All the surface and pulleys are frictionless in the shown arrangement. Pulleys `P` and `Q` are massless .The force applied by clamp on pulley `P` is

Options

A. `(mg)/(6) (- sqrt(3) hati - 3hatj)`
B. `(mg)/(6) ( sqrt(3) hati + 3hatj)`
C. `(mg)/(6) sqrt2`
D. None of these

Answer

B

Explanation

`a = ("Net pulling force")/("Total mass")`
`= (2 mg sin 30^(@))/(2 m + m) = (g)/(3)`

`FBD of m`
`T = ma = mg//3`
Resultant of tensions
`R = - (T cos 30^(@)) hati - (T sin 30^(@) + T) hatj`
Putting `T = mg//3`
`R = - (sqrt3)/(6) mg hati - (mg)/(2) hatj`
Since pulley `P` is in equilibrium. Therefore,
`F + R = 0`
when , `F =` force applied by clamp on pulley
`:. F = - R = (mg)/(6) (sqrt3 hati + 3 hatj)`