How Does a Movable Pulley Illustrate the Trade-off Between Power and Distance?

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Homework Help Overview

The discussion revolves around the mechanics of a single movable pulley and the relationship between power and distance in lifting loads. Participants are exploring the implications of the statement, "What we gain in power, we lose in distance," in the context of physics principles related to work and mechanical advantage.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are attempting to understand the meaning of "gain in power" and how it relates to the distance moved by the effort compared to the load. Some are questioning whether the term "power" is being used correctly in this context, suggesting it may refer more accurately to "load." Others draw parallels to levers to illustrate similar principles of mechanical advantage.

Discussion Status

The discussion is active, with participants providing insights and questioning the terminology used in the problem. There is a recognition of the potential confusion surrounding the term "power," and some participants are exploring the concept of mechanical advantage in relation to the pulley system. No explicit consensus has been reached, but various interpretations and clarifications are being explored.

Contextual Notes

There is a noted ambiguity in the use of the term "power," with suggestions that it may be misleading in the context of the problem. Participants are also considering the implications of energy losses in the system and how they affect the relationship between input and output work.

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Homework Statement


For a single movable pulley explain the truth of the statement, "What we gain in power, we lose in distance."



The Attempt at a Solution



I know that the effort needs to move twice the distance as that of the load. So is this the loss??

Please do explain to me "gain in power"!

Thanks in advance!
 
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Pulleys are used to lift objects. "Power" in this case refers to the weight that the pulley is lifting (the load suspended from it). The product of the load and the distance lifted is the work done by the lifting force. This energy is supplied by the mechanism hauling the rope (chain) in.
 
Think of a lever. It's easy to see that the gain in leverage is offset by a loss of distance.
 
andrevdh said:
Pulleys are used to lift objects. "Power" in this case refers to the weight that the pulley is lifting (the load suspended from it). The product of the load and the distance lifted is the work done by the lifting force. This energy is supplied by the mechanism hauling the rope (chain) in.

Are you talking about mechanical advantage? ratio of load to effort is 2?
 
No, instead of power you should read load, that is the actual weight that the pulley is lifting. The way that the term power is used in this problem is very misleading (and confusing especially in a Physics course as you have discovered) and I suspect that the author ment to use the term load. It would have been closer to the truth if the term "lifting or output power" was used. The reason why I suspect that this was his intention is that if one neglect energy losses the work input should be equal to the work output by the system:

[tex]f_i\ s_i = f_o\ s_o[/tex]

This means that if the "output power" (load / output force) is four times the "input power" (input force) then the load can be lifted through only a quarter of the distance (speed) that the input force moves.
 
Last edited:
andrevdh said:
No, instead of power you should read load, that is the actual weight that the pulley is lifting. The way that the term power is used in this problem is very misleading (and confusing especially in a Physics course as you have discovered) and I suspect that the author ment to use the term load. It would have been closer to the truth if the term "lifting or output power" was used. The reason why I suspect that this was his intention is that if one neglect energy losses the work input should be equal to the work output by the system:

[tex]f_i\ s_i = f_o\ s_o[/tex]

This means that if the "output power" (load / output force) is four times the "input power" (input force) then the load can be lifted through only a quarter of the distance (speed) that the input force moves.

Thanks for the help! I can understand your point of view! Thanks a lot.
 

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