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

A rod of weight W is supported by two parallel knife edge 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

`(Wx)/(d)`

`(Wd)/(x)`

`(W(d-x))/(x)`

`(W(d-x))/(d)`

Text Solution

Verified by Experts

The correct Answer is:

Let the normal reactions on A and B be `N_(A)` and `N_(B)`. At equilibrium `N_(A)+N_(B)`= W
Again when the resultant torques with respect to the centre of mass is zero then
`N_(A)x+W.0-N_(B)(d-x)=0 or, N_(A)(x)/(d-x)-N_(B) = 0`
So, `N_(A)+N_(A)(x)/(d-x)=Wor,N_(A)(1+(x)/(d-x)) = W`
or, `N_(A)(d)/(d-x) = W:.N_(A) = (W(d-x))/(d)`

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