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

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Homework Help Overview

The discussion revolves around the relationship between the acceleration of a block and a pulley system, specifically analyzing the forces acting on both the block and the pulley when a horizontal force is applied. The context is within classical mechanics, focusing on dynamics and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the block and pulley, with one participant attempting to derive a relationship between their accelerations. There are considerations of equilibrium and the implications of distances moved by the block and pulley during motion.

Discussion Status

Some participants have offered insights into the motion of the pulley and the block, while others reflect on misunderstandings regarding the equilibrium of the pulley. The discussion is ongoing, with various interpretations and approaches being explored.

Contextual Notes

There is mention of assumptions such as the pulley being frictionless and the smoothness of the horizontal surface. Additionally, the concept of the "no-stretch assumption" is introduced, indicating a potential gap in textbook explanations.

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!
 
Last edited:

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