Block `A` of mass `m` and block `B` of mass `2m` are placed on a fixed triangular wedge by means of a light and inextensible string and a frictionless pulley as shown in fig . The wedge is inclined at `45^(@)` to the horizontal on both sides . The coefficient of friction between the block `A`and the wedge is `2//3` and that between the block `B` and the wedge is `1//3` .If the system of `A and B` is released from rest then find .
a. the acceleration of `A`
b. tension in the string
c.the magnitude and direction of the frictional force acting on `A`

Explanation

In this problem as `(m_(B) - m_(A)) "g sin" theta ( = "mg"//sqrt2)` is lesser than `(mu_(B) m_(B) + mu_(A) m_(A)) "g cos" theta ( = 4 "mg"// 3sqrt2)` the masses will not move and hence ,
(i) Acceleration of B = acc`.` of A = 0
(ii) For equilibrium of B
2 mg sin `theta = T + mu_(B) 2 "mg cos" theta`
i.e., `T = (2"mg")/(sqrt2)[1 - (1)/(3)] = (4 "mg")/(3sqrt2)`

(iii) As for mass A , T = ( 4 mg / `3sqrt2`) up the plane and mg sin `theta = ("mg" //sqrt2)` up the plane and mg sin `theta = ("mg" //sqrt2)` down the plane , for its equilibrium , friction of value `f = ( 4 "mg")/(3sqrt2) - ("mg")/(sqrt2) = ("mg")/(3 sqrt2)` must act down the plane .