Find the acceleration of the hanging block

In summary: Emily!In summary, the conversation discusses a physics problem involving a two-block/two-pulley setup with no friction. The goal is to find the acceleration of the hanging block, and the conversation includes various equations and diagrams to help solve the problem. The conversation also mentions a hint from the professor to draw three body diagrams. Ultimately, the conversation ends with a proposed solution for the equations, but it is uncertain if it is correct.
  • #36
Emily_20 said:
Do you mean T2=T1? Thanks for help I appreciate it.

No that's not correct.
As the pulley is mass less so no net force should act on it.
Now can you tell me what are the forces acting on the pulley?
Emily_20 said:
Thanks for help I appreciate it.
You are welcome.
 
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  • #37
So Fnet for pulley equals zero? So no tension?
 
  • #38
Emily_20 said:
So Fnet for pulley equals zero? So no tension?

##F_{net}=0##
This does not means that no force acts on it.
Think of a situation in which a block is kept on the floor and it is being pushed in right direction by the force of ##20N## and in the left direction by the force of ##20N##. The two forces cancels each other so ##F_{net}=0## .
But two forces are still acting on it however they are canceled by each other.

So for pulley
There is ##2T_{1}## is acting in upward direction and ##T_{2}## is acting in downward direction.
So ##2T_{1}-T_{2}=0##
So ##2T_{1}=T_{2}##

Now you need to find relation between ##a_{1}## and ##a_{2}##.
 
  • #39
So now I have:

Fx T1=m1a1

For the second block (m2):
Fy T2-m2g=-m2a2

For the pulley:
2T1=T2

So:

2T1-m2g=-m2a2

2(m1a1)-m2g=-m2a2

Am I on the right track? Thank you!
 
  • #40
Emily_20 said:
So now I have:

Fx T1=m1a1

For the second block (m2):
Fy T2-m2g=-m2a2

For the pulley:
2T1=T2

So:

2T1-m2g=-m2a2

2(m1a1)-m2g=-m2a2

Am I on the right track? Thank you!

Correct.
Go ahead.
 
  • #41
Actually I'm stuck now because I don't know what to do with a1 and a2. Can you guide me please.
 
  • #42
Emily_20 said:
Actually I'm stuck now because I don't know what to do with a1 and a2. Can you guide me please.
Do you know calculus?
I have labelled the three parts of the string (let it's length be ##l##).
1418ljc (1).jpg


From the figure

##x+2z+y=l##

If you differentiate this equation with respect to time (two times) you will get the relation between ##a_{1}## and ##a_{2}##.
 
  • #43
Sorry I don't know calculus and my physics course is based on Algebra could you help me with this one?
 
  • #44
I need to find n terms of m1,m2, and g. find the acceleration of the hanging block in the two-block/ two-pulley setup shown in the diagram
 
  • #45
Emily_20 said:
Sorry I don't know calculus and my physics course is based on Algebra could you help me with this one?

You don't need to know very much about the calculus.
Do you know that ##\frac{dx}{dt}=v## and ##\frac{dv}{dt}=a##

Any way, can you tell me if you pull ##m_{2}## by ##x## then how much distance will ##m_{1}## covers?
 
  • #46
I think x/2?
 
  • #47
Emily_20 said:
I need to find n terms of m1,m2, and g. find the acceleration of the hanging block in the two-block/ two-pulley setup shown in the diagram

By this method you will find the accelerations in terms if m1,m2 and g.
If you differentiate the equation two times wrt time you will see that all those variable would vanish and you will get a relation between ##a_{1}## and ##a_{2}##.

You can do it without calculus too. :rolleyes:
Just try to find the answer of the question in #post45.
 
  • #48
I really do not know how to get from 2(m1a1)-m2g=-m2a2 . I am sorry, its difficult for me Could you give me more hints
 
  • #49
Emily_20 said:
I think x/2?

Yes.:)
So if the acceleration of ##m_{2}## is ##a_{2}## then acceleration of ##m_{1}## is ##\frac{a_{2}}{2}##

Now can you find relation between a1 and a2?
 
  • #50
2(m1a2/2)-m2g=m2a2
 
  • #51
2m12a2/2-m2g=m2a2
 
  • #52
I got a2=m2g/(2m1-m2)
 
  • #53
No.
I think I lead you on a wrong path. Sorry for that.:s
If m2 is pulled by 'x' distance then m1 is pulled by '2x' distance.

So ##a_{1}=2a_{2}##.
Just use this value and find the answer.
 
  • #54
2(m12a2)-m2g=-m2a2
 
  • #55
Emily_20 said:
2(m12a2)-m2g=-m2a2

Correct. Go ahead.
 
  • #56
4m1a2-m2g=-m2a2

a2=m2g/(4m1+m2)
 
  • #57
Emily_20 said:
4m1a2-m2g=-m2a2

a2=m2g/(4m1+m2)

That looks good to me.:)
 
  • #58
Thank you so much! you're my genius hero lol. Thank you everyone else who helped me too I appreciate it very much you saved me!
 
  • #59
I understand everything so far but I am confused about this part:" If m2 is pulled by 'x' distance then m1 is pulled by '2x' distance.

So a1=2a2." how did you know that the distance is 2x?
 
  • #60
a2=m2g/(4m1+m2) is the final answer correct?
 
  • #61
Emily_20 said:
Thank you so much! you're my genius hero lol. Thank you everyone else who helped me too I appreciate it very much you saved me!

You are welcome.:D
 
  • #62
Emily_20 said:
I understand everything so far but I am confused about this part:" If m2 is pulled by 'x' distance then m1 is pulled by '2x' distance.

So a1=2a2." how did you know that the distance is 2x?

This part is confusing too.
1418ljc (2).jpg


If m2 is lowered down by 'x' then string on the left side of the pulley is increased by 'x' and string on the right side of the pulley is also increased by 'x'.
Hence, m1 covers distance 2x if m2 is lowered down by distance 'x'.

There are two pulleys here I am talking about the movable pulley.
 
  • #63
Emily_20 said:
a2=m2g/(4m1+m2) is the final answer correct?

Yes, ##a_{2}=\frac{m_{2}g}{4m_{1}+m_{2}}##
 
  • #64
Thank you, your awesome!
 
  • #65
Emily_20 said:
Thank you, your awesome!

You are welcome.:)
 

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