A uniform rod of mass `M = 2 kg` and length `L` is suspended by two smooth hinges `1` and `2` as shown in Fig. A force `F = 4 N` is applied downward at a distance `L//4` from hinge `2`. Due to the application of force `F`, hinge `2` breaks. At this instant, applied force `F` is also removed. The rod starts to rotate downward about hinge `1`. (`g = 10 m//s^(2)`)

The reaction at hinge `1`, just after breaking of hinge `2`, is

Options

A. `20N`
B. `10N`
C. `5N`
D. `0`

Answer

C

Explanation

Now rod is not in equilibrium ltbgt `:. F=Ma_(CM)`
`implies Mg-N=Ma`……..i
and `tau=Ia`
`impliesMg(L/2)(ML^(2))/3 a/((L/2))` (`:'(ML^(2))/3, a=alpha(L/2)`)
`gta=3/4g`……..ii
From eqn i and ii
`N=(Mg)/4=5N`