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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

`(w x)/(d)`

B

`(w d)/(x)`

C

`(w(d-x))/(x)`

D

`(w(d-x))/(d)`

Text Solution

Verified by Experts

As the weight w balances the normal reaction.

So, `w = N_(1) + N_(2)` ...(i)
Now balancing torque about the COM,
i.e., anti-clockwise momentum
= clockwise momentum
`rArr N_(1)x = N_(2)(d-x)`
Putting the value of `N_(2)` from Eq. (i), we get
`N_(1)x = (w-N_(1))(d-x)`
`rArr N_(1)x = wd-wx -N_(1)d + N_(1)x`
`rArr N_(1)D = w(d-x)`
`rArr N_(1) = (w(d-x))/(d)`
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