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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Laws of Motion

NEET 2015 · Physics · Laws of Motion

A rod of weight W is supported by two parallel knife edges A and B...

A rod of weight W\text{W} is supported by two parallel knife edges A\text{A} and B\text{B} and is in equilibrium in a horizontal position. The knives are at a distance d\text{d} from each other. The centre of mass of the rod is at distance x\text{x} from A\text{A} . the normal reaction on A\text{A} is:A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The

Options

  1. AWxd\dfrac{Wx}{d}
  2. BWdx\dfrac{Wd}{x}
  3. C\dfrac{W\left(d - x\right)}{x}
  4. D\dfrac{W\left(d - x\right)}{d}

Correct answer

D. \dfrac{W\left(d - x\right)}{d}

Step-by-step solution

A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The
Equation for force balance
NA+NB=W{N}_{A} + {N}_{B} = W ...(i)
NB=W−NA{N}_{B} = W - {N}_{A}
Equation for torque balance about centre of mass of rod.
NAx=NB(d−x){N}_{A}x = {N}_{B}\left(d - x\right)
NAx=(W−NA)(d−x){N}_{A}x = \left(W - {N}_{A}\right)\left(d - x\right)
NAx=Wd−Wx−NAd+NAx{N}_{A}x = Wd - Wx - {N}_{A}d + {N}_{A}x
NAd=W(d−x){N}_{A}d = W\left(d - x\right)
NA=W(d−x)d{N}_{A} = \dfrac{W\left(d - x\right)}{d}
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