Equilibrium of a Rigid Body: Force and Torque Conditions
Physics · System Of Particles And Rotational Motion · NEET
A rigid body is in equilibrium only when TWO conditions are both true at the same time: the net external force on it is zero (no linear acceleration) AND the net external torque about any point is zero (no angular acceleration). Force zero alone is not enough. Memory hook: "Two locks, one door" - both the force lock (ΣF = 0) and the torque lock (Στ = 0) must click before the body truly stays still.
Left: a rigid rod in true equilibrium needs both ΣF = 0 and Στ = 0. Right: a couple has zero net force yet a non-zero net torque, so it spins the body and is never in equilibrium.
Your doubts, answered
If the net force on a body is zero, is it always in equilibrium?
No. Zero net force gives only translational equilibrium (the centre of mass does not accelerate). The body can still spin up because of an unbalanced torque. Full mechanical equilibrium of a rigid body needs BOTH ΣF = 0 and Στ = 0. Example from NCERT: two equal and opposite forces applied at the two ends of a rod, both pointing the same way, give ΣF = 0 but a non-zero net torque, so the rod turns. That is a couple.
What is the difference between translational and rotational equilibrium?
Translational equilibrium means ΣF = 0, so the centre of mass stays at rest or moves with constant velocity. Rotational equilibrium means Στ = 0, so the body does not gain any angular acceleration. A body can have one without the other. A rigid body is in MECHANICAL (complete) equilibrium only when both hold together.
About which point should I take torque when solving an equilibrium problem?
When a body is truly in equilibrium, Στ = 0 about EVERY point, so you may pick any point you like. Smart choice: take torques about the point where an unknown force acts (a hinge, pivot, or the floor contact). That force has zero moment arm there, so it drops out of the equation and you solve for the remaining unknown directly. This is exactly why the NEET rod-on-wedge and leaning-rod problems take torque about the pivot or the base.
Can a couple ever hold a body in equilibrium?
No. A couple is two equal, opposite, parallel forces with different lines of action. Its net force is zero, but its net torque is NOT zero. So a couple gives pure rotation - it can never produce equilibrium. To balance a couple you need an equal and opposite couple, not a single force.
Why do NEET problems say 'take g downward at the centre of mass'?
For calculating torque due to gravity you treat the whole weight mg as acting at the centre of gravity, which coincides with the centre of mass in a uniform gravitational field. For a uniform rod this is the midpoint. So the weight of a uniform rod contributes a torque mg times the horizontal distance from the pivot to the rod's midpoint.
⚠️ The NEET trap ✗ The body has zero net force, so it must be in equilibrium. ✓ Equilibrium of a rigid body needs BOTH ΣF = 0 and Στ = 0. A couple has ΣF = 0 yet still spins the body. Check torque separately every time. 🧠 NTA loves the couple. Net force zero + still rotating = not in equilibrium.
Real NEET questions
2017
Which of the following statements are correct?
(a) The centre of mass of a body always coincides with its centre of gravity.
(b) The centre of mass is the point where the total gravitational torque on the body is zero.
(c) A couple on a body produces both translational and rotational motion.
(d) Mechanical advantage greater than one means a small effort can lift a large load.
A · A. (b) and (d) ✓
B · B. (a) and (b)
C · C. (b) and (c)
D · D. (c) and (d)
Solution: Test each statement against the equilibrium and couple rules.
(a) FALSE: centre of mass and centre of gravity coincide only in a uniform gravitational field, not always.
(b) TRUE: the centre of gravity is defined as the point about which the total gravitational torque is zero (the weight acts through it).
(c) FALSE: a couple has zero net force, so it produces ONLY rotation, never translation.
(d) TRUE: mechanical advantage = load/effort; MA greater than 1 means a small effort lifts a large load.
Correct statements: (b) and (d) -> Option A.
2021
A uniform rod (200 cm, 500 g) is balanced on a wedge at the 40 cm mark. A 2 kg mass hangs at the 20 cm mark and an unknown mass m hangs at the 160 cm mark. For equilibrium, m is (g = 10 m/s^2):
A · A. 1/6 kg
B · B. 1/12 kg ✓
C · C. 1/2 kg
D · D. 1/3 kg
Solution: Use rotational equilibrium: take torques about the wedge (pivot) at the 40 cm mark, so the normal reaction has zero moment arm and drops out.
Step 1 - locate each weight relative to the pivot (in metres):
- 2 kg at 20 cm -> distance = 40 - 20 = 20 cm = 0.20 m, on the LEFT of pivot -> anticlockwise.
- Rod weight 0.5 kg acts at its centre = 100 cm mark -> distance = 100 - 40 = 60 cm = 0.60 m, on the RIGHT -> clockwise.
- Mass m at 160 cm -> distance = 160 - 40 = 120 cm = 1.20 m, on the RIGHT -> clockwise.
Step 2 - Στ = 0 (g cancels on both sides):
Anticlockwise = Clockwise
2 x 0.20 = 0.5 x 0.60 + m x 1.20
0.40 = 0.30 + 1.20 m
Step 3 - solve: 1.20 m = 0.10 -> m = 1/12 kg. Answer B.
2025
A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall, making 60 degrees with the wall; its lower end rests on a rough horizontal floor. The friction force exerted by the floor is (g = 10 m/s^2):
A · A. 200 N
B · B. 200√3 N
C · C. 100 N
D · D. 100√3 N ✓
Solution: Apply BOTH equilibrium conditions.
Forces on the rod: weight mg = 20 x 10 = 200 N down at the midpoint; wall is SMOOTH so its reaction N_w is horizontal; floor gives a vertical normal N_f and a horizontal friction f.
Step 1 - vertical force balance (ΣF_y = 0): N_f = mg = 200 N.
Step 2 - horizontal force balance (ΣF_x = 0): f = N_w (friction balances the wall push).
Step 3 - torque about the base (floor contact) so N_f and f drop out. The rod makes 60 degrees with the wall, so it makes 30 degrees with... take angle with the wall = 60 degrees.
Moment of N_w = N_w x (L cos60) [vertical height of top end].
Moment of weight = mg x (horizontal distance to midpoint) = mg x (L sin60)/2.
Set equal: N_w x L cos60 = mg x (L sin60)/2
N_w = mg x tan60 / 2 = 200 x √3 / 2 = 100√3 N.
Step 4 - since f = N_w, friction f = 100√3 N. Answer D.
Solved System Of Particles And Rotational Motion NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What are the two conditions for equilibrium of a rigid body?
(1) Net external force zero: ΣF = 0 (translational equilibrium). (2) Net external torque zero about any point: Στ = 0 (rotational equilibrium). Both must hold at the same time for complete mechanical equilibrium.
How many equations do I need for a coplanar (2D) problem?
Three scalar equations: ΣF_x = 0, ΣF_y = 0, and Στ = 0. NCERT states that for coplanar forces these three conditions are enough for mechanical equilibrium.
What is partial equilibrium?
When only one condition holds. If only ΣF = 0 the body has translational equilibrium but may rotate (a couple). If only Στ = 0 the body may still translate. Complete equilibrium needs both.
What is a couple and why can't it give equilibrium?
A couple is a pair of equal, opposite, parallel forces with different lines of action. Its net force is zero but its net torque is not, so it produces pure rotation and can never keep a body in equilibrium.
Does the centre of gravity always match the centre of mass?
Only in a uniform gravitational field. In everyday NEET problems (small bodies near Earth) they coincide, so you take the weight acting at the centre of mass when computing torque.
