Question

An iron sphere weighing `10N` rests in a `V` shaped smooth trough whose sides form an angle of `60^(@)` as shown in the Then the reaction forces are
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A. `R_(A) =10N` and `R_(B) = 0` in case (i)
B. `R_(A) =10N` and `R_(3) =10N` in case (ii)
C. `R_(A) = (20)/(sqrt3)N` and `R_(B) = (10)/(sqrt3)N` in case (iii)
D. `R_(A) =10N` and `R_(B) =10N` in all the `3` cases

Answer 1

Correct Answer - A::B::C
Since the sphere is not moving `sumF_(H) =0`
`R_(B) sin 60 =0`
`:. R_(B) =0`
`& R_(A) =W =10N`
`sum F_(H) =0`
`:.R_(A) sin 60 = R_(B) sin 60 rArr R_(A) =R_(B) =R`
Now `sum F_(U) =0 :. 2R cos 60 -W =0`
`R =W =10N`
`sum F_(V) =0 :. R_(A) sin 60 =W`
`rArr R_(A) = (20)/(sqrt3) N` also `sumF_(H) =0,`
`R_(A/2) -R_(B) =0` So `R_(B) = (10)/(sqrt3) N`


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