# Single movable pulley!

by ritwik06
Tags: movable, pulley, single
 P: 586 1. The problem statement, all variables and given/known data For a single movable pulley explain the truth of the statement, "What we gain in power, we lose in distance." 3. 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!
 HW Helper P: 1,446 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.
 PF Patron P: 3,921 Think of a lever. It's easy to see that the gain in leverage is offset by a loss of distance.
P: 586

## Single movable pulley!

 Quote by andrevdh 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.
 HW Helper P: 1,446 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: $$f_i\ s_i = f_o\ s_o$$ This means that if the "output power" (load / output force) is four times the "input power" (input force) then the load can be lifted throught only a quarter of the distance (speed) that the input force moves.
 Quote by andrevdh 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: $$f_i\ s_i = f_o\ s_o$$ This means that if the "output power" (load / output force) is four times the "input power" (input force) then the load can be lifted throught only a quarter of the distance (speed) that the input force moves.