In the figure a truck is moving on a horizontal surface with acceleration `a`. Two blocks of equal masses `m` are supported on the truck as shown in figure. Given that when the block at the top surface is just about to slide, other block remains hanging at `30^(@)` from the vertical. In this system.

Options

A. `a=(g)/(sqrt(3))`
B. `T=(2)/(sqrt(3))mg`
C. `mu=(5-sqrt(3))/(3sqrt(3))`
D. `T=(sqrt(3))/(2)mg`

Answer

A, B

Explanation

`T sin 30^(@)=ma …..(1)`
`T cos 30^(@)=mg …….(2)`

dividing equation `(1)` by equation `(2)`
`tan 30^(@)=(a)/(g)`
`rArr a=gtan 30^(@)`
`rArr a=(g)/(sqrt(3))` Ans.
From `(2)T=(mg)/(cos 30^(@))=(2mg)/(sqrt(3))`Ans.
and `mu mg-T=ma`
`rArr mu mg=T+ma=(2mg)/(sqrt(3))+ma`
`=(2mg)/(sqrt(3))+(mg)/(sqrt(3))`
`rArr mumg=(3mg)/(sqrt(3))=sqrt(3)mg`
`=mu=sqrt(3) Ans.`