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A rod of weight w is supported by two pa...

A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is.. And on B is......

A

`d / d - x`

B

`d - x/d`

C

`d - x/x`

D

`x/ d - x`

Text Solution

Verified by Experts

The correct Answer is:
c
A and B denote knife edges, separated by distance d, C denotes center of mass of the rod.

Let `R_(A)` =Normal reaction at A
`R_(B)` =Normal reaction of b
For the rod to be in equilibrium in horizontal position, the moment of all the forces about A or about B should be zero
`therefore` Take moment of forces about A,
`xW = R_(B)d`, where W denotes weight of rod or `R_(B) = (xW)/(d)`
Again, take moment of force about B,
`(d-x)W = R_(A)d` or `R_(A) =(d-x)/(d) W`
`therefore R_(A)/R_(B) = (d-x)/x`
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