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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

A

`N_(A) = 2v (lx //d), N_(B) = wx//d`

B

`N_(A) =v(l-x//d), N_(B) = w x //d`

C

`N_(A) = 2v(l-x//d), N_(B) = 2w x //d`

D

`N_(A) = w x //d, N_(B) = w(1-x/d)`

Text Solution

Verified by Experts

The correct Answer is:
B
`N_(A) + N_(B) = W`…….(1)
Net torque about point C=0
`N_(A) x = N_(B0 (d-x)`………(2)
Solving Eq.s (1) and (2) , we get
`N_(A) = (d-x)/d w` and d`N_(B) = x/d w`
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