# Solving the Mystery of a Mechanical Device: Where is the Error?

• Gh778
In summary: I don't want to talk about this because it's not the problem.In summary, the device doesn't seem to work correctly because the square object with balls inside receives a net torque on it.
Gh778
Hi,

I imagined this device and I don't understand where is the error. I posted here because it's a simple mechanical device, I have trouble with the sum of torque. The potential energy must be constant. I think I forgot a torque, if you see where ?

A torus turns counterclockwise around the center C1. C1 is fixed to the ground. The center C2 is fixed to the torus and a square object can turn around C2. I put inside the square Object a lot of small balls where there isn't the torus. Each ball is attracked by a spring (not drawn). The springs attrack balls like gravity can do, but Gravity1 is higher than Gravity2. I called it gravity but it's not a gravity it's just for have a pressure with balls like water can do in a recipient. The springs are attached on the green points, and the green points are on the square Object not on the torus. I drawn all forces I see and I can have the sum of forces on the center C2 like the yellow force so the torus don't receive a torque from C2. In the contrary the square Object with balls inside receives a net torque on it. The square Object will turn counterclockwise BUT I turn the torus counterclockwise more and more in the time for have the same angular velocity for the square Object and the torus. The angular velocity of the torus "follows" the angular velocity of the square Object for keep constant the angle. I need an energy for accelerate the torus but this energy can be recover later. In the contrary, I don't give an energy for rotate the square Object but with time it turns more and more, and the potential energy seems to increase. I need to accelerate more and more the torus but I don't think the problem come from this particulary. What's wrong in this device ?

The device:

The forces:

Where the balls are inside the square object:

Several positions of the device, for show the angle:

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I'm not trying to be rude, but your diagrams and your descriptions are very convoluted. I don't think anyone can give you a straight answer until you simplify some of this.

You're right, I drawn so many forces. Maybe, first, just study the square Object without the torus. I drawn 3 springs for show where they are, one end of a spring is attached to a green point and another end to a ball. I don't drawn all green points. There is one spring for one ball. The forces F4, F5 and F6 come from the pressure of balls, it's like the pressure in water with gravity. The forces F1, F2 and F3 come from the forces on the green points from springs, all springs attrack the balls to the "bottom" (bottom on the image). I drawn red forces for show how a spring attrack. F1+F2+F3 is lower than F4+F5+F6 because the torus has a curved shape. The forces F7, F8 and F9 come from the pressure in another part of the square Object, it's not the same force of attraction so forces are smaller than the right part. The square Object with the forces of pressure and force from springs are like that:

Tell me if that first step is ok for you please ? Maybe I'm wrong in these forces.

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With gravity + water, there is forces from pressure at left, at right and at bottom, I have this : (I did not draw all forces)

If I want to have the same pressure at left and at right but now with balls and springs without gravity (I can't draw all springs, but there is one spring for each ball). Note the forces from springs are to up and there is forces of pressure at bottom too. I have this:

Now, with the square object I want 2 areas (areas are separated) with 2 differents attraction (attraction1 > attraction2) but it's the same, I use springs and balls (balls are compressible), like that do you understand my forces ?

Note there is the torus so some forces are on the torus not on the square object.

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I changed my diagram maybe like that it's clear enough ?

I noted the pressure at right from 0 to 10 and the pressure at left from 0 to 2 (it can be bars for example). So, I drawn only the forces from the springs. The pressure gives forces at right and at left that I symbolized with 3 different forces in my last messages. The balls are compressible. It's possible to have a torque on the object and not on the support.

With only the Attraction2 :

There is not a up force from the springs because balls pressure to the bottom in the same time. There are forces from pressure at left.

With only the Attraction1:

Here, there are up forces from the springs.

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I understood my error, I forgot the pressure at bottom higher than the forces from the springs. So I calculated the torque on the object and the torque on the support in this easier device. I increase the height of the object. The sum of torque on the support is 0 but the sum of torque on the red object is not zero. I use the springs for create the pressure. The springs are fixed on the bottom, on the green points.

I noted:

'c' the length of the "square"
'R' the outer radius of the torus
'd' the distance c1c2 x or y-axis (look at the second image)

I fixed values: c=3 and R=6.5, d=4.34 for have an idea of the datas

I consider the pressure like the height of the balls because I can choose with springs.

**************************************************************************CALCULATION*********************************************************************

******************************************Springs**************************************
Start integration x:
s1=d-c/2=2.84

Middle integration x:
sm=d=4.34

End integration x:
s2=d+c/2=5.84

Height:
H=d+c/2=5.84

Integration1:
$$\frac{1}{c/2}\int_{s1}^{sm} (H-\sqrt(R^2-x^2))(d-x) dx$$

Integration2:
$$\frac{1}{c/2}\int_{sm}^{s2} (H-\sqrt(R^2-x^2))(d-x) dx$$

Values: -0.20+1.63

Sum of forces:
$$\frac{1}{c}\int_{s1}^{s2} (H-\sqrt(R^2-x^2)) dx$$

Value: 1.15

******************************************Right side**************************************

$$\frac{1}{c/2}\int_{0}^{c/2} x(-x+\frac{c}{2}) dx$$

$$\frac{1}{c/2}\int_{c/2}^{c} x(x-\frac{c}{2}) dx$$

Values: -0.37+1.87

Sum of forces:
Value: c/2=1.5

******************************************Left side**************************************

I want to cancel the sum of forces on C2, so I need to have Fx at 1.5-1.15=0.35

The additionnal clockwise torque on the object will be:

Value: -0.35*4.34=-1.5

*******************************************Results**************************************

Torque on the object: -0.20+1.63-0.37+1.87-1.5=1.42

Torque on the support: 0

If you could help me to find my error ?

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## 1. What is the first step in solving a mechanical device error?

The first step in solving a mechanical device error is to carefully examine the device and identify any visible defects or abnormalities. This can include loose or broken parts, missing components, or signs of wear and tear.

## 2. How do you determine which part of the mechanical device is causing the error?

To determine which part of the mechanical device is causing the error, it is important to understand the function of each component and how they work together. Analyzing the device's design and consulting its manual can help narrow down the potential problem areas.

## 3. What tools and techniques can be used to troubleshoot a mechanical device error?

Some common tools and techniques used to troubleshoot a mechanical device error include visual inspection, measuring devices, testing equipment, and diagnostic software. It may also be helpful to consult with other experts or refer to documentation from the device's manufacturer.

## 4. How can you prevent future errors from occurring in a mechanical device?

To prevent future errors, it is important to regularly maintain and service the mechanical device. This includes cleaning and lubricating parts, replacing worn or damaged components, and performing routine checks to identify any potential issues before they become major problems.

## 5. Are there any common mistakes that should be avoided when solving a mechanical device error?

One common mistake to avoid when solving a mechanical device error is jumping to conclusions or assuming the cause of the error without properly assessing and diagnosing the problem. It is also important to follow proper safety protocols and procedures when working with mechanical devices to avoid injury or further damage.

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