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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

`(wx)/d`

B

`(wd)/x`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
As the weight w balances the normal reactions.

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 vluie 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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Knowledge Check

  • 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 a distance x from A .

    A
    the normal reaction at `A` is `(wx)/(d)`
    B
    the normal reaction at `A` is `(w(d - x))/(d)`
    C
    the normal reaction ar `B` is `(wx)/(d)`
    D
    the normal reaction at `B` is `(w(d - x))/(d)`.

  • A uniform metre rod is bent into L shape with the bent arms at 90^@ to each other. The distance of the centre of mass from the bent point is

    A
    `L/(4sqrt(2))m`
    B
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    C
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  • A uniform rod AB hinged about a fixed point P is initially vertical. A rod is released from vertical position. When rod is in horizontal position: The acceleration of the centre of mass of the rod is

    A
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    B
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    C
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    D
    `-(9g)/7hati-(3g)/7hatj`

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