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            

A)  \[\frac{wx}{d}\]                              

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

C)  \[\frac{w(d-x)}{x}\]      

D)  \[\frac{w(d-x)}{d}\]

Correct Answer: D

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