Question

A rod of weight $W$ is supported by two 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. the normal reaction on A is
(A) $\dfrac{{W(d - x)}}{x}$
(B) $\dfrac{{W(d - x)}}{d}$
(C) $\dfrac{{Wx}}{d}$
(D) $\dfrac{{Wd}}{x}$

Answer 1

Hint : For equilibrium, upward forces must balance downward forces. Anticlockwise moment must also be equal to the clockwise moment. In the rod, the weight of the body must be equal to the sum of the normal reaction at A and B.

Formula used: In this solution we will be using the following formula;
$\Rightarrow {F_{NET}} = ma$ where ${F_{NET}}$ is the net force on a body, $m$ is the mass of the body, and $a$ is the acceleration of the body.
$\Rightarrow T = Fd$ where $T$ is the moment of a force acting on a body, $F$ is the force acting of the body, and $d$ is the distance from the axis of rotation.

Complete step by step answer
The forces acting in this system are the weight of the body pushing it downwards and the normal reaction force of the knife edge acting upwards. Hence total force on the rod is
${F_{NET}} = W - {N_A} - {N_B} = ma$ (assuming downward is positive) where $W$ is the weight of the rod, ${N_A}$ is the normal reaction on knife edge A, and ${N_B}$ is the normal reaction on knife edge B, $m$ and $a$ are the mass and acceleration respectively (from Newton’s second law). But since the rod is said to be in equilibrium, then $\Rightarrow a = 0$
Then
$\Rightarrow {F_{NET}} = W - {N_A} - {N_B} = 0$
$\Rightarrow W = {N_A} + {N_B}$
Now the centre of mass is at distance $x$ from A. Hence, the distance from the B is $d - x$ , since the distance between them is $d$ .
Thus, taking the moment of the normal reactions about the centre of mass, we have
${T_{NET}} = {N_A}x - {N_B}(d - x) = 0$ (assuming clockwise moment is positive), since moment is given by $T = Fd$ where $T$ is the moment of a force acting on a body, $F$ is the force acting of the body, and $d$ is the distance from the axis of rotation.
To find the reaction on the knife edge A.
From $W = {N_A} + {N_B}$
$\Rightarrow {N_B} = W - {N_A}$
Then by substituting into ${N_A}x - {N_B}(d - x) = 0$ , we have
$\Rightarrow {N_A}x - \left( {W - {N_A}} \right)(d - x) = 0$
$\Rightarrow {N_A}x = \left( {W - {N_A}} \right)(d - x)$
By expansion, we have
$\Rightarrow {N_A}x = Wd - Wx - {N_A}d + {N_A}x$
$\Rightarrow {N_A}x$ cancels out, as in
$\Rightarrow 0 = Wd - Wx - {N_A}d$
hence, making ${N_A}$ subject
$\Rightarrow {N_A}d = W(d - x)$
$\Rightarrow {N_A} = \dfrac{{W(d - x)}}{d}$
Hence, the correct answer is option B.

Note
For circumvention of any confusion, the net torque on a body is defined as
$\Rightarrow {T_{NET}} = I\alpha$ , where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. But since the rod was in equilibrium, then $\alpha = 0$ .
Also, when a body is in equilibrium the moment of the force can be taken about any point, (since any point can be said to be the centre of the (non-existent) rotation). However, the point of choice must be the same for all the forces acting on the body.
For example, alternatively, taking the moment about the knife edge B, we have,
$\Rightarrow {N_A}d = W(d - x)$ , hence
$\Rightarrow {N_A} = \dfrac{{W(d - x)}}{d}$