Find the mass `M` so that it remains at rest in the abjoining figure. Both the pulley and string are light and friction is absent everywhere. `(g = 10 m//s^(2))`

Explanation

In the figure shown,
`a = ("Net pulling force")/("Total mass") = (30 - 20)/(3 + 2)`
`= 2 m//s^(2)`

`2 kg` `T - 20 = 2xxa = 4`
`:. T = 24N`
Now `Mg = 2T`
`M = (2T)/(g) = (48)/(10)`
`= 4.8 kg`