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`, before hinge `2` breaks, is

Options

A. `24N`
B. `12N`
C. `11N`
D. `10N`

Answer

C

Explanation

As the rod is in equilibrium
`:. SigmaF_(x)=0`
`SigmaF_(y)=0`
`Sigmatau=0`
Taking torque about hinge `2`, we get
`N_(1)xxL=Mgxx1/2+FxxL/4`
`:. N_(1)=(Mg)/2+F/4=11N`