Double pulley Atwood machine (with 3 masses)

[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)

 

[ehild]

Pull out g, it is nicer.
You get the result expected. m3 does not accelerate. It can be in rest, but then the hanging pulley is in rest, too, and your system is equivalent with a single pulley-two mass system. The "driving force" is the difference of the weights, the total mass is m1+m2.
I go to sleep...

 

[Sho Kano]

Thanks haruspex, and ehild!

 

[Sho Kano]

No, a3=A

On second thought, why is a3 = A?
a3 is down, while A is upwards

 

[haruspex]

On second thought, why is a3 = A?
a3 is down, while A is upwards

Depends on conventions! If all accelerations are positive up then a3 and A will have opposite signs, but if A is positive up and a3 is positive down they will have the same sign.

 

[Sho Kano]

Depends on conventions! If all accelerations are positive up then a3 and A will have opposite signs, but if A is positive up and a3 is positive down they will have the same sign.

So because originally, I specified downwards as positive, that's why a3 is positive.

 

[ehild]

On second thought, why is a3 = A?
a3 is down, while A is upwards

Look at the figure in Post # 56, 'A' was how the rope around the upper pulley accelerated along its length. It was said that the pulleys move clockwise. The rope moves with the pulley, so the right piece of it moves downward, and the left piece moves upward. The rope must keep its length! A is the magnitude of acceleration of all points of the rope.
The block m3 is attached to the right end of the rope, which moves downward, so the acceleration of m3 is equal to A. The hanging pulley is connected to the other end, so it accelerates upward, so its acceleration with respect to the ground is -A.
When ropes are involved in a problem, we can speak of the acceleration along their length .

 

[Anjum S Khan]

What are the answers ?

 

[Sho Kano]

What are the answers ?

See post 70, a1 = -A-B, a2 = B-A, a3 = A

These are the answers to part 1