Is it possible for a feather to lift a car?

[CharlohAwk]

I'm no physics major... And I have no formal education... I just have a task that I wonder if could be possible... Could a feather lift a car? Could you use the weight of a feather and the proper lever system or systems to lift a standard sized car off the ground?

 

[phinds]

I'm no physics major... And I have no formal education... I just have a task that I wonder if could be possible... Could a feather lift a car? Could you use the weight of a feather and the proper lever system or systems to lift a standard sized car off the ground?

In an ideal situation, yes but in reality no. You would need a MASSIVELY leveraged system and in any real such system, there would be friction and losses that would utterly swamp the weight of a feather.

 

[jbriggs444]

In an ideal situation, yes but in reality no. You would need a MASSIVELY leveraged system and in any real such system, there would be friction and losses that would utterly swamp the weight of a feather.

While each stage of a multi-stage lever/gear/pulley/whatever system will drain a fraction of the input power due to frictional losses, the output never becomes non-zero. The losses are multiplicative, not additive.

As a practical matter, you're going to have to lift that feather up to the top of your contraption and harvest its potential energy hundreds of thousands of times to get the car to lift off of the ground.

 

[phinds]

The losses are multiplicative, not additive.

I've been away from classical mechanics of this sort for so many decades that I totally forgot that (if I ever knew it in the first place). It seems counter-intuitive but I'll take your word for it.

 

[jbriggs444]

I've been away from classical mechanics of this sort for so many decades that I totally forgot that (if I ever knew it in the first place). It seems counter-intuitive but I'll take your word for it.

It is not something that they explicitly teach in school. It is what I would consider an obvious consequence. Power out = power in times efficiency. If you want the efficiency of a chain of devices, you take the product of the efficiencies at each step in the chain.

 

[DrClaude]

Don't you need to first overcome static friction?

 

[jbriggs444]

Don't you need to first overcome static friction?

Sure, so?

You size each step in the chain to deal with forces of the magnitude that are expected. Static or dynamic friction will be a fraction of the input forces in that range.

Edit: To your point, this friction-based efficiency model is not strictly linear. Near the limit of low force inputs for a fixed mechanism, efficiency will drop to zero. Hence the requirement to properly size the mechanism.

 

[joel59]

In a windless environment with precisly the right size lever and fulcrum it seems the added weight on the lever should work - a tipping point. Maybe this is cheating since the extremely long lever weights the action.

 

[jbriggs444]

In a windless environment with precisly the right size lever and fulcrum it seems the added weight on the lever should work - a tipping point. Maybe this is cheating since the extremely long lever weights the action.

Sounds fair to me. Make sure the lever is balanced by itself before you put the car and feather on. However...

With a single stage mechanism, practicalities can get in the way. The longer the lever, the more massive it will need to be. That calls for bigger bearings and more torque from static friction at the bearings. One would have to get clever to build a lever long enough with bearings good enough. A multi-stage force multiplier design is one way to evade that problem.

 

[CWatters]

The lever would also have to be very accurately balanced. Just a few mm too long and the extra weight that side would mean a feather wasn't needed :-)

 

[sophiecentaur]

The losses are multiplicative, not additive.

Isn't that only in the case of a linear system?
As I learned at School:
Efficiency is Mechanical Advantage / Velocity Ratio and the MA is not independent of load. I may be too intuitive about this, though.
But 'stiction' is a limiting factor in all machines, isn't it?

 

[jbriggs444]

Isn't that only in the case of a linear system?

You are correct. The friction-based loss model that I have in mind is one that is approximately linear over the expected operating conditions.

Efficiency is Mechanical Advantage / Velocity Ratio and the MA is not independent of load.

Sure. But MA is roughly independent of load under the expected operating conditions for devices with a fixed gear ratio.

 

[LaplacianHarmonic]

Yeah... don't we all wish to be experimenting in the vacuum of space... however, there is this concept called inertia. Look up inertia and tell me what you have learned.

 

[nasu]

I don't think the geologists, for one, will really wish this.