Additive levers?

[Mokele]

Ok, I have a question about a simple, but not so simple, system.

 

Let's say I have two rigid, massless poles of length L, joined by a massless, infinitesimally small motor that exerts torque T. If the poles are parallel, and I stick one into a concrete wall for anchoring, obviously the motor will move the free pole. Plus, if I push down on the tip of the free pole, I should need TxL newtons of force to prevent movement. That's simple enough, good old levers and such.

 

But what if I attach another motor to the tip of the free pole, and another pole to that motor, so I have 3 poles in a line, joined by two motors? (another way of imagining it is like your arm, bending at the wrist and elbow) Do I have Tx2L + TxL acting on my hand at the tip of the furthest pole? Put another way, if there's a lever at the end of the lever, do you add the effects?

 

Or, hopefully better, Here is a very crappy drawing (since I suck at art) of the system, and I'm basically trying to find F in terms of T and L that keeps the system static

 

Is F = Tx2L + TxL correct? Am I phrasing this horribly?

 

Mokele

[J.C.MacSwell]

Ok' date=' I have a question about a simple, but not so simple, system.

 

Let's say I have two rigid, massless poles of length L, joined by a massless, infinitesimally small motor that exerts torque T. If the poles are parallel, and I stick one into a concrete wall for anchoring, obviously the motor will move the free pole. Plus, if I push down on the tip of the free pole, I should need [b']TxL newtons [/b] of force to prevent movement. That's simple enough, good old levers and such.

 

 

 

Mokele

 

Try T divided by L

[Mokele]

Duh, my mistake. So that final formula possible formula would be F = T/2L + T/L, i suppose.

 

That'll teach me to go from memory rather than my notes, while posting on an empty stomach...

 

Mokele

[J.C.MacSwell]

Duh' date=' my mistake. So that final formula possible formula would be F = T/2L + T/L, i suppose.

 

That'll teach me to go from memory rather than my notes, while posting on an empty stomach...

 

Mokele[/quote']

 

A force of T/2L will stop the rotation of the first motor but fail to stop the rotation of the second (unless you set it the second torque to T/2)

[Mokele]

So is my revised formula right?

[J.C.MacSwell]

So is my revised formula right?

 

No. In your massless system, there is no solution as the second torque could not equal the first.

 

An instantaneous snapshot would give the second torque to be at maximum T/2 which would put the force at F= T/2L.

 

If you want more force from the given torques then shorten the levers. (and get rid of the second)

 

What are you trying to accomplish?

[Mokele]

No. In your massless system, there is no solution as the second torque could not equal the first.

 

Can you explain why not? I'm a bit lost at this point...

 

What are you trying to accomplish?

 

It's actually a model (in a way) of what goes on in the arm of a brittlestar (they move by means of muscles in the arms, as opposed to the usual starfish use of tube-feet). The arm is a series of disks, acted upon by peripherally arranged muscles which only span the distance between successive disks.

 

So basically, imagine this as the last two joints of the brittlestar arm. I'm trying to figure out, if the muscles on one side contract simultaneously (generating two equal torques about the joints), what the force would be on an object at the very, very tip (like a pebble the animal was bracing against).

 

Mokele

[J.C.MacSwell]

Can you explain why not? I'm a bit lost at this point...

 

 

 

It's actually a model (in a way) of what goes on in the arm of a brittlestar (they move by means of muscles in the arms' date=' as opposed to the usual starfish use of tube-feet). The arm is a series of disks, acted upon by peripherally arranged muscles which only span the distance between successive disks.

 

So basically, imagine this as the last two joints of the brittlestar arm. I'm trying to figure out, if the muscles on one side contract simultaneously (generating two equal torques about the joints), what the force would be on an object at the very, very tip (like a pebble the animal was bracing against).

 

Mokele[/quote']

 

Try working it out backwards. Assume the force at the outer end and solve for the torques T1 and T2 and any reaction forces at each joint. You will find that if torque T2 is anything but half of torque T1 there will be a net force upwards or downwards at the second joint (F/2 downwards for T1=T2). Since it is massless it's instantaneous acceleration would be infinite.

 

If you assume a mass or masses somewhere between the first joint and the outer end then you can get a dynamic result with T1=T2. The acceleration/motion will depend on your choices of placement of masses and the reaction forces which change as the joint angles change.

 

But the only static solution will be with T2 =T1/2 or less if you account for gravitational forces at the masses.

 

I guess if this thing is underwater you could assume a buoyant force at the second joint of F/2 and get a static result with T2 =T1. You would need something to restrain the second joint.

 

Does T1 have to equal T2? (Don't just tell me God made it that way! )

[Mokele]

Try working it out backwards. Assume the force at the outer end and solve for the torques T1 and T2 and any reaction forces at each joint. You will find that if torque T2 is anything but half of torque T1 there will be a net force upwards or downwards at the second joint (F/2 downwards for T1=T2). Since it is massless it's instantaneous acceleration would be infinite.

 

If you assume a mass or masses somewhere between the first joint and the outer end then you can get a dynamic result with T1=T2. The acceleration/motion will depend on your choices of placement of masses and the reaction forces which change as the joint angles change.

 

But the only static solution will be with T2 =T1/2 or less if you account for gravitational forces at the masses.

 

So, do you think the problem might just be that I'm simplifying too much, and that I need to include more realistic stuff like masses, resistances to movement due to elastic and damping tissues, etc?

 

Does T1 have to equal T2? (Don't just tell me God made it that way! )

 

I'm not sure. From quick inspection, it looks like the muscle cross-sectional areas are the same for every segment, but I don't know about muscle composition of neural control. Still, if both weren't generating equal forces, I'd expect the less-used muscle to be smaller via both atrophy from lack of use and selective pressures for less wasted muscle mass.

 

Seems like I need to work on this some more... Bloody organism systems being so complicated...

 

Thanks very much for your help and patience with me!

 

Mokele

[J.C.MacSwell]

So' date=' do you think the problem might just be that I'm simplifying too much, and that I need to include more realistic stuff like masses, resistances to movement due to elastic and damping tissues, etc?

 

Mokele[/quote']

 

Yeah. The trick is to keep it as simple as you reasonably can unless you have a good computer model or are really good at calculus (that's when my practical side/laziness/sloping forehead usually settles for a good first order appproximation)

 

I'm not sure. From quick inspection' date=' it looks like the muscle cross-sectional areas are the same for every segment, but I don't know about muscle composition of neural control. Still, if both weren't generating equal forces, [b']I'd expect the less-used muscle to be smaller via both atrophy from lack of use and selective pressures for less wasted muscle mass[/b].

 

Mokele

 

This seems right. So if it's not that way maybe there is a reason. If they are used in a curve rather than straight possibly, but you would still expect some "tapering off" at least toward the outer end.

 

If they break off do they regenerate?

[Mokele]

Yeah. The trick is to keep it as simple as you reasonably can unless you have a good computer model or are really good at calculus (that's when my practical side/laziness/sloping forehead usually settles for a good first order appproximation)

 

Yeah, thing is I was hoping to make derivations from theory, and get equations, rather than from results of computers and tests. I guess I still could, just with a *lot* more work.

 

This seems right. So if it's not that way maybe there is a reason. If they are used in a curve rather than straight possibly, but you would still expect some "tapering off" at least toward the outer end.

 

Oh, they're usually wiggled all which ways, and the arms do taper at the tips.

 

If they break off do they regenerate?

 

Yep, I'm pretty sure they do. In fact, in some species, if a portion of the body comes with the arm, it can grow into a whole new starfish/brittlestar.

 

Mokele