Can Conservation of Energy Explain Pulley and Inclined Plane Mechanics?

AI Thread Summary
The discussion focuses on applying the conservation of energy principle to analyze a pulley system and inclined plane mechanics. Initially, the user calculated the relationship between the weights and movements in the system but expressed uncertainty about the accuracy of their answer. After further consideration, they realized that moving the pulley down affects the weight's position more significantly than initially thought, leading to a corrected formula. The final conclusion indicates that the weight W is actually four times the weight w divided by the sine of the angle theta. The conversation emphasizes the importance of understanding mechanical advantage in pulley systems.
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Homework Statement


feynman1-8.jpg


Homework Equations


Conservation of Energy / Virtual Work
$$\sum_i m_i g h_i = 0$$

The Attempt at a Solution


Moving the pulley I labeled A down by ##\delta y## moves ##w## down by ##\delta y##, and moves ##W## up by ##\frac{1}{2} sin\theta##.

So by conservation of energy I have $$-w\delta y + \frac{1}{2}W \delta y sin\theta = 0$$ leading to ##W = \frac{2w}{sin\theta}##.

I feel uncertain about this answer however, because it seems like the mechanics would be the same if you removed Pulley A and replaced it with ##w##.
 
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showzen said:
Moving the pulley I labeled A down by δy moves w down by δy
Are you sure about that?
 
Not sure why you are using virtual work either. I would attack this problem with mechanical advantage of a system of pulleys.

Regards,
KQ6UP
 
haruspex said:
Are you sure about that?

My reasoning here is that if pulley A is moved down to be even with the anchor point, ##w## will have dropped by the initial length between the anchor and the pully.
feynman1-8b.jpg
 
showzen said:
My reasoning here is that if pulley A is moved down to be even with the anchor point, ##w## will have dropped by the initial length between the anchor and the pully.View attachment 104192

I see my mistake now! ##\delta y## moves ##w## by ##2\delta y##. So ##W=\frac{4w}{sin\theta}##, which is what I get from force analysis as well.
 
showzen said:
I see my mistake now! ##\delta y## moves ##w## by ##2\delta y##. So ##W=\frac{4w}{sin\theta}##, which is what I get from force analysis as well.
That looks good.
 
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