Question

Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration a< g. Repeat parts (a) and (b).

Answer 1

$R_1+ma-mg=0$

$\Rightarrow R_1=m(g-a)=mg-ma$
Again $F-T-\mu R_1=0\Rightarrow T-\mu R_1=0\Rightarrow F-\left[\mu \right(mg-ma\left)\right]-\mu (mg-ma)=0$
$\Rightarrow T=\mu (mg-ma)\Rightarrow F-\mu mg-\mu ma-\mu mg+\mu ma=0\Rightarrow F=2\mu mg-2\mu ma=2\mu m(g-a)$
(b) Let acceleration of the blocks be $a_1$
$R_1=mg-ma$
and $2F=T-\mu R_1-m{a}_1-m{a}_1=0\Rightarrow 2F=T-\mu R_1+M{a}_1\text{Now}T=\mu R_1+M{a}_{1}T=\mu R_1-M{a}_1=0=\mu (mg-ma)+M{a}_1=\mu mg-\mu ma+M{a}_1$
Substracting the value of F and $T_1$ we get, $2\left[2\mu m\right(g-a)-2(\mu mg-\mu ma+M{a}_1)-\mu mg+\mu ma-m{a}_1=0\Rightarrow 4\mu mg-4\mu ma-2\mu ma=m{a}_1+M{a}_1\Rightarrow a_1=\frac{2\mu m(g-a)}{M+M}$
Both blocks move with this accelelration but, in opposite directions.