What force F must be applied so that `m_(1)` and `m_(2)` are at rest on `m_(3)` in

Answer

`F=(m_(1)+m_(2)+m_(3))m_(2)g//m_(1)`

Explanation

`T=m_(2)g`
Now, `T=m_(1)a`
`m_(2)g=m_(1)a`
`a=((m_(2)g)/m_(1))`
Common acceleration of the system is:
`a=F/((m_(1)+m_(2)+m_(3)))`
`F=((m_(1)+m_(2)+m_(3))m_(2)g)/m_(1)`