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

Explanation

`R_1=ma-mg=0`

`rarr R_1=m(g-a)`
`=mg-ma`…….i
Again `F-T-muR_1=0`
`rarr T-mu R_1=0`
`rarr F-[mu(mg-ma)]-mu(mg-ma)=0`
`rarr T=mu(mg-ma)`
`rarrF-mu mg-mu ma -mu mg+mu ma =0`
`rarr F=2mu mg-2muma` ltbr. `=2mu m(g-a)`
b. Let acceleration of the blocks be `a_1`
`R_1=mg-ma`..............i
`and 2F=T-muR_1ma_1=0`
`- 2F=T-muR_1+Ma_1`
Now `T=muR_1+Ma_1`
`T=muR_1-Ma_1=0`
`=mu(mg-ma)+Ma_1`
`=mumg-muma+Ma_1`
`Substracting value of F and T we get
`2[gmum(g-a)-2(mumg-muma+Ma_1)-mumg+muma-ma_1=0`
`rarr 4mumg-4muma-2muma=ma_1+Ma_1 `
`rarr a_1=(2mum(g-a))/(M+m)`
Both blocks move with this acceleratio but in opposite directions.