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`

Answer

(i) zero (ii) `(2sqrt(2))/(2) mg` (iii) `(mg)/(3sqrt(2))`

Explanation

(i) For finding direction of friction first assume there is no friction anywhere. In the absence of friction the block B will move down the plane and the block A will move up the plane. Frictional force opposes this motion.
F.B.D. of the block A
`implies T = m g sin 45^(@) - f_(1) = ma`
From F.B.D of B
and `2m g sin 45^(@) - f_(2) - T = 2ma`
Adding (1) and (2), we get
`mg sin 45^(@) - (f_(1) + f_(3)) = 2ma`
For a to be non-zero mg sin 45° must be greater than the maximum value of `(f_(1) + f_(2))`
`:' (f_(1) + f_(2))_("max") = mu_(1) N_(1) + mu_(2) N_(2) = (mu_(1) m_(1) + 2 mu_(2) m_(2)) g cos 45^(@) = (4)/(5) m g cos 45^(@)`
`implies m g sin 45^(@) lt (f_(1) + f_(2))_("max")`, Hence block will remain stationary
(ii) F.B.D. of the block B
`f_(2("max")) = mu_(2) N_(2) = (1)/(3) 2m g cos 45^(@) = (2)/(3 sqrt(2)) mg`
`(W)_(B) = 2m g sin 45^(@) = (2mg)/(sqrt(2))`
`:' 2m g sin 45^(@) gt f_(2("max"))`, therefore block B has tendency to slide down the plane.
For block B to be at rest
`T + f_(2("max")) = 2m g sin 45^(@)` `implies T = (mg)/(sqrt(2)) (2 - (2)/(3)) = (4 mg)/(3sqrt(2)) implies T = (2sqrt(2))/(3) mg`
(iii) Component of weight of A parallel to inclined plane
`(W)_(A) = m g sin 45^(@) = (mg)/(sqrt(2))`
As `T gt (W)_(A)`
Hence block A has tendency to move up the plane, therefore frictional force on the block A will be down the plane.
For A to be at rest
F.B.D of A `m g sin 45^(@) + f = T`
`implies f = T - m g sin 45^(@) = (2 sqrt(2m) g)/(3) - (mg)/(sqrt(2)) implies f = (mg)/(3sqrt(2))`