Block A of mass m and block B of mass `2m` are placed on a fixed traingular wedge by means of a massless inextensible string and a frictionless pulley as shown in figure The wedge is inclined at `45^(@)`to the horizontal on both sides .The coefficient of friction between blocks A and the wedge is `2//3` and that between block B and is released from rest find the following

The acceleration of A is

Options

A. `(g)/(3sqrt(2))`
B. Zero
C. `(g)(sqrt(7))`
D. `(g)/(2sqrt(3))`

Answer

b

Explanation

`T - mg sin 45^(@) - f_(l_1) = ma , 2mg sin 45^(@) - T-f_(l_2) = 2ma`

where `f_(l_1) = mu_(1)mg cos 45^(@)= 2 mg cos 45^(@)//3`
and `f_(l_2) = mu_(2)2mg cos 45^(@)= 2 mg cos 45^(@)//3`
Now we get `a = (g)/(9sqrt(2))`
This is negative, which is not possible Hence `a= 0`