question_answer

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 [NEET 2015 ]

A)       \[\frac{wx}{d}\]

B) \[\frac{wd}{x}\]

C) \[\frac{w(d-x)}{x}\]

D) \[\frac{w(d-x)}{d}\]

Correct Answer: C

Solution :

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

\[\Rightarrow \]   \[{{N}_{1}}x={{N}_{2}}(d-x)\]

Putting the value of \[{{N}_{2}}\] from Eq. (i), we get

\[{{N}_{1}}x=(w-{{N}_{1}})(d-x)\]

\[\Rightarrow \]   \[{{N}_{1}}x=wd-wx-{{N}_{1}}d+{{N}_{1}}x\]

\[\Rightarrow \]   \[{{N}_{1}}d=w(d-x)\]

\[\Rightarrow \]   \[{{N}_{1}}=\frac{w(d-x)}{d}\]