Rotate pulleys using gravity.

Jigar111

Cross section of pulley1, pulley2 , pulley3 and pulley4 are as shown in figure. Centres of Pulley 1 and pulley 2 are placed on same horizontal line. Centres of pulley 1 and pulley 2 are fixed. Pulley 1 and pulley 2 can rotate about their centre. Centres of pulley3 and pulley4 are also placed on one horizontal line. Pulley 3 and pulley 4 can also rotate about their centre. Centre of pulley 3 and pulley4 are joined with one stationary plate.Pulley 3 and pulley4 are simply supported on pulley1 and pulley2.Now we apply some load or weight on stationary plate. Finaly this force is applied on pulley 1 and pulley 2 in vertical downward direction because pulley 3 and pulley4 are simply supported on pulley 1 and pulley2. Tengential componants of these forces helps to rotate pulley1 , pulley2 , pulley3 and pulley4.Pl visit [crackpot link deleted]

russ_watters

If you are here to learn, great, but be advised that we will not support crackpottery here.

The flaw in your reasoning is simple: force on a single contact point of a pulley can only be along the line connecting their centers. To figure out what the forces are everywhere, draw a free body diagram at each pulley center.

FredGarvin

I can't even tell what he is trying to do here.

russ_watters

Quote by FredGarvin

I can't even tell what he is trying to do here.

It appears to me that he thinks that since weight acts straight down while the line of force between the touching pulleys is at an angle that there is a force component at the contact point of the pulleys, tangential to them. The error is drawing a fbd (if you can call it that) that shows the weight acting from the edge of the pulley instead of its center.

Regardless, any 2nd-grader playing with Legos could tell you this won't do anything.

RainmanAero

Hi Guys,

Quote by russ_watters

The error is drawing a fbd (if you can call it that) that shows the weight acting from the edge of the pulley instead of its center.

And that ain't the only error! He also labels those forces as each mg/2 (so he is obviously not accounting for the weight of pulleys 3 & 4 and the stationary plate that conjoins them).

Regardless, any 2nd-grader paying with Legos could tell you this won't do anything.

Sure enough... but I too would like to know just exactly what problem he thinks he is solving with this Rube Goldberg special?

Rainman

Danger

It looks like the basis for a fairly decent printing press...

kleinjahr

Just another version of the classic counterweighted perpetual motion machine. Works as well as any other.
See Gardner D. Hiscox's "970 Mechanical Appliances and Novelties of Construction" Copyright 1904.

Mech_Engineer

This machine won't work any better than the first one you proposed... http://www.physicsforums.com/showthread.php?p=1155813

The fact remains you cannot escape one simple law: conservation of energy. No matter how many pulleys you add in.

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russ_watters Mentor	If you are here to learn, great, but be advised that we will not support crackpottery here. The flaw in your reasoning is simple: force on a single contact point of a pulley can only be along the line connecting their centers. To figure out what the forces are everywhere, draw a free body diagram at each pulley center.

FredGarvin Recognitions: Science Advisor	I can't even tell what he is trying to do here.

Danger Recognitions: Gold Member	It looks like the basis for a fairly decent printing press...

kleinjahr	Just another version of the classic counterweighted perpetual motion machine. Works as well as any other. See Gardner D. Hiscox's "970 Mechanical Appliances and Novelties of Construction" Copyright 1904.

Mech_Engineer Blog Entries: 2 Recognitions: Gold Member Science Advisor	This machine won't work any better than the first one you proposed... http://www.physicsforums.com/showthread.php?p=1155813 The fact remains you cannot escape one simple law: conservation of energy. No matter how many pulleys you add in.