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