A rod of weight w is supported by two paralle | vedclass

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(D) Let $N_A$ and $N_B$ be the normal reactions at knife edges $A$ and $B$ respectively.Since the rod is in equilibrium,the net force is zero: $N_A +

Vedclass · Accepted Answer

Both $(A)$ and $(B)$ are correct.

Vedclass · Answer

(D) Let $N_A$ and $N_B$ be the normal reactions at knife edges $A$ and $B$ respectively.Since the rod is in equilibrium,the net force is zero: $N_A + N_B = w$.Taking torque about point $A$,the clockwise torque due to weight $w$ must equal the counter-clockwise torque due to $N_B$: $w \cdot x = N_B \cdot d$.Thus,$N_B = \frac{wx}{d}$.Substituting $N_B$ into the force equation: $N_A = w - \frac{wx}{d} = w(1 - \frac{x}{d}) = \frac{w(d-x)}{d}$.Therefore,the normal reaction at $A$ is $\frac{w(d-x)}{d}$ and at $B$ is $\frac{wx}{d}$.Comparing this with the options,both $(A)$ and $(B)$ are correct.

$A$ 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 a distance $x$ from $A$.

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