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For the mass $m_1$, $$m_1\textbf{a}_1 = \textbf{T}_1$$

For the mass $m_2$, $$m_2\textbf{a}_2 = \textbf{T}_2 + (-\textbf{T}_1)$$

For the mass $m_3$, $$m_3\textbf{a}_3 = T_2 \hat{y} - m_3 g \hat{y}$$

Note that the magnitude of the tension force acting between $m_1$ and $m_2$ is $T_1$ and, the magnitude of the tension force acting between $m_1$ and $m_2$ is $T_2$.


In this elementary problem it is easy to calculate the acceleration of the blocks and the tension forces acting in between them. Before doing any calculation, it is assumed that $T_1 \neq T_2$ and then calculation shows that indeed this is the case. My question is what the general principle is for the prior assumption: $T_1 \neq T_2$?


closed as off-topic by John Rennie, stafusa, sammy gerbil, Jon Custer, Daniel Griscom Nov 3 '17 at 22:45

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  • 1
    $\begingroup$ If they were the same, the resultant force on $m_2$ would be zero so the whole systems, starting from rest, would not move. Besides, when solving this sort of exercise, the question is why must some forces be the same, not why can't they. $\endgroup$ – stafusa Nov 3 '17 at 8:15
  • $\begingroup$ -1. Not clear. Why do you need to know beforehand whether $T_1=T_2$? $\endgroup$ – sammy gerbil Nov 3 '17 at 9:40
  • $\begingroup$ I have edited it. $\endgroup$ – omehoque Nov 7 '17 at 3:46

$$\mathbf{T_1} = \mathbf{T_2} \Rightarrow m_2\mathbf{a_2} = \mathbf{0} \Rightarrow \mathbf{a_2} = \mathbf{0} $$

Assuming ideal strings, the acceleration on all of the blocks is the same. Therefore, $\mathbf{a_1} = \mathbf{a_2} = \mathbf{a_3} = \mathbf{0}$.

However, if I pick $m_1 = m_2 = 0$, I would still have that $\mathbf{a_3} = \mathbf{0}$. That would mean that the third block is not accelerating even if it is alone. If the masses are zero, then connecting the block to then using ideal strings and an ideal pulley would be the same as not to connect it at all, since that the forces exerted by any of them would be zero as well due to Newton's Second Law of Motion.

Well, gravity does exist. Therefore, we have reached an absurd affirmation and the original assumption that $\mathbf{T_1} = \mathbf{T_2}$ has to be false.


We know that $T_1\ne T_2$ (unless the mass of one of the blocks is $0$) because each of the strings, which pull the blocks with $T_1$ and $T_2$, must pull all of the blocks down the chain from itself, not just the individual block it is attached to.

For example, say you have blocks $A, B, C$ attached by strings $1,2,3$, such that string $3$ is extending out from $C$. If $3$ were pulled such that the blocks had some acceleration $a$, then you can easily see how the tension between the three string is unequal.

  1. $T_1=m_A\cdot a$
  2. $T_2=(m_A+m_B)\cdot a$
  3. $T_3=(m_A+m_B+m_C)\cdot a$


  1. $T_1=m_A\cdot a$
  2. $T_2=T_1+m_B\cdot a$
  3. $T_3=T_2+m_C\cdot a$

We can also show that $T_1\ne T_2$ by examining what it would mean if $T_1=T_2$:

If $T_1=m_1\cdot a$ then

$$T_1=T_2\Rightarrow m_1\cdot a=(m_1+m_2)\cdot a\Rightarrow m_1=m_1+m_2$$

Which is false unless, as previously stated, $m_2=0$.


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