Question

In the arrangement shown in figure,find the tensions in the rope and accelerations of the masses $m_1$ and $m_2$ and pulleys $p_1$ and $p_2$ when the system is set free to move. Assume the pulleys to be massless and strings are light and inextensible.

Answer 1

Applying the equations of motion for pulleys and blocks

For $m_1$ : $T-m_{1}g=m_1{a}_1$........(i)

For $m_2$ : $m_{2}g-T=m_2{a}_2$.......(ii)

For pulley $P_1$: $2T-T=0\times a_{p_1}$......(iii)

For pulley $P_2$ : $T_2-2t=0\times a_{p_2}$

From Eq (iii), $T=0$; from (iv), $T_2=0$, and from (i) and (ii), $a_1=-g$ also $a_2=g$. Hence, blocks $m_1$ and $m_2$ both will move down with acceleration g downward direction.

Calculating $a_{p_1}$ and $a_{p_2}$

Constraint relations : consider the reference line and the position vectors of the pulleys and masses. Write the length of the rope in terms of position vectors and differentiate it to obtain the relations between accelerations of the masses and pulleys.

$\bullet$ For the length of the string connecting $P_2$ and $m_2$ not to change and for this rope not to slacken, $a_{p_2}=a_2=g$.

$\bullet$ Length of the string connecting

$p_1$ to $m_1$ not to change and for this rope not to slacken:

$l=(x_1-x_{p_1})+x_{p_1}+(x_{p_2}-x_{p_1})+x_{p_2}$

$=x_1-x_{p_1}+x_{p_2}$

Differentiationg this equation w.r.t. t twice, we get

$\frac{d^{2}l}{{dt}^2}=\frac{d^2{x}_1}{{dr}^2}-\frac{d^2{a}_{p_1}}{{dt}^2}2+2\frac{d^2{x}_{p_3}}{{dt}^2}\Rightarrow 0=a_1-a_{p_1}-2a_{p_2}$

$\Rightarrow$ $0=g-a_{p_1}+2g\Rightarrow a_{p_1}=3g$