In fig.block A and B are connected together by a string and placed on a smooth inclined plane. B is connected to C(which is suspended vertically) by another string which passes over a smooth pulley fixed to th plane. The mass A is `m_(A)=1` kg and mass of B is `m_(B)=2 kg`.

(a) If the system is at rest, find the mass of C.
(b) If the mass of C is twice the mass calculate in (a) then find the acceleration of the system.

Explanation

(a) From the free-body diagram of A :
`T_(1)`: Tension in string betwwne A and B
`N_(A)`: Normal reaction between A and incline
As A is at rest, net force parallel and perpendivular to inclined should be balanced.

`N_(A)=m_(A)g cos theta = 10 cos 30^(@) = 5 sqrt(3)N`
`T_(1)=m_(A)g sin theta = 10 sin 30^(@) = 5N`
From the free-body diagram of B:
As B is connected to both strings, two tensions `T_(1)` and `T_(2)` will act on it.
`T_(2)` is the tension (force) of string between B and C acting upwards.
`T_(1)` is the tension of string between A and B acting downwards
Balancing forces:
`N_(B)=m_(B)g cos 30^(@)=20 cos 30^(@)=10sqrt(3)N`
`T_(2)=T_(1)+m_(B)g sin 30^(@)=5+20sin30^(@)`
`=5+20 xx 1/2 = 15N`
Force diagram of C:
`T_(2)` is the pulling force of string on block C, therefore, `m_(C )g=T_(2)`, hence `M_(C)=1.5kg`
(b) In this case, mass of `C=3 kg`
Let the acceleration of the system be a. We can assume arbitrary direction of motion. Let the blocks move up the incline and block C move downward.
From FBD of A: `T_(1)-10 sin 30^(@)=1a` ...(i)
From FBD of : `T_(2)-T_(1)-20 sin 30^(@)=2a` ...(ii)
From FBD of C: `30-T_(2)=3a` ...(iii)
Solving (i), (ii) and (iii) , `a=2.5 ms^(-2)` .