A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal…

1. $\frac{\mathrm{W}\mathrm{x}}{\mathrm{d}}$
2. $\frac{\mathrm{W}\mathrm{d}}{\mathrm{x}}$
3. $\frac{\mathrm{W}\left(\mathrm{d}-\mathrm{x}\right)}{\mathrm{x}}$
4. $\frac{\mathrm{W}\left(\mathrm{d}-\mathrm{x}\right)}{\mathrm{d}}$

Solution

Equation for force balance
${\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}}=\mathrm{W}$ ...(i)
${\mathrm{N}}_{\mathrm{B}}=\mathrm{W}-{\mathrm{N}}_{\mathrm{A}}$
Equation for torque balance about centre of mass of rod.
${\mathrm{N}}_{\mathrm{A}}\mathrm{x}={\mathrm{N}}_{\mathrm{B}}\left(\mathrm{d}-\mathrm{x}\right)$
${\mathrm{N}}_{\mathrm{A}}\mathrm{x}=\left(\mathrm{W}-{\mathrm{N}}_{\mathrm{A}}\right)\left(\mathrm{d}-\mathrm{x}\right)$
${\mathrm{N}}_{\mathrm{A}}\mathrm{x}=\mathrm{W}\mathrm{d}-\mathrm{W}\mathrm{x}-{\mathrm{N}}_{\mathrm{A}}\mathrm{d}+{\mathrm{N}}_{\mathrm{A}}\mathrm{x}$
${\mathrm{N}}_{\mathrm{A}}\mathrm{d}=\mathrm{W}\left(\mathrm{d}-\mathrm{x}\right)$
${\mathrm{N}}_{\mathrm{A}}=\frac{\mathrm{W}\left(\mathrm{d}-\mathrm{x}\right)}{\mathrm{d}}$

Asked in: NEET 2015 (Phase 1)

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