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(A) Let $N_1$ be the normal reaction on $A$ and $N_2$ be the normal reaction on $B$.For vertical equilibrium,the sum of upward forces must equal the d

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$\frac{W(d - x)}{d}$

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(A) Let $N_1$ be the normal reaction on $A$ and $N_2$ be the normal reaction on $B$.For vertical equilibrium,the sum of upward forces must equal the downward force:$N_1 + N_2 = W$  --- $(i)$For rotational equilibrium,we take the torque about the centre of mass of the rod (where the weight $W$ acts). The torque due to $W$ is zero.Sum of torques about the centre of mass = $0$$N_1 \cdot x = N_2 \cdot (d - x)$  --- $(ii)$From equation $(i)$,$N_2 = W - N_1$.Substitute this into equation $(ii)$:$N_1 x = (W - N_1)(d - x)$$N_1 x = W(d - x) - N_1(d - x)$$N_1 x = W(d - x) - N_1 d + N_1 x$$N_1 d = W(d - x)$$N_1 = \frac{W(d - x)}{d}$

$A$ 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

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