Question

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

Answer 1

Explanation:

(a) From the diagram, we get:

$R_{1}-m g+m a=0$

$\Rightarrow R_{1}=m g-m a$             …(i)

Now,

$F-\mu R_{1}-T=0 \text { and }-\mu R_{1}+T=0$

$F=2 \mu m(g-a)$

(b) The blocks have the acceleration as a1.

$R_{1}=-m a+m g$              ….(i)

And,

$m a_{1}=2 F-T-\mu R_{1}$    …(ii)

Now,

$T=\mu m g+M a_{1}-\mu m a$

In equation (ii), when the value of F and T is substituted, we get:

$\Rightarrow 4 \mu m g-4 \mu m a-2 \mu m g+2 \mu m a=m a_{1}+M a_{1}$

$\Rightarrow a 1=m+2 \mu m g-a M$

Thus, it is proven that two blocks move with identical acceleration a1 but in reverse directions.