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Units And Measurement

Question
CBSEENPH11020540

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 distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

  • wx over straight d
  • wd over straight x
  • fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight x end fraction
  • fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight d end fraction

Solution

D.

fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight d end fraction

As the weight w balances the normal reactions.

So, 
w= N1 +N2   ----- (i) 
Now balancing torque about the COM,
i.e. anti -clockwise momentum= clockwise momentum

rightwards double arrow space straight N subscript 1 straight x space equals space straight N subscript 2 space left parenthesis straight d minus straight x right parenthesis
Putting space the space value space of space straight N subscript 2 space from space equation space left parenthesis straight i right parenthesis comma space we space get
straight N subscript 1 straight x space equals space left parenthesis straight W minus straight N subscript 1 right parenthesis left parenthesis straight d minus straight x right parenthesis
rightwards double arrow space straight N subscript 1 straight x space equals space wd minus space wx minus straight N subscript 1 straight d space plus straight N subscript 1 straight x
straight N subscript 1 straight d space equals negative space straight w space left parenthesis straight d minus straight x right parenthesis
straight N subscript 1 space equals space fraction numerator straight w left parenthesis straight d minus straight x right parenthesis over denominator straight d end fraction