Finding Forces and Work in a Pulley System

AI Thread Summary
To determine the force required to pull a box with weight WA up at a constant speed in a frictionless pulley system, the relationship between the forces and tensions is established, revealing that T1 equals 4F. The work done by the tractive force F is calculated as Fs, while the work done by T1 is also Fs, confirming the correctness of the approach. Since the mass moves without acceleration, the force needed to maintain constant speed is F = WA/4. The discussion emphasizes the importance of understanding the constraints of motion and the relationship between the displacements of the ropes and pulleys. Overall, the calculations and reasoning provided are validated by the principles of physics involved in the pulley system.
gunni91
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Homework Statement



Need some serious help with this one :O

A box with a weight WA hangs in pulleys ,(see figure 1). Find the force to pull A up with constant speed and calculate the work which the tractive force F produces ( with the movement of s) and also the work and tractive force T1 produce .
Pulleys and strings are frictionless and both without mass!.

pulley.jpg


Homework Equations






The Attempt at a Solution

 
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What have you tried so far?

ehild
 
gunni91 said:

Homework Statement



Need some serious help with this one :O

A box with a weight WA hangs in pulleys ,(see figure 1). Find the force to pull A up with constant speed and calculate the work which the tractive force F produces ( with the movement of s) and also the work and tractive force T1 produce .
Pulleys and strings are frictionless and both without mass!.

pulley.jpg


Homework Equations






The Attempt at a Solution


taikng tensions ,T1=4F
and taking constrained motion,
acc. of pulling end=4 times acc. of wA
differentiating that we get s=4s1
therefore work done by F=Fs
work done by T1=T1s1=4F x s/4=Fs

correct me if i m wrong,i just tried.
 
The result is correct. Balance of forces gives that WA=T1=4F. The question is the force needed to move the load with constant speed, so write out that F=WA/4.
The mass moves with constant speed so there is no acceleration. The constraints apply for the displacements and changes of length of the pieces of the ropes, but not for the non-existent accelerations. You can say that moving the end of the left rope by s will cause the other two pieces shorten by s/2 so the middle pulley raises by s/2. In the same way, raising the end of the right rope by s/2 lifts the right pulley by s/4.

ehild
 
thank you echild.
 
sorry didn't see the name
properly,very sorry.thanks ehild!
 
ErwinMoses said:
sorry didn't see the name
properly,very sorry.thanks ehild!

You are welcome :smile:

ehild
 

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