(a) In this case net pulling force
`=m_(A)g sin 60^(@)+m_(B)g sin 60^(@)-m_(C)g sin30^(@)`
`=(1)(10)(sqrt(3))/(2)+(3)(10)((sqrt3)/(2))-(2)(10)((1)/(2))`
=24.64 N
Total mass being pulled = 1+3+2=6 kg
`therefore` Acceleration of the system a `=(24.64)/(6)=4.10 ms^(-2)`
(b) For the tension in the string between A and B.
FBD of A
`m_(A)g sin 60^(@)-T_(1)=(m_(A))(a)`
`therefore T_(1)=m_(A)g sin 60^(@)-m_(A)a`
`=m_(A) ( g sin 60^(@)-a)`
`therefore T_(1)=(1) (10)xx(sqrt(3))/(2)-4.56 N`
For the tension in the string between B and C.
FBD of C
`T_(2)-m_(C)g sin 30^(@)= m_(C)a`
`therefore T_(2)=m_(C)(a+g sin 30^(@))`
`therefore T_(2)=2[4.10+10((1)/(2))]=18.2N`