Question

In the arrangement shown in $Fig.6.144$ block B starts form rest and moves towards right with a constant acceleration. After time t, the velocity of A with respect to B becomes V. Determine the acceleration of A.

Answer 1

Taking reference line through support, S let $x_A$ and $x_B$ are the distances of blocks A and B, respectively, from S. The total length of the string.

$l=4x_A+5x_B+l_0$, where $l_0$ is some part of string which is over pulley and somewhere else which remains constant. Differentiating l w.r.t. time, we get

$\frac{dl}{dt}=\frac{d}{dt}(4x_A+5x_B+l_0)$

or $o=4\frac{{dx}_A}{dt}+5\frac{{dx}_B}{dt}\Rightarrow \frac{{dx}_A}{dt}=\frac{5}{4}\frac{{dx}_B}{dt}$

$\frac{{dx}_A}{dt}=-v_A$ (-because $x_a$ is decreasing with time)

$\frac{{dx}_B}{dt}=-v_B$ (+because $x_b$ is increasing with time)

or $v_A=\frac{5}{4}{v}_B$; also $a_A=\frac{5}{4}{a}_B$

Method 2:

Charge in the length of segment I, $\Delta l_{1,2}=0+\left(x_B\right)$

Segment II

$\Delta l_{3,4}=\left({-x}_A\right)+\left(x_B\right)$

Segment III

$\Delta l_{5,6}=\left({-x}_A\right)+\left(x_B\right)$

Segment IV

$\Delta l_{7,8}=\left({-x}_A\right)+\left(x_B\right)$

Segment V

$\Delta l_{9,10}=\left({-x}_A\right)+\left(x_B\right)$

Total charge in segment lengths should be zero.Therefore,

$\Delta l={\Delta l}_{1,2}+{\Delta l}_{3,4}+{\Delta l}_{5,6}+{\Delta l}_{7,8}+{\Delta l}_{9,10}=0$

$=x_B+[{-x}_A+x_B]+[{-x}_A+x_B]+[{-x}_A+x_B]+[{-x}_A+x_B]=0$

$=-4x_A+5x_B=0$

or $x_A=\frac{5}{4}{x}_B\Rightarrow v_A=\frac{5}{4}{v}_B\Rightarrow a_A=\frac{5}{4}{a}_B$

Equations of motion

For A :

$4T=m_A{a}_A={Ma}_A$..........(i)

For B :

$F-5T=m_B{a}_B=3{Ma}_B$.......(ii)

$a_B=\frac{16F}{73}$

and

$a_A=\frac{5}{4}\times \left(\frac{16F}{73}\right)=\frac{20F}{73}$