Question

A rod of weight $w$ is supported by two parallel knife edges $A$ & $B$ and is in equilibrium in a horizontal position. The knives are at a distance $d$ from each other. The center of mass of the rod is at a distance $x$ from $A.$ The normal reactions at $A$ and $B$ will be

• A. $N_A=2w(1-x/d),N_B=wx/d$
• B. $N_A=w(1-x/d),N_B=wx/d$
• C. $N_A=2w(1-x/d),N_B=2wx/d$
• D. $N_A=w(2-x/d),N_B=wx/d$

Answer 1

Answer: (B)

$N_A=w(1-x/d),N_B=wx/d$

Rod is in the equilibrium,

$\sum F_Y$=0 $\Rightarrow N_A+N_B=w$ ......1

Taking moments about point A,

Moments clockwise = Moments Anticlockwise

$w\times x=N_B\times d$

$\therefore N_B=\frac{wx}{d}$ .......2

Solving equations 1 and 2,

$N_A=w(1-\frac{x}{d})$, $N_B=\frac{wx}{d}$