(a) `m_(A)g sin60^(@)-T_(1)=m_(A)a`..........(1)
`T_(1)-T_(2)+m_(B)gsin60^(@)=m_(B)a` .............(2)
`T_(2)-m_(C)gsin30^(@)=m_(C)a`.........(3)
`therefore` Acceleration of the system `a=(24.64)/(6)=4.1 m//s^(2)`
(b) For the tension `T_(1)` in the string between A and B, `m_(A)g sin60^(@)-T_(1)=(m_(A))(a)`
(b) For the tension `T_(1)` in the string between A and B, `m_(A)gsin60^(@)-T_(1)=(m_(A))(a)`
`rArrT_(1)=(1)(10xxsqrt(3)/(2)-4.1)=4.56N`
For the tension `T_(2)` in the string between B and C
`T_(2)-m_(C)g sin30^(@)=m_(C)a`
`rArr T_(2)=m_(C)(a+g sin30^(@))=2[4.1+10((1)/(2))]=18.2N`