(a) In this case net pu lling force
`= m_(A)g sin 60^(@) + m_(B) sin 60^(@) - m_(c) g sin 30^(@)`
`= (1)(10) sqrt(3)/(2) + (3)(10) (sqrt((3)/(2))) - (2)(10) ((1)/(2))`
`= 24.64N`
Total mass being pulling`= 1 + 3 + 2 = 6kg`
:. Accelleration of the system `a = (24.64)/(6) = 4.1m//s^(2)`
For the tension in the string between `A` and `B` .
`FBD` of `A`
`m_(A)g sin 60^(@) - T_(1) = (m_(A)) (a)`
`T_(1) = m_(A) g sin 60^(@) - m_(A)`
`= m_(A) (g sin 60^(@)-a)`
`T_(1) = (1) (10 xx sqrt(3)/(2)-4.1)`
`= 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`
`T_(2) = m_(C) (a+g sin 30^(@))`
`T_(2) = 2 [4.1 + 10 ((1)/(2))]`
`= 18.2 N`