plate?

 

[thevinciz]

What angle does N_A make to the plate?
How far along the plate is its mass centre from B?
What angle does the weight of the plate make to the plate?

(1) I think it is α
(2) Isn't it L/2?
(3) α ?

 

[haruspex]

(1) I think it is α

That would make N_A horizontal.

(2) Isn't it L/2?

So why no /2 in your equation?

(3) α ?

I see no reason for it to do so. Start with this question: what angle does the plate make to the horizontal?

 

[thevinciz]

That would make N_A horizontal.

So why no /2 in your equation?

I see no reason for it to do so. Start with this question: what angle does the plate make to the horizontal?

(1) I think the angle that NA makes to plate is 90 because it's normal force.
(2) I see.
(3) I really can't find the angle which the plate make to the horizontal. Could you give me a hint please?

 

[haruspex]

I think the angle that NA makes to plate is 90 because it's normal force.

The end of the plate rests on the slope. To decide where the normal is you have to find the contact plane, i.e. a flat plane that fits between them without intersecting either.

can't find the angle which the plate make to the horizontal.

It is unknown. The question is whether there is an angle which makes the system stable.

 

[thevinciz]

The end of the plate rests on the slope. To decide where the normal is you have to find the contact plane, i.e. a flat plane that fits between them without intersecting either.

It is unknown. The question is whether there is an angle which makes the system stable.

(1) Isn't NA perpendicular to the plate? It looks same as a typical rectangular object on a inclined plane problems

(2) I think I found the angle which plate makes to the horizontal now, it is 90-β.
But I don't know how to find the answer to your latest questions.

 

[haruspex]

1) Isn't NA perpendicular to the plate? It looks same as a typical rectangular object on a inclined plane problems

In a typical object on an inclined plane problem, the normal is normal to the plane. If it is a rectangular object flush to the plane then it will be normal to that too, but here it is not.
See section 1 in https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/

2) I think I found the angle which plate makes to the horizontal now, it is 90-β.

Reread what I wrote in post #6. The slope of the plate is unknown. You can see this by drawing the diagram with the two slopes unchanged but sliding the plate to a different angle.

 

[thevinciz]

In a typical object on an inclined plane problem, the normal is normal to the plane. If it is a rectangular object flush to the plane then it will be normal to that too, but here it is not.
See section 1 in https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/

Ah, I see. So the NA makes α angle with vertical direction or 90° to the left inclined plane.

with the two slopes unchanged

(2) I understand that you meant slope of both planes, if I understand correctly I think there is a angle which make this system possible. Because there are normal forces to make it become steady.

 

[haruspex]

So the NA makes α angle with vertical direction or 90° to the left inclined plane.

Yes.

I think there is a angle which make this system possible.

That is what the question is asking, and determining it is not trivial.

Can you answer this: if three forces act on a body in equilibrium, what can you say about the lines of action of those forces?

 

[thevinciz]

Can you answer this: if three forces act on a body in equilibrium, what can you say about the lines of action of those forces?

The sum of vectors is zero
The Vectors cross or intersect at same point and they are in the same plane.

 

[haruspex]

The Vectors cross or intersect at same point

That's the one.
You know α and β add up to π/2 in the first case. Draw yourself a diagram that represents that better, showing the three forces intersecting.
It will help to create an unknown, θ, for the slope of the plate.

 

[thevinciz]

That's the one.
You know α and β add up to π/2 in the first case. Draw yourself a diagram that represents that better, showing the three forces intersecting.
It will help to create an unknown, θ, for the slope of the plate.

The angle which the plate makes to horizontal, θ, is 90°-β. How could this help to solve problems?

 

[haruspex]

The angle which the plate makes to horizontal, θ, is 90°-β.

For the third time, at least, θ is not known.
Draw the two slopes at some angles α and β. Mark a point A for one end of the plate on the left hand slope. You can now put B anywhere you like on the right hand slope, so θ is any value from -α (i.e. sloping down to the right, as shown) to +β.

Your task is to say whether, for the given α and β, there is some θ for which the arrangement is stable.

 

[thevinciz]

For the third time, at least, θ is not known.
Draw the two slopes at some angles α and β. Mark a point A for one end of the plate on the left hand slope. You can now put B anywhere you like on the right hand slope, so θ is any value from -α (i.e. sloping down to the right, as shown) to +β.

Your task is to say whether, for the given α and β, there is some θ for which the arrangement is stable.

there is

 

[haruspex]

there is

You are only guessing. You have not shown that.
I should clarify for you that this is a question about stability, which is not the same as equilibrium.
A pencil on its point could, in principle, be in equilibrium, but it cannot be stable. The slightest perturbation will make it fall over.

Draw the diagram I described in post #12.
Call the point at the base where the slopes meet O and centre of the plate G. Show the three intersecting forces, with the normals at right angles to the slopes and α+β=π/2. You should then be able to answer this key question: what is the distance OG?

 

[thevinciz]

You are only guessing. You have not shown that.
I should clarify for you that this is a question about stability, which is not the same as equilibrium.
A pencil on its point could, in principle, be in equilibrium, but it cannot be stable. The slightest perturbation will make it fall over.

Draw the diagram I described in post #12.
Call the point at the base where the slopes meet O and centre of the plate G. Show the three intersecting forces, with the normals at right angles to the slopes and α+β=π/2. You should then be able to answer this key question: what is the distance OG?

I have drawn my diagram, it looks like rectangle which has L as diagonal. I think OG is L/2.

 

[haruspex]

I have drawn my diagram, it looks like rectangle which has L as diagonal. I think OG is L/2.

Right!
What is the orientation of the other diagonal through the mass centre?
If the plate were to slip a little, with one end going down its slope and the other end up its slope, what path would the mass centre move along? What would happen to the height of the mass centre?

 

[thevinciz]

Right!
What is the orientation of the other diagonal through the mass centre?
If the plate were to slip a little, with one end going down its slope and the other end up its slope, what path would the mass centre move along? What would happen to the height of the mass centre?

The other diagonal passes through the mass centre of plate.

I have drawn the left end plate going down and the other going up. The mass centre of the new diagram is located at same position of the first diagram. Right?

 

[haruspex]

The other diagonal passes through the mass centre of plate.

Sure, but I asked what its orientation is, e.g. what angle does it make to the vertical?

The mass centre of the new diagram is located at same position of the first diagram.

My question was what path would the mass centre take if the plate were to slip. For that, remember what you found in post #17.

 

[thevinciz]

Sure, but I asked what its orientation is, e.g. what angle does it make to the vertical?

My question was what path would the mass centre take if the plate were to slip. For that, remember what you found in post #17.

(1) From AG=OG since they are diagonals, so AOG = GOA = 45° because AGO = 90° which is caused by intersecting of two diagonals. So OG makes 45° angle to the left inclined plane.

(2) Circular path?

 

[haruspex]

because AGO = 90° which is caused by intersecting of two diagonals

The diagnals of a rectangle only intersect at 90° if it is a square.
Remember the line of action of gravity. What points does that go through?

Circular path?

Yes.

 

[thevinciz]

The diagnals of a rectangle only intersect at 90° if it is a square.
Remember the line of action of gravity. What points does that go through?

Intersecting point of diagonals and 3 forces.

 

[thevinciz]

I think I got the angle which OG makes to left inclined plane, it is 90°-α

 

[haruspex]

I think I got the angle which OG makes to left inclined plane, it is 90°-α

Yes, or more simply, it is vertical.
So at all times G lies on a circle centred on O and in equilibrium it is vertically above O. If the plate slips, one way or the other, will G move higher or lower?

 

[thevinciz]

If the plate slips, one way or the other, will G move higher or lower?

no

 

[haruspex]

no

You wrote, correctly, that if it slips G will follow the arc of a circle centred at O, and that at equilibrium it is vertically above O. So if it moves left or right it will also move down, right? It is like a pebble at the top of a perfectly smooth sphere.

 

[thevinciz]

You wrote, correctly, that if it slips G will follow the arc of a circle centred at O, and that at equilibrium it is vertically above O. So if it moves left or right it will also move down, right? It is like a pebble at the top of a perfectly smooth sphere.

I got it

 

[aang]

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.

 

[BvU]

​

Please read the PF guidelines

Better to start a new thread that pick up one from three years ago. You seem to have a different problem statement, too.

$\$

 

[aang]

Well the first problem is α+β = 90.
IT is the exact same problem.
Do i really need to start a new thread for the same question.

 

[haruspex]

Well the first problem is α+β = 90.
IT is the exact same problem.
Do i really need to start a new thread for the same question.

and where α=45,β=60.

Those don't add to 90.
But more importantly, as with a new thread in a homework forum, you should show some attempt. A diagram employing the advances already made in this thread would be a good start.

Edit: aang has now started a new thread at https://www.physicsforums.com/threads/equilibrium-of-a-stiff-plate-on-inclined-planes.1004932/