Find the acceleration of the hanging block

[Emily_20]

Homework Statement

In terms of m1,m2, and g. find the acceleration of the hanging block in the two-block/ two-pulley setup shown in the diagram. The stands holding the string rests on the floor. All pulleys are massless, and there is no friction anywhere in the system.

Homework Equations

F=ma

The Attempt at a Solution

First I drew free body diagrams of both blocks and of the lower pulley. I numbered the tensions in the pictures. The equations that I got for m1:
Fy= N=mg
Fx= T1=m1a
For m2 I got:
Fy= T2-m2g=-a*m2
For pulley I got:
T1+T3=T2

Then I substituted T2 Into the equation for m2 and I got:

T1+T3-m2g=-a*m2
m1*a+T3-m2g=-a*m2
Now I am confused but I decided to call T3 just T because T is same in the rope.
As a result my answer is a= (m2g-T)/(m2+m1), but it is probably wrong. Could you please check if I am on track here?
Thanks for the help!

 

[_N3WTON_]

It is sort of hard to tell if you are on the right track (imo) because the picture you posted is basically impossible to read

 

[Emily_20]

Sorry.. Here is a link for it http://tinypic.com/r/1418ljc/8

 

[Emily_20]

Also ignore what it says on the picture it is not part of the question.

 

[_N3WTON_]

Edit: nvm, I can't read XD

 

[Emily_20]

No, all it says "the lab stand holding the string at the upper right rests on the floor. Thanks for help I appreciate it.

 

[NascentOxygen]

Hi Emily. Did you take into consideration their different accelerations? Determine their ratio this way― if m1 slides 1cm to the right, m2 will drop by what distance?

BTW, the pic in your first post is perfectly fine on my Android.

 

[Emily_20]

Would it be 1 cm too?

 

[NascentOxygen]

Do you have half a metre of string and a wheel or small bottle that can take the place of the pully? Arrange something over the edge of your table and you can try it and see!

m1 can be a heavy book or something.

Would it be 1 cm too?

 

[Emily_20]

Ok, you have a few mistakes:
you don't need a free body diagram for the hanging pulley (think of it and the hanging block as a system)
also, both tensions on the hanging block are in the positive y-direction

The professor gave a hint to draw 3 body diagrams. One for block 1 second for block 2 and last one for the lower hanging pulley.

 

[Emily_20]

Do you have half a metre of string and a wheel or small bottle that can take the place of the pully? Arrange something over the edge of your table and you can try it and see!

m1 can be a heavy book or something.

I think m2 will fall faster than m1 will so m1 will still stay on the table…

 

[_N3WTON_]

The professor gave a hint to draw 3 body diagrams. One for block 1 second for block 2 and last one for the lower hanging pulley.

ok then disregard my comment, sorry

 

[Emily_20]

ok then disregard my comment, sorry

Its ok ! I appreciate your help.

 

[NascentOxygen]

I think m2 will fall faster than m1 will so m1 will still stay on the table…

I didn't mean falling. oo)

As you slide the book 10cm along the table, take notice how far your improvised "pulley" is allowed to descend.

No falling!

 

[Emily_20]

I didn't mean falling. oo)

As you slide the book 10cm along the table, take notice how far your improvised "pulley" is allowed to descend.

No falling!

All the top would get "down" and the pulley will hang but the block would stay on the table. Anyways this problem seems to be difficulties and I am not very good with pulleys so sorry if I do not get what your saying. This problem is very important for me it worths 22 pts. Thanks for your time.

 

[_N3WTON_]

Its ok ! I appreciate your help.

no problem

 

[Emily_20]

Here are my new equations for the FBD:

For the first block (m1):
Fx T=m1a

For the second block (m2):
Fy T-m2g=-m2a

For the pulley:
2T-T-m2g=0 Not sure for this one if m2 is right and not sure that ma=0

 

[Emily_20]

And for the pulley its for Fy cause no Fx

 

[_N3WTON_]

not sure that ma=0

Well remember that the pulley is without mass..

 

[Emily_20]

For the first block (m1):
Fx T=m1a

For the second block (m2):
Fy T-m2g=-m2a

For the pulley:
2T-T=0 Is it right now?

 

[_N3WTON_]

For the first block (m1):
Fx T=m1a

For the second block (m2):
Fy T-m2g=-m2a

For the pulley:
2T-T=0 Is it right now?

that looks ok to me, I'll try and solve it sometime tomorrow and get back to you...in the mean time I'm going to have a few beers :)

 

[Emily_20]

So T=0?
If it is so Then

For the first block (m1):
Fx 0=m1a

For the second block (m2):
Fy 0-m2g=-m2a
a=m2*g/m2
so a=g?

 

[_N3WTON_]

So T=0?
If it is so Then

For the first block (m1):
Fx 0=m1a

For the second block (m2):
Fy 0-m2g=-m2a
a=m2*g/m2
so a=g?

That doesn't look right. Let me take a look..

 

[Emily_20]

Ok thank you!

 

[_N3WTON_]

I think one problem may be that you are assuming that the acceleration of both masses is the same

 

[Emily_20]

For the first block (m1):
Fx T=m1a1

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

For the pulley:
T=0

Is it right now (I put a1 and a2)?

 

[_N3WTON_]

For the first block (m1):
Fx T=m1a1

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

For the pulley:
T=0

Is it right now (I put a1 and a2)?

it still doesn't seem right, I'm sorry but I have to go now...I'm sure someone will be by to offer help that is better in every way than my help...if not i'll try to work through the problem tomorrow

 

[Emily_20]

Ok, thank you! Hope someone will help me if not I am relaying on you ;)

 

[_N3WTON_]

Ok, thank you! Hope someone will help me if not I am relaying on you ;)

lol that's not very wise :P...but I guarantee someone will be able to help more than I can

 

[Emily_20]

alright lol