# Mechanical question -- Mechanical arm with special constraints

## Main Question or Discussion Point

Here in this case an arm in special constrain.
The arm is capable to rotate around the main center if force
is applied to the center. The torque practically generated by
the force that is pushing Tarm. Tamr rotates and pulls with more or less the same force at the top that horizontal force then converted to rolling on the surface. So that would be a normal force and the remaining component that would roll the whole system.
The Two red arrow means that the force is created at the bottom and the top "exactly" the same would be

My question would be that how much is the torque from C point perspective?

I wasn't able to calculate out because PC is a constraint that is not stationary. Can anybody help on this?

Green parts are stationary.

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jack action
Gold Member
because PC is a constraint that is not stationary.
It is irrelevant whether it is stationary or not.

Isolate Tarm and then solve for $\overrightarrow{F} = m\overrightarrow{a}$ and $\overrightarrow{T} = I\overrightarrow{\alpha}$ for all forces acting on it.

What you mean its irrelevant?

I have found one law, that if the system have torque then it would be stoppable by an opposing torque only. This somewhat means the same as you said that it doesn't matter.
However if the constraint is not stationary then the whole system would move together and the result of the calculus would be different than stationary (not calculating friction)
Do you have a system to describe graphically force vectors on this system? I would love to see what you put on PC, please if you have time. Because what force will act to what direction there?
Also why solve for accel if the force is available and the armlengths and angles are too? OR you mean in finite element method?

It is irrelevant whether it is stationary or not.

Isolate Tarm and then solve for $\overrightarrow{F} = m\overrightarrow{a}$ and $\overrightarrow{T} = I\overrightarrow{\alpha}$ for all forces acting on it.

jack action
Gold Member
However if the constraint is not stationary then the whole system would move together and the result of the calculus would be different than stationary
"Stationary" means no acceleration or velocities are involved (i.e. = 0), that's it. It's the difference between dynamics and statics.
Do you have a system to describe graphically force vectors on this system?
Do you have one? Separate the components from your assembly and do a free body diagram on each one. Including velocity and acceleration. You should be able to create equations with those that will give you a system with X equations and X unknown variables. Then you can solve this system of equations.

This is a small presentation on how it is done.

Hej thanks for the answer but please no... You seems to be off the subject, thanks again anyways.

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billy_joule
Hej thanks for the answer but pls no... You seems to be off the subject, thanks again anyways.

Somebody else pls?
Jack actions suggestion is the approach taught to every mechanical engineering student the world over. It is definitely not 'off the subject'.

Jack actions suggestion is the approach taught to every mechanical engineering student the world over. It is definitely not 'off the subject'.
Please, is this because of summer? Read the post line by line. See how he mixes up elements please. this is why I said off the subject.

But BTW I found a solution to this using this vector approach which is fine for this linear forces..
Interestingly the main torque is zero at this angle, but the torque for the rotating arm is the original torque that is being generated at the bottom 500N force. So that approach that ISOLATES will not work... because it doesnt tell the "double nature" of this torque only the torque it would if it does.

If anybody like I can explain its much simpler solution to this...

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billy_joule
If anybody like I can explain its much simpler solution to this...
I'm interested in seeing your solution.

Its this simple. LOL.
Because the p1-p2 work like an arm it would try to lift the pc to perpendicular direction. Obviously the distances are telling that it will be double force there. On the top the single stays. So the self torque would be 500*0.1= 50Nm and the torque from the center it would be zero so the system doesnt move, even though it seems, because as much force on the top double in the center that generates opposing torque but with half arm lenght so it is the same results in zero.
Here is the solution graphically for two problems.

The purple force just in case you need it to know how much pull it have there.
I'm interested in seeing your solution.

jack action
Gold Member
Your analysis is difficult to follow because of the language barrier (maybe math equations would more appropriate), but something seems to be missing from your analysis:
• Where are the acceleration and angular acceleration of the arm and Tarm? If there are no accelerations, no motions will be initiated.
Also, I still don't understand why there seems to be a reaction force on your roller in a direction where there are no constraints (tangent to the big circle).

you figured right, there is no otherwise it would be a perpetual machine LOL!
:D

jack action
Gold Member
you figured right, there is no otherwise it would be a perpetual machine LOL!
:D
????

A perpetual motion machine is not one who has acceleration. A perpetual motion machine is a hypothetical machine that can do work indefinitely without an energy source.

We assumed there is no friction to simplify your problem. That makes your machine a perpetual motion machine. Of course we know it's not true and once you can figure out the solution to your problem, it will be necessary for you to modify that solution to include friction to do a full analysis of the machine (i.e. identifying the speed limit versus the power input).

Edit: Actually, with or without friction, it is still possible to put enough power to have a constant acceleration. The machine will have its speed constantly increasing, until the stresses are so large that something will break.

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Obviously I thought about the problem I put up.
Dont make problem out of it, sorry.