Question

Consider a rod of weight W that is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance a from each other. The center of mass of the rod is at a distance x from A. The normal reaction on A is?

Answer 1

Step 1: Given and assume

${\mathrm{N}}_1$ is the normal reaction on A.

${\mathrm{N}}_2$ is the normal reaction on B.

$\mathrm{a}$ is the distance from each other.

$\mathrm{W}$ is the weight of the rod.

Step 2: Calculating normal reaction

By equating the forces and taking the vertical equilibrium by the normal reaction, we get

${\mathrm{N}}_1+{\mathrm{N}}_2=\mathrm{W}$…(equation $1$)

In the above diagram, torque balance about the center of mass of the rod, we get

$\mathrm{Nx}={\mathrm{N}}_2\left(\mathrm{a}-\mathrm{x}\right)$…(equation $2$)

By converting the equation $1$ in terms of the normal reaction on B, we get

${\mathrm{N}}_2=\mathrm{W}-{\mathrm{N}}_1$

By substituting the value of $N_2$ in the equation $2$, we get

$\mathrm{Nx}=\left(\mathrm{W}-{\mathrm{N}}_1\right)\left(\mathrm{a}-\mathrm{x}\right)$

By simplifying the above equation, we get

$\mathrm{Nx}=\left(\mathrm{Wa}-\mathrm{Wx}-{\mathrm{N}}_1\mathrm{a}+\mathrm{Nx}\right)$

By taking the common terms and simplifying the equation, we get

$\mathrm{Nd}=\mathrm{W}\left(\mathrm{a}-\mathrm{x}\right)$

By converting the equation in terms of the normal reaction on A, we get

${\mathrm{N}}_1=\frac{\mathrm{W}(\mathrm{a}-\mathrm{x})}{\mathrm{a}}$

Therefore, the normal reaction on A is $\frac{\mathrm{W}\left(\mathrm{a}-\mathrm{x}\right)}{\mathrm{a}}$.