Equilibrium of a Homogeneous Dice on Horizontal Rods: Finding the Critical Angle

In summary, the conversation discusses a problem involving a homogeneous dice placed on two horizontal rods with a coefficient of friction of 0.2. The goal is to find the angles at which the dice is in a state of equilibrium. One angle, 45°, is already known but the conversation explores other angles and the forces involved. A fellow user, Bartholomew, offers some preliminary geometry and equations to help solve the problem. Sebastian expresses gratitude and plans to continue working on the problem.
  • #1
jcdenton
Hi !

I've been working on the following problem for quite some time now and I just can't figure out how to do it:
A homogeneous dice, whose edges have a length of a, is situated on two horizontal, thin, rigid rods. The rods are parallel to each other and are installed on the same height above the ground. Each of the rods is also parallel to one side of the dice. The coefficient of friction, μ, equals 0.2. For which angles α is the dice in a state of equilibrium ?
It is clear that the dice is in such a state for an angle α of 45°. However, due to friction, there must be additional angles, which are slightly bigger and smaller than 45°. But I do not know how to find these angles. I have tried to draw a diagram of all the forces involved in the problem, but I don't know which torques I need to consider and at which points I need to draw them.
It would be very nice if someone could help me solve this problem.

Thank you very much

Sebastian
 

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  • #2
Some preliminary geometry

I did a bunch of triangulation and found two things you need:

1.) If the center of mass of the square is traced down to where it crosses the line between the two pegs, the distance from the left peg to that crossing is

(L + (L - L COS a) COS a) - (2 ^ .5 * L / 2 * SIN (a + PI/4))

And the distance from the right peg to that crossing is (L - distance to left peg).

I don't know how much that expression can be simplified--I didn't even try.

You need those distances to find how much of the weight rests on each peg--it's not an equal split except when "a" is 45 degrees.

I'm going to call the distance from the left peg to the crossing X and the distance from the right peg to the crossing L - X so that I don't have to say that whole thing out again.

2.) First, the distance from the center of the square to the crossing is

(2 ^ .5 * L / 2 * SIN (a + PI/4)) - L * COS a SIN a

I will call that distance Y.

The distance from the left peg to the center of mass is therefore

(X ^ 2 + Y ^ 2) ^ .5

and the distance from the right peg to the center of mass is

((L - X) ^ 2 + Y ^ 2) ^ .5


Those two distances are your lever arms.
 
  • #3
Hi !

@Bartholomew: Thank you very much for your efforts. Even though I haven't yet completely understood your calculations, they surely have provided me with a clou of how this problem might be solved. I'll continue working on it.

Thank you again

Sebastian
 

1. What is the purpose of studying equilibrium of a homogeneous dice on horizontal rods?

The purpose of this study is to determine the critical angle at which a homogeneous dice will remain in equilibrium on horizontal rods. This can help us understand the factors that affect equilibrium and stability in physical systems.

2. How is the critical angle of equilibrium calculated for a homogeneous dice on horizontal rods?

The critical angle of equilibrium is calculated by using the principle of moments, where the sum of the clockwise moments must equal the sum of the counterclockwise moments. This can be represented by the equation ΣMcw = ΣMccw, where Mcw is the clockwise moment and Mccw is the counterclockwise moment.

3. What factors affect the critical angle of equilibrium for a homogeneous dice on horizontal rods?

The critical angle of equilibrium for a homogeneous dice on horizontal rods is affected by the weight and distribution of the dice, as well as the length and angle of the rods. Other factors such as surface friction and air resistance may also play a role in the equilibrium of the dice.

4. Can the critical angle of equilibrium be greater than 90 degrees?

Yes, the critical angle of equilibrium can be greater than 90 degrees. This means that the dice can still remain in equilibrium even if it is tilted at an angle greater than 90 degrees from the horizontal rods. However, this may not be a stable equilibrium and the dice may easily fall off if disturbed.

5. How can the results of this study be applied in real-life situations?

The results of this study can be applied in various fields such as engineering, architecture, and physics. Understanding the factors that affect equilibrium and stability can help in designing structures and systems that can withstand external forces and maintain their balance. This can also aid in predicting and preventing potential accidents or failures in physical systems.

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