# Double pulley Atwood machine (with 3 masses)

• Sho Kano
In summary: No.It might help to think about the three heights from the ground. Let these be y1, y2 and y. If the length of the string is s, can you write a relationship between y1, y2, y and...There is no relationship between y1, y2, y and a.
A is correct. Check the signs in the numerator for B. I got the opposite. I will check my derivation.

ehild said:
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

Sho Kano said:
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.

ehild said:
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

Sho Kano said:
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

ehild said:
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

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

Sho Kano said:
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.

Sho Kano said:
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?

ehild said:
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?

Sho Kano said:
A = a3 = 0
thus, m1m3g+m2m3g−4m1m2g = 0
Yes, what does it mean for m3?

ehild said:
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?

Sho Kano said:
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.

ehild said:
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)

Sho Kano said:
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?

ehild said:
Yes, what do you get?
m3 = 4m1m2/m1+m2

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

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)

Last edited:
PARENTHESES!

SammyS
ehild said:
PARENTHESES!
B = (m2g - m1g) / (m1 + m2)
a1 = (-m2g + m1g) / (m1 + m2)
a2 = (m2g - m1g) / (m1 + m2)

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...

Thanks haruspex, and ehild!

ehild said:
No, a3=A
On second thought, why is a3 = A?
a3 is down, while A is upwards

Sho Kano said:
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.

haruspex said:
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.

Sho Kano said:
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 said:
See post 70, a1 = -A-B, a2 = B-A, a3 = A

These are the answers to part 1

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