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 \[{{N}_{1}}x=(w-{{N}_{1}})(d-x)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{N}_{1}}x=wd-wx-{{N}_{1}}d+{{N}_{1}}x\] \[\Rightarrow \,\,\,\,\,{{N}_{1}}d=w(d-x)\] \[\Rightarrow \,\,\,\,{{N}_{1}}=\frac{w(d-x)}{d}\]You need to login to perform this action.
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