# Equilibrium of a stiff plate on inclined planes

• thevinciz
In summary, a thin stiff uniform rectangular plate is lying on two inclined surfaces with angles α and β between the horizontal surface and the left and right inclined surfaces respectively. In the case of α+β = 90°, the static balance shown in the figure is always possible. The problem can be solved by calculating the horizontal and vertical components of the normal forces at points A and B, and finding the contact plane between the plate and the inclined surfaces. The angle that NA makes to the plate is 90° and the angle that the plate makes to the horizontal is unknown. The question is whether there is an angle that makes the system stable.
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

## 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?

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thevinciz said:
moment at point B, so NA*L = mg*Lcosα.
What angle does NA 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?

haruspex said:
What angle does NA 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 ?

thevinciz said:
(1) I think it is α
That would make NA horizontal.
thevinciz said:
(2) Isn't it L/2?
So why no /2 in your equation?
thevinciz said:
(3) α ?
I see no reason for it to do so. Start with this question: what angle does the plate make to the horizontal?

haruspex said:
That would make NA 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?

thevinciz said:
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.
thevinciz said:
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.

haruspex said:
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.

thevinciz said:
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.
thevinciz said:
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.

haruspex said:
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.

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

haruspex said:
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.

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thevinciz said:
So the NA makes α angle with vertical direction or 90° to the left inclined plane.
Yes.
thevinciz said:
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?

haruspex said:
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.

thevinciz said:
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.

haruspex said:
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?

thevinciz said:
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.

haruspex said:
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

thevinciz said:
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?

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haruspex said:
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.

thevinciz said:
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?

haruspex said:
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?

thevinciz said:
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?
thevinciz said:
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.

haruspex said:
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?

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thevinciz said:
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?
thevinciz said:
Circular path?
Yes.

haruspex said:
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.

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

thevinciz said:
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?

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

no

thevinciz said:
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.

haruspex said:
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

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.

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

##\ ##

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.

aang said:
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.
aang said:
and where α=45,β=60.
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.

## 1. What is the concept of equilibrium in relation to a stiff plate on inclined planes?

The concept of equilibrium in this context refers to the state in which the plate is not moving or rotating, and all the forces acting on it are balanced.

## 2. How is the equilibrium of a stiff plate on inclined planes calculated?

The equilibrium of a stiff plate on inclined planes can be calculated using the principles of statics, which involve analyzing the forces and moments acting on the plate and ensuring that they are balanced.

## 3. What factors affect the equilibrium of a stiff plate on inclined planes?

The equilibrium of a stiff plate on inclined planes can be affected by the angle of inclination, the weight and dimensions of the plate, and the surface friction between the plate and the inclined plane.

## 4. What are the different types of equilibrium that can be achieved for a stiff plate on inclined planes?

The different types of equilibrium that can be achieved for a stiff plate on inclined planes include stable equilibrium, where the plate remains stationary even when slightly disturbed, and unstable equilibrium, where the plate will move if even a small force is applied.

## 5. How is the equilibrium of a stiff plate on inclined planes used in real-world applications?

The concept of equilibrium for a stiff plate on inclined planes is used in various engineering and construction applications, such as designing stable structures on sloped surfaces or calculating the forces on a plane during takeoff and landing for aircrafts.

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