Tension in a string which connects 3 pulleys

[danut]

Homework Statement

In the system bellow, there are 3 ideal pulleys. m₁ = 1.00 kg, m₂ = 1.00 kg, m₃ = 2.00 kg.
Find the tension in the string/wire that connects said pulleys.

Relevant Equations

-

I'm struggling to get to the correct answer, which I posted down bellow.
The pulleys are ideal, so I figured that m₁ and m₂ will both move upwards (towards the ceiling?) with the acceleration a, while m₃ will move downwards with the acceleration -a.
Let T be the tension in the string which connects the 3 pulleys, and T' the tension of the string that holds the object m₃.
Using the law of cosines, T' = 2T.
T - m₁g = m₁a (1)
T - m₂g = m₂a (2)
m₃g - 2T = m₃a (3)

adding up the first 2 equations, I got that
a = [g(m₁ + m₂) - 2T]/(m₁+m₂)

and replacing a in the 3rd equation, I finally got that T = [gm₃(m₁ + m₂)]/(m₁ + m₂ + m₃) which does (I think) coincidentally get me to 9,81 N.
Please help a girl out!!

 

[kuruman]

adding up the first 2 equations, I got that
a = [g(m₁ + m₂) - 2T]/(m₁+m₂)

How about adding all three equations together. What do you get then?

 

[danut]

How about adding all three equations together. What do you get then?

I did think about it, because the T would cancel out.
a = g(m₃ - m₁ - m₂)/(m₁ + m₂ + m₃)

(1) => T = m₁(a + g)

T = m₁g[(m₃ - m₁ - m₂)/(m₁ + m₂ + m₃) + 1]
T = m₁g[2m₃/(m₁ + m₂ + m₃)]

T = g[2m₁m₃/(m₁ + m₂ + m₃)] which ultimately is 9,81N. Is however the solution that I thought of correct? Because it is very different from the equation that the author gets, it's the only hint given for this problem. I also apologize for my poor formatting, I don't know how to use Latex!!

 

[SammyS]

Homework Statement: In the system bellow, there are 3 ideal pulleys. m₁ = 1.00 kg, m₂ = 1.00 kg, m₃ = 2.00 kg.
Find the tension in the string/wire that connects said pulleys.
Relevant Equations: -

View attachment 348640I'm struggling to get to the correct answer, which I posted down bellow.
The pulleys are ideal, so I figured that m₁ and m₂ will both move upwards (towards the ceiling?) with the acceleration a, while m₃ will move downwards with the acceleration -a.
Let T be the tension in the string which connects the 3 pulleys, and T' the tension of the string that holds the object m₃.
Using the law of cosines, T' = 2T.
T - m₁g = m₁a (1)
T - m₂g = m₂a (2)
m₃g - 2T = m₃a (3)

adding up the first 2 equations, I got that
a = [g(m₁ + m₂) - 2T]/(m₁+m₂)

and replacing a in the 3rd equation, I finally got that T = [gm₃(m₁ + m₂)]/(m₁ + m₂ + m₃) which does (I think) coincidentally get me to 9,81 N.
Please help a girl out!!
View attachment 348641

Yes, your result does give that T = 9,81 N .
This result does appear to depend on having $m_1+m_2=m_3$ as well as having $m_3=2\,\rm{kg}$

With the values given in the problem, what is your result for acceleration, ?

 

[danut]

Yes, your result does give that T = 9,81 N .
This result does appear to depend on having $m_1+m_2=m_3$ as well as having $m_3=2\,\rm{kg}$

With the values given in the problem, what is your result for acceleration, ?

Oh, I see that acceleration is 0 in my case, could that be right?

 

[SammyS]

Oh, I see that acceleration is 0 in my case, could that be right?

Can you see why ?

 

[kuruman]

Oh, I see that acceleration is 0 in my case, could that be right?

Yes.

 

[danut]

Can you see why ?

I think I do now Could it be that a₁ = a₂ = a, and a₃ = a₁ + a₂ = 2a (of course, in absolute value)? I really hope I'm not mistaken!!

 

[kuruman]

It is true by symmetry that $\mathbf{a}_1=\mathbf{a}_2=\mathbf{a}$. To find the relation of the magnitude $a$ to the magnitude of $a_3$, think of it this way. If each of the masses $m_1$ and $m_2$ rises by amount $y$, by what fraction (or multiple) of $y$ does $m_3$ descend?

 

[danut]

It is true by symmetry that $\mathbf{a}_1=\mathbf{a}_2=\mathbf{a}$. To find the relation of the magnitude $a$ to the magnitude of $a_3$, think of it this way. If each of the masses $m_1$ and $m_2$ rises by amount $y$, by what fraction (or multiple) of $y$ does $m_3$ descend?

My best guess is that the multiple is 2, like adding up the distances that the 2 masses travel: y' = y + y = 2y? I use the word guess because I feel like there's so much nuance to this sort of problems and I'm definitely seeming to be missing some of it

 

[Orodruin]

The system of equations you have arrived at in the OP is incorrect in the case of general masses as it assumes all masses have the same acceleration $a$. This can only be argued in the symmetric case where $m_1 = m_2$. In the general case, you will need an additional equation relating the accelerations of the masses which will essentially follow from the constant length of the string.

However, in your particular case, there is symmetry and you can get by with the equations you stated.

 

[Orodruin]

My best guess is that the multiple is 2, like adding up the distances that the 2 masses travel: y' = y + y = 2y? I use the word guess because I feel like there's so much nuance to this sort of problems and I'm definitely seeming to be missing some of it

How much more string is available between the pulleys (ie, how much more length of string will the middle pulley be hanging from) if one of the outer masses rises by y? How is this length distributed among the two string segments the middle pulley is hanging from?

 

[Lnewqban]

The pulleys are ideal, so I figured that m₁ and m₂ will both move upwards (towards the ceiling?) with the acceleration a, while m₃ will move downwards with the acceleration -a.

Welcome, @danut !

It is always convenient to draw free body diagrams of the involved pulleys in order to determine balance of forces (no movement or no acceleration), or lack of it.