Question

A rod of weight $\omega$ 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. Then,

• A. the normal reaction at A is $\frac{\omega x}{d}$
• B. the normal reaction at A is $\frac{\omega (d-x)}{d}$
• C. the normal reaction at B is $\frac{\left(\omega x\right)}{d}$
• D. the normal reaction at B is $\frac{\omega (d-x)}{d}$

Answer 1

Answer: (B), (C)

The correct options are
B the normal reaction at A is $\frac{\omega (d-x)}{d}$
C the normal reaction at B is $\frac{\left(\omega x\right)}{d}$


$\sum T_B=0$
$\Rightarrow -N_{1}d+W(d-x)=0$
$N_1=\frac{W(d-x)}{d}$
$Since{N}_1+N_2=W$
$\Rightarrow N_2=\frac{Wx}{d}$