A uniform square plate of mass 'm'is supported as shown. If the cable suddenly breaks, assuming centre of mass is on horizontal line passing through `A` determine, the reaction at `A` is `(mg)/n` that `n` is

Answer

4

Explanation

`I=(mb^(2))/6+(mb^(2))/2=(mb^(2))/2(1+1/3)`

`I=(2mb^(2))/2 g, ` Hence, `(mgb)/(sqrt(2))=(2mb^(2))/2alpha=(3g)/(2sqrt(2)b)`
Velocity of `O` is zero so `N_(x)=0`
`mg-N_(y)=mb/(sqrt(2))alpha=mb/(sqrt(2))((3g)/(2sqrt(2)b))=(3mg)/4 :. N_(y)=(mg)/4`