Some one me with the derivation

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Homework Help Overview

The discussion revolves around a physics problem involving pulleys and blocks, specifically focusing on deriving an expression for the acceleration of a block with mass m2 in relation to another block with mass m1. The setup includes light pulleys, frictionless surfaces, and a non-stretching cord.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the accelerations of the two blocks, with one participant attempting to derive an expression for the acceleration of block 2 based on the mass of block 1. Questions arise regarding the correctness of the equations used and the forces acting on the blocks.

Discussion Status

There is an ongoing exploration of the derivation process, with some participants providing equations and others questioning the assumptions made. A participant expresses uncertainty about the correctness of their derivation, while another suggests that the expressions may be equivalent despite differences in presentation.

Contextual Notes

Participants note the importance of accurately identifying forces acting on the blocks, and there is a mention of a diagram that was initially missing. The discussion also highlights a potential misunderstanding regarding terminology related to the cord.

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Homework Statement


In figure, the pulleys and the cod are light, all surfaces are frictionless, and the cord does not
stretch.
a) How does the acceleration of block 1 compare with the acceleration of block 2?
The mass of block 2 is m2 = 1.3 kg. Derive an expression for the acceleration of the
block having mass m2 as a function of the mass of block 1, m1.


Homework Equations


mg-T=m2a2
T=m1a1=m12a2



The Attempt at a Solution



The first one (a) i have done.
Which is that due to the pulley the acceleration of a2=1/2 a1

Can some one help me with the b part
 

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Hi lioric. Is there supposed to be a diagram to accompany this?
 


sorry i was in a rush and forgot
now its there
 


This is what i got for the derivation. But something is not right
 


Mass 2 equation
The pulley give 2 tensions so 2T
m2g-2T=m2a2

Mass 1 equation

T=m1a1
since a2 = 1/2a1
a1= 2a2

T=m12a2

substitute mass 1 equation in mass 2 equation as T

m2g-2(2m1a2)=m2a2
m2g-4m1a2=m2a2

make subject a2

-4m1a2-m2a2=-m2g
change negative sign by multiplying by -1

4m1a2+m2a2=m2g

a2(4m1+m2)=m2g

a2=m2g/4m1+m2

But i feel that this is not right.
 


lioric said:
In figure, the pulleys and the cod are light, all surfaces are frictionless, and the cord does not stretch.
It's cog, not cod.
mg-T=m2a2
That equation is not correct. What are all the forces on m2?
 


m2 would have gravity and tension
 


haruspex said:
It's cog, not cod.
I believe it's c-o-r-d
 


ok enough about the "CORD"

m2 would have gravity and tension those are the forces on M2

so what's next?
 
  • #10


lioric said:
The pulley give 2 tensions so 2T
m2g-2T=m2a2
That's better (your post crossed with mine).
a2=m2g/4m1+m2
Looks ok to me, if parentheses inserted:
a2=m2g/(4m1+m2). What bothers you?
 
  • #12
lioric said:
It shows the answer where g is out side a bracket. mine is in the numerator
The two are equivalent.
 
  • #13
ok thank you
 

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