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In my textbook there is a case given that is to find acceleration and tension produced in a string passed over a frictionless pulley and attached with bodies of different masses at its ends. Now below I am showing you some of the derivation given in my textbook and then in last I will tell what my question exactly is.

Derivation:

(In diagram, the mass at right is $m_1$ and at left is $m_2$)

Let $m_1>m_2$ then body at right will accelerate down with acceleration $a$ and body at left will move up with same acceleration.

The net force acting on body at right is given by, $m_1g$ $-$ $T$ (since $W_1$ $>$ $T$) and the net force acting on the body at left is given by, $T$ $-$ $m_2g$ (since $T$ $>$ $m_2g$).

Now come to my question. We know that,

$a=\frac{Net-Force}{m}$

Acceleration for body at right is given by,

$a=\frac{m_1g-T}{m_1}$

Acceleration for body at left is given by,

$a=\frac{T-m_2g}{m_2}$

Since it is said that acceleration produced in both bodies is same then,

$\frac{m_1g-T}{m_1}$ $=$ $\frac{T-m_2g}{m_2}$ but it is written that,

$m_1>m_2$ and therefore,

$\frac{m_1g-T}{m_1}\neq \frac{T-m_2g}{m_2}$

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I ask that Can two bodies move with equal magnitude of acceleration if forces acting on both are unequal? If not then is that all given in my textbook is wrong? If yes then how?

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    $\begingroup$ There's no problem. Take your last equality, and solve for T. $\endgroup$ – garyp Feb 23 '15 at 15:17
  • $\begingroup$ Could you expand on the very final step (last two lines)? $\endgroup$ – DJohnM Feb 23 '15 at 15:18
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    $\begingroup$ Your last step is incorrect. It's not clear to me why you conclude that $m_1>m_2$ implies inequality. Perhaps it would help to consider the case $m_1=m_2$. $\endgroup$ – garyp Feb 23 '15 at 15:29
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Firstly, yes two bodies with different masses can experience the same acceleration, precisely in fact by varying the force with which each is pushed. Newton's got our back on this one with: $$\mathbf{F} = m \mathbf{a}$$ which for $m_1 \ne m_2$ yields: $$\mathbf{F}_1 = m_1 \mathbf{a} \qquad \mathbf{F}_2 = m_2 \mathbf{a}$$ I'll cite this specific case as an example of this very phenomena! in fact, the equation you wrote down equating the acceleration of the two masses is correct and can be solved to give an expression for $T$. \begin{align*} & \dfrac{m_1g -T}{m_1} = \dfrac{T - m_2g}{m_2}\\ \Rightarrow &\ m_2m_1g - m_2T = m_1T - m_1m_2 g\\ \Rightarrow &\ 2m_1 m_2 g = T(m_1 + m_2)\\ \Rightarrow &\ T = \dfrac{2m_1 m_2 g}{m_1 + m_2}\\ \end{align*} So all is right with the world and you don't have to throw out your textbook!

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