A livotad beam of negligible mass has mass suspended from one end and an Atwood's machine suspended from the other. The frictionless pulley has negligible mass and dimension. Gravity is directed downwards and `M_(2) = 3 M_(3)`,`l_(2) = 3 l_(1)` . Find the retio `M_(1)//M_(2)` which will ensure that the beam has no tedency to rotate just after the masses are released.

Options

A. `(M_(1))/(M_(2)) = 2`
B. `(M_(1))/(M_(2)) = 3`
C. `(M_(1))/(M_(2)) = 4`
D. None of these

Answer

B

Explanation

`For M_(2) and M_(3)`

`a = (M_(2)g - M_(3)g)/(M_(2) + M_(3)) = (3 M_(3)g - M_(3)g)/(3 M_(3) + M_(3))`
`= (g)/(2)`
Now `FBD of M_(2)` given the equation,
`M_(2)g - T = M_(2). a = (M_(2)g)/(2)`
`:. T = (M_(2)g)/(2)`
or `2 T = M_(2)g`
Now taking moment of forces about support point
` M_(1)g (l_(1)) = (2 T) l_(2) = (M_(2)g) (3 l_(1))`
` :. (M_(1))/((M_(2)) = 3`