Single movable pulley!


by ritwik06
Tags: movable, pulley, single
ritwik06
ritwik06 is offline
#1
May31-07, 02:18 AM
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!
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
andrevdh
andrevdh is offline
#2
May31-07, 05:49 AM
HW Helper
P: 1,449
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.
Mentz114
Mentz114 is offline
#3
May31-07, 08:42 AM
PF Gold
P: 4,081
Think of a lever. It's easy to see that the gain in leverage is offset by a loss of distance.

ritwik06
ritwik06 is offline
#4
Jun1-07, 02:29 AM
P: 586
Question

Single movable pulley!


Quote Quote by andrevdh View Post
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???
andrevdh
andrevdh is offline
#5
Jun1-07, 04:19 AM
HW Helper
P: 1,449
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 throught only a quarter of the distance (speed) that the input force moves.
ritwik06
ritwik06 is offline
#6
Jun1-07, 11:58 PM
P: 586
Quote Quote by andrevdh View Post
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 throught 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.


Register to reply

Related Discussions
block on movable wedge Introductory Physics Homework 16
Pulley/Torque problem where pulley has mass Introductory Physics Homework 4
Single Mode to Multi-mode back to Single Mode Electrical Engineering 0
Single slit single electron Advanced Physics Homework 1
Can you solve the "particle sliding on a movable inclinded plane" problem... Classical Physics 1