Find the acceleration of the hanging block

[Satvik Pandey]

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?

Thanks for help I appreciate it.

You are welcome.

 

[Emily_20]

So Fnet for pulley equals zero? So no tension?

 

[Satvik Pandey]

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 $a_{2}$.

 

[Emily_20]

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!

 

[Satvik Pandey]

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.

 

[Emily_20]

Actually I'm stuck now because I don't know what to do with a1 and a2. Can you guide me please.

 

[Satvik Pandey]

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 $l$).

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 $a_{2}$.

 

[Emily_20]

Sorry I don't know calculus and my physics course is based on Algebra could you help me with this one?

 

[Emily_20]

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

 

[Satvik Pandey]

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?

 

[Emily_20]

I think x/2?

 

[Satvik Pandey]

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.
Just try to find the answer of the question in #post45.

 

[Emily_20]

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

 

[Satvik Pandey]

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?

 

[Emily_20]

2(m₁a₂/2)-m₂g=m₂a₂

 

[Emily_20]

2m₁2a₂/2-m2g=m₂a₂

 

[Emily_20]

I got a₂=m₂g/(2m₁-m₂)

 

[Satvik Pandey]

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.

 

[Emily_20]

2(m₁2a₂)-m₂g=-m₂a₂

 

[Satvik Pandey]

2(m₁2a₂)-m₂g=-m₂a₂

Correct. Go ahead.

 

[Emily_20]

4m₁a₂-m₂g=-m₂a₂

a₂=m₂g/(4m₁+m₂)

 

[Satvik Pandey]

4m₁a₂-m₂g=-m₂a₂

a₂=m₂g/(4m₁+m₂)

That looks good to me.:)

 

[Emily_20]

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!

 

[Emily_20]

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?

 

[Emily_20]

a2=m2g/(4m1+m2) is the final answer correct?

 

[Satvik Pandey]

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

 

[Satvik Pandey]

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.

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.

 

[Satvik Pandey]

a2=m2g/(4m1+m2) is the final answer correct?

Yes, $a_{2}=\frac{m_{2}g}{4m_{1}+m_{2}}$

 

[Emily_20]

Thank you, your awesome!

 

[Satvik Pandey]

Thank you, your awesome!

You are welcome.:)