What is the Relationship Between Acceleration of a Block and a Pulley?

AI Thread Summary
The discussion focuses on the relationship between the acceleration of a block and a pulley in a frictionless system. It is established that the acceleration of the block (m1) is twice that of the pulley (m2), represented by the equation a1 = 2a2. Participants analyze the forces acting on both the block and the pulley, noting that the pulley is not in equilibrium due to the applied force and tension. A method involving the distances moved by the block and the pulley is suggested to derive the acceleration relationship. The conversation highlights a common misunderstanding in introductory mechanics regarding the equilibrium of the pulley.
imatreyu
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Homework Statement



A horizontal force F is applied to a frictionless pulley of mass m2. The horizontal surface is smooth. Show that the acceleration of the block of mass m1 is twice the acceleration of the pulley.

LOOKS LIKE THIS: http://cnx.org/content/m14060/latest/npq1.gif
But WITHOUT block B and its string.

Homework Equations


F=ma


The Attempt at a Solution


I drew separate force diagrams for m1 (the block) and m2 (the pulley. In the x direction, the block is only being acted on by T1 going in the pos. x direction. In the x direction, the pulley is being acted on by 2T1 and F. 2T1 is going in the neg. x direction. F, the opposite.

I have to show that a1= 2a2

So:

The pulley is in equilibrium:
F-2T1 = 0
m2a2 - 2(m1a1)=0

. . .and I don't know where to go from here. . . .I can't eliminate mass. . .


Thank you in advance!
 
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It is easy to see that the pulley is moving. Let s_1 be the distance between pulley and A, s_2 be the distance between pulley and B. During the pull, s_2 stays constant. For s_1, it is decreasing right? But that section goes to the upper side of the pulley, so we can set 2s_1 equals also a constant. the reason of 2s_1 comes from the initial situation, we ignore the upper portion that is left of the originial position of A.

Hence 2s_1+s_2=constant, differentiate twice yields your desired result.
 
Oh. IDK why I thought the pulley was in equilibrium.

Thank you so much!
 
imatreyu said:
Oh. IDK why I thought the pulley was in equilibrium.

Thank you so much!

I made the same mistake as you when I was having a introductory mechanics class.
The method I present here is sometimes referred to no-stretch assumption. I don't know why the method is always not mentioned in the textbooks. Is it too obvious for the authors?
 
I suppose they think so! The textbook (College Physics, 3rd ed. Serway & Faughn) says nothing about using distances and time derivatives . . . xD

Thank you!
 
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