# Stability of a given structure composed of three rods and a string

• JohanL
In summary, the problem involves two prisms, one on top of the other, with no friction present. The top prism glides down to the ground, causing the bottom prism to move to the left. The goal is to determine how far the bottom prism moves. The lagrangian for the system is given by L = Tp1 + Tp2 - mgh, where h is the height to the center of mass of the top prism. A constraint is needed to solve the problem, but it is unclear how to set up the equation for the constraint.
JohanL
Three rods with length l and mass m are arranged like the figure in the attached file. There connections are without friction.
One rod is horizontal and the other two parallell and in the same plane.
An elastic string connects point A and B.
The tension in the string is proportional to its length.

Determine the equillibrium angle a and investigate the stability.

Any ideas on how to solve this problem?

Thank you.

#### Attachments

• problem.GIF
1.1 KB · Views: 318
I don't even know how to start on this one.
Maybe someone can give some hints?

Is the elastic coefficient k of the string also given as data in the problem?

only that

tension in the string = k * length of string

There must be something more to this. What is holding these rods together? Are they hinged? Is the system subject to gravity? Is some part of this apparatus being held in place so it cannot free fall?

Wait. I think I get it.

See enhanced diagram.

"What is holding these rods together? Are they hinged?"
Yes, as stated: "...[their] connections are without friction..."

Last edited:
I think you've got it!

Let's try this for starters

For equilibrium, neglect the thickness of the rods. Reference the zero of gravitational potential energy U to the top mount. Then

$$U = -2mg \l cos a$$

The string is stretched to a length x given by the law of cosines

$$x^2 = \l^2 + \l^2 -2 \l^2cos(90 - a)$$

$$x^2 = \l^2 + \l^2 -2 \l^2sina$$

$$x^2 = 2 \l^2 (1 - sina)$$

The potential energy of the string equals the lost gravitational potential energy

$$2mg \l cos a = \frac{1}{2}kx^2 = k \l^2 (1 - sina)$$

$$\frac{1 - sina}{cos a} = \frac{2mg}{k \l}$$

Last edited:
OlderDan said:
I think you've got it!

You are using an energy conservation approach. Remember that we don't know the initial height. Even if we did, we would only find the point at which the system will temporarily stop (still no equilibrium). What we are asked is to find the equilibrium angle.

OlderDan said:
For equilibrium, neglect the thickness of the rods. Reference the zero of gravitational potential energy U to the top mount. Then

$$U = -2mg \l cos a$$

Wrong.

JohanL. Use the equilibrium condition with forces that apply on the bottom rod.

ramollari said:
You are using an energy conservation approach. Remember that we don't know the initial height. Even if we did, we would only find the point at which the system will temporarily stop (still no equilibrium). What we are asked is to find the equilibrium angle.

You are quite right. So my erroneous calculation, if done correctly, should be the minimum angle if the system were released from rest with an angle of a = 90, assuming the elastic constant of the string is sufficient to keep the thing from going past vertical. That assumption is of course not justified- My bad.

Also another hint: the absence of friction at the hinges makes the reaction force of a hinge colinear with the rod to which it joins (see why). With this, it should be very easy to set up the equilibrium conditions for the horizontal and vertical directions and express the angle a in terms of m, l, and k.

Let me see if I can salvage a bit of dignity after making my earlier sophomoric error. An energy approach is valid if done correctly, so let's see if I can do it right. Again, referencing gravitational potential energy to zero at a = 90

$$U = -2mg \l cos a$$

The string is stretched to a length x given by the law of cosines

$$x^2 = \l^2 + \l^2 -2 \l^2cos(90 - a)$$

$$x^2 = \l^2 + \l^2 -2 \l^2sina$$

$$x^2 = 2 \l^2 (1 - sina)$$

The total potential energy combining gravity and the spring is the sum of the two

$$U_{total} = -2mg \l cos a + \frac{1}{2}kx^2$$

$$U_{total} = -2mg \l cos a + k \l^2 (1 - sina)$$

Equilibrium is at the minimum of the potential energy (not the zero, DOH! as Homer would say). The zero of the potential energy will give the limiting value for a. As I think more one it, there will be a limiting value somewhere, though it might well be at a negative angle between zero and 90.

$$\frac {dU_{total}}{da} = 2mg \l sin a + -k \l^2 cosa = 0$$

$$2mg sin a = k \l cosa$$

$$tan a = \frac{k \l}{2mg }$$

Working out the result directly in terms of the forces involved as recommended by ramollari is a worthwhile exercise. If I have done my work right, the two results should agree.

Thank you!

The answer for the stability is
stable if 0 < a < pi/2
unstable if pi < a < 3pi/2

And its unstable in that range because the potential energy don't have a minimum there.

I have another problem i haven't been able to solve. Maybe you can give me some hints?

If you have two prisms on top of each other, one smaller than the other, and the one on top glids down to the ground the lower prisms moves to the left.
The problem is to find how far the lower prism moves.

There is no friction between the prisms or between the lower prism and the ground.

If you choose an orthogonal coordinate system with the x-axis parallell to the ground and if Tpi is the kinetic energy for the i:th prism you get for the lagrangian

L = Tp1 + Tp2 - mgh

where h is the height to the center of mass of prism 2, the prism on top of the other.
Then i guess you need a constraint. Here I am not sure about how to set up the equation for the constraint. The prisms are in contact all the time but i can't see an obvious relation between their center of masses or something like that.
Im not very good at mechanics!

#### Attachments

• prisms.GIF
989 bytes · Views: 456
JohanL said:
If you have two prisms on top of each other, one smaller than the other, and the one on top glids down to the ground the lower prisms moves to the left.
The problem is to find how far the lower prism moves.
...

The prisms are in contact all the time but i can't see an obvious relation between their center of masses or something like that.
Hint: What's the net horizontal force on the system (both prisms)?

## What is the stability of a structure composed of three rods and a string?

The stability of a structure depends on various factors such as the material of the rods, the strength of the string, and the angle at which the rods are positioned. It is difficult to determine the stability of a structure without knowing these factors.

## How do the materials of the rods affect the stability of the structure?

The materials of the rods play a crucial role in determining the stability of the structure. If the rods are made of a strong and rigid material such as steel, the structure will be more stable compared to if the rods are made of a weaker material such as plastic.

## What is the maximum angle at which the rods can be positioned for the structure to be stable?

The maximum angle at which the rods can be positioned for the structure to be stable depends on the length and strength of the rods, as well as the tension in the string. It is important to consider these factors when determining the maximum angle for stability.

## How does the strength of the string affect the stability of the structure?

The strength of the string is a crucial factor in determining the stability of the structure. If the string is too weak, it may not be able to hold the weight of the rods and the structure may collapse. It is important to use a strong and durable string for a stable structure.

## Can the stability of the structure be improved?

Yes, the stability of the structure can be improved by using stronger materials for the rods, increasing the tension in the string, and adjusting the angle at which the rods are positioned. It is important to carefully consider these factors to ensure the stability of the structure.

• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
972
• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
62
Views
10K
• Introductory Physics Homework Help
Replies
13
Views
4K