Double pulley Atwood machine (with 3 masses)

[Sho Kano]

Not quite. I suggest assigning all forces and accelerations as positive in the same direction. If that direction is down, it means you treat the pull of a rope on one of the masses as being down, but expect the value to turn out to be negative, i.e. it is really up.

Now, I would not have gone this route in the first place, but when I asked you what your convention was you said it was positive down everywhere, but it turns out that was not really true. Anyway, I think you will have gained something out of all this.

How would YOU do this problem convention wise?

 

[haruspex]

How would YOU do this problem convention wise?

It varies according to whim.

 

[Sho Kano]

It varies according to whim.

Can you give me an example of the non-naive way?

 

[Sho Kano]

I deleted my previous post and changed to downward as positive.

View attachment 98882

The masses are connected to the strings, and the strings move on the rotating pulleys clockwise. A is the acceleration of the upper string, and B is the acceleration of the lower string with respect to the moving pulley, you can write the acceleration of the masses and that of the moving pulley in terms of A and B.
(a3=A, a_pulley=-A, a1=-(A+B), a2=B-A.The tensions act with upward forces on the masses, but T3 is upward and both T12 are downward on the moving pulley. Try to write your equations again.

-T12 + m1g = m1a1
-T12 + m2g = m2a2
-T3 + 2T12 = 0

a1 = -A-B
a2 = -A+B
-2A = a1 + a2
A = -a3
2a3 = a1 + a2

-T3 + m3g = m3a3

 

[haruspex]

-T12 + m1g = m1a1
-T12 + m2g = m2a2
-T3 + 2T12 = 0

a1 = -A-B
a2 = -A+B
-2A = a1 + a2
A = -a3
2a3 = a1 + a2

-T3 + m3g = m3a3

Those equations are all correct if we take all accelerations as positive down but all tensions as positive up forces on the masses. So let's go with that.

Edit: sorry, there is still a sign error in the accelerations. I didn't notice, partly because it is in an equation that you previously had correct. See e.g. the last line of your post #45.

 

[Sho Kano]

Those equations are all correct if we take all accelerations as positive down but all tensions as positive up forces on the masses. So let's go with that.

${ a }_{ 2 }\quad =\quad \frac { { m }_{ 1 }{ m }_{ 3 }g\quad -\quad 4{ m }_{ 1 }{ m }_{ 2 }g\quad +\quad { m }_{ 2 }{ m }_{ 3 }g }{ { m }_{ 2 }{ m }_{ 3 }\quad +\quad { m }_{ 1 }{ m }_{ 3 }\quad -\quad 4{ m }_{ 1 }{ m }_{ 2 } }$
The negative in the denominator again, but I can't find any error in my work. Can you verify if it's wrong?

 

[ehild]

Use A and B in the equations instead of a1, a2, a3. Eliminate the T-s and solve for A and B. You get the accelerations from them. There will be no negative term in the denominator.

 

[ehild]

-
A = -a3

No, a3=A

 

[Sho Kano]

No, a3=A

Oh that's right,
-A+a3=0

 

[Sho Kano]

Use A and B in the equations instead of a1, a2, a3. Eliminate the T-s and solve for A and B. You get the accelerations from them. There will be no negative term in the denominator.

Here are my results so far, are these right?
$A\quad =\quad \frac { { m }_{ 1 }{ m }_{ 3 }g\quad +\quad { m }_{ 2 }{ m }_{ 3 }g\quad -\quad 4{ m }_{ 1 }{ m }_{ 2 }g }{ 4{ m }_{ 1 }{ m }_{ 2 }\quad +\quad { m }_{ 1 }{ m }_{ 3 }\quad +\quad { m }_{ 2 }{ m }_{ 3 } } \\ B\quad =\quad \frac { 2{ m }_{ 2 }{ m }_{ 3 }g\quad -\quad 2{ m }_{ 1 }{ m }_{ 3 }g }{ 4{ m }_{ 1 }{ m }_{ 2 }\quad +\quad { m }_{ 2 }{ m }_{ 3 }\quad +\quad { m }_{ 1 }{ m }_{ 3 } }$

 

[ehild]

A is correct. Check the signs in the numerator for B. I got the opposite. I will check my derivation.

 

[Sho Kano]

A is correct. Check the signs in the numerator for B. I got the opposite. I will check my derivation.

I didn't find any error upon looking at my derivation- I'll check it once more

 

[ehild]

I didn't find any error upon looking at my derivation- I'll check it once more

It is OK, the mistake was at my side.

 

[Sho Kano]

It is OK, the mistake was at my side.

Got it. So now I just plug A and B into these equations and that is the answer right?
a1 = -A-B
a2 = -A+B
2a3 = a1+a2

 

[ehild]

Got it. So now I just plug A and B into these equations and that is the answer right?
a1 = -A-B
a2 = -A+B
2a3 = a1+a2

Remember, a3=A. See the picture in post #56

 

[Sho Kano]

Remember, a3=A. See the picture in post #56

Gotcha, won't believe how happy I was to see that the denominators of A and B were the same so I could easily put them together

 

[Sho Kano]

For the 3rd question, would this be the right answer?
m3 = m1 + m2

 

[ehild]

Gotcha, won't believe how happy I was to see that the denominators of A and B were the same so I could easily put them together

Congratulation! Now it is easy to answer the other questions.

 

[ehild]

For the 3rd question, would this be the right answer?
m3 = m1 + m2

No, why do you think it? You have the expression for a3=A. What follows from it?

 

[Sho Kano]

No, why do you think it? You have the expression for a3=A. What follows from it?

A = a3 = 0
thus, m1m3g+m2m3g−4m1m2g = 0
and solve for m3
is this right?

 

[ehild]

A = a3 = 0
thus, m1m3g+m2m3g−4m1m2g = 0

Yes, what does it mean for m3?

 

[Sho Kano]

Yes, what does it mean for m3?

After solving for m3 for that equation, that would be the answer for question 3.
Then for question 4, I'd just plug in that into the m3's of a1 and a2 right?

 

[ehild]

After solving for m3 for that equation, that would be the answer for question 3.
Then for question 4, I'd just plug in that into the m3's of a1 and a2 right?

Use that A=0, so a1 = -B and a2 = B. Plug in the expression for m3 into B.

 

[Sho Kano]

Use that A=0, so a1 = -B and a2 = B. Plug in the expression for m3 into B.

Gotcha, so it goes like this right?

3) m1m3g+m2m3g−4m1m2g = 0 ; solve for m3
4) a1 = -B ; a2 = +B ; plug in m3 from (3)

 

[ehild]

Gotcha, so it goes like this right?

3) m1m3g+m2m3g−4m1m2g = 0 ; solve for m3
4) a1 = -B ; a2 = +B ; plug in m3 from (3)

Yes, what do you get?

 

[Sho Kano]

Yes, what do you get?

m3 = 4m1m2/m1+m2

 

[ehild]

m3 = 4m1m2/m1+m2

It would be good if you had not forgot parentheses. m3 = 4m1m2/(m1+m2).

 

[Sho Kano]

B = 8m1m2m2g - 8m1m1m2g / (4m1m1 + 5m1m2 + m2m2 + 4m1m2m2 + 4m1m1m2)
a1 = -8m1m2m2g + 8m1m1m2g / (4m1m1 + 5m1m2 + m2m2 + 4m1m2m2 + 4m1m1m2)
a2 = 8m1m2m2g - 8m1m1m2g / (4m1m1 + 5m1m2 + m2m2 + 4m1m2m2 + 4m1m1m2)

edit: something went horribly wrong in my calculations. Here is my revised answer
B = m1m2m2g - m1m1m2g / (m1m1m2 + m1m2m2)
B = m2g - m1g / (m1 + m2)
a1 = -m2g + m1g / (m1 + m2)
a2 = m2g - m1g / (m1 + m2)

 

[ehild]

PARENTHESES!

 

[Sho Kano]

PARENTHESES!

B = (m2g - m1g) / (m1 + m2)
a1 = (-m2g + m1g) / (m1 + m2)
a2 = (m2g - m1g) / (m1 + m2)