RBD O12

Solution Q12

Solution: Question 12

Diagram: Pseudo Force Action

COM a (Frame) F_pseudo = -ma Line of action
Step 1: Translational Equilibrium

In the inertial frame, the body has an acceleration \( a \). Therefore, the net external force is not zero (\( \Sigma F = ma \)).
(a) The body is NOT in translational equilibrium.

Step 2: Rotational Equilibrium

In the non-inertial frame, the body is in total equilibrium. This means the sum of real torques plus the pseudo-force torque is zero.
The pseudo force \( -m\vec{a} \) acts effectively at the Center of Mass (COM).

Since the pseudo force acts through the COM, its torque about the COM is zero. Consequently, the sum of real external torques about the COM must also be zero. This holds true regardless of the frame.

(b) Net torque about mass centre is a null vector.

Step 3: Torque about Collinear Points

The torque of the pseudo force is zero about any point lying on its line of action (the line passing through COM parallel to acceleration). Thus, the net torque of real forces about any such point is also zero.

(c) Net torque about any point collinear with the line of acceleration is a null vector.

Conclusion: Correct Options are (a), (b), and (c).