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A uniform rod of mass M = 2 kg and lengt...

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

A

`24N`

B

`12N`

C

`11N`

D

`10N`

Text Solution

Verified by Experts

The correct Answer is:
C

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`
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Knowledge Check

  • 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

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