What are the two conditions for complete mechanical equilibrium of a rigid body?

A rigid body is in complete mechanical equilibrium when: (1) Translational equilibrium is satisfied (the vector sum of all external forces is zero: sum F_ext = 0), and (2) Rotational equilibrium is satisfied (the vector sum of all external torques about any arbitrary reference point is zero: sum tau_ext = 0).

When do the Centre of Mass (CM) and Centre of Gravity (CG) not coincide?

CM and CG coincide whenever the gravitational field g is uniform throughout the volume of the body. For astronomical or gigantic bodies (like a mountain or tall space elevator) where g varies measurably across the body, CG lies slightly lower than CM.

What is Mechanical Advantage (MA) of a lever?

Mechanical Advantage is the ratio of Load to Effort (MA = Load / Effort = Effort arm / Load arm). If the effort arm is longer than the load arm, MA > 1, meaning a small effort lifts a heavier load.

§ 6.4NCERT Class 11 · Physics · Chapter 6

Equilibrium of Rigid Bodies & Principle of Moments

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 10 min read

A point particle is in equilibrium whenever the net external force is zero. However, for an extended rigid body, zero net force guarantees only that the body does not accelerate translationally; it may still spin vigorously under an unbalanced torque. Complete mechanical stability requires both translational and rotational equilibrium.

1. The Two Conditions of Mechanical Equilibrium

A rigid body is said to be in complete mechanical equilibrium if both its linear momentum and angular momentum are constant in time:

Condition 1: Translational Equilibrium (Zero Net Force)

The vector sum of all external forces acting on the rigid body must be zero:

∑ F_ext = 0 ⇒ ∑ F_x = 0, ∑ F_y = 0, ∑ F_z = 0

Consequence: P_total = constant and the linear acceleration of the centre of mass is zero (A_cm = 0).

Condition 2: Rotational Equilibrium (Zero Net Torque)

The vector sum of all external torques acting on the rigid body about ANY arbitrary reference point must be zero:

∑ τ_ext = 0 ⇒ ∑ τ_x = 0, ∑ τ_y = 0, ∑ τ_z = 0

Consequence: L_total = constant and the angular acceleration is zero (α = 0).

INVARIANCE THEOREM: If a body is in translational equilibrium (∑ F = 0), the net torque is completely independent of the choice of reference origin. Therefore, in solving static equilibrium problems, you may choose the pivot point anywhere — strategically choosing a point through which unknown reaction forces pass eliminates them from the torque equation!

2. Principle of Moments and Levers

A lever is an ideal light rigid rod pivoted at a fixed point called the fulcrum. Two forces act on it: the Effort F₁ applied at effort arm d₁ and the Load F₂ acting at load arm d₂.

Load × Load arm = Effort × Effort arm ⇒ F₂ d₂ = F₁ d₁

Mechanical Advantage (MA)

The efficiency of force amplification is quantified by the Mechanical Advantage:

Mechanical Advantage (MA) = Load / Effort = F₂ / F₁ = d₁ / d₂ = (Effort Arm) / (Load Arm)

MA > 1: Effort arm > Load arm (e.g., crowbar, bottle opener, car jack) — a small effort lifts a large load.
MA < 1: Effort arm < Load arm (e.g., pair of tongs, human forearm lifting weights) — requires greater force but amplifies speed and displacement.
MA = 1: Physical beam balance (equal arms).

3. Centre of Gravity (CG) vs Centre of Mass (CM)

Attribute	Centre of Mass (CM)	Centre of Gravity (CG)
Definition	Point where the entire mass of the body can be assumed to be concentrated (∑ m_i r_i = 0).	Point through which the resultant gravitational force (weight) acts, and where net gravitational torque is zero (∑ τ_g = 0).
Dependency	Depends solely on the spatial distribution of mass.	Depends on mass distribution AND the local gravitational field g.
Uniform g Field	Identical and coincides exactly with CG.	Identical and coincides exactly with CM.
Non-Uniform g Field	Remains at the geometric mass centroid.	Shifts towards the region of stronger gravitational field (lies lower than CM).

4. Visualising Ladder Equilibrium on a Wall

Figure 6.4 — Free-Body Diagram of a Ladder Leaning Against a Smooth Wall

           | Wall (Smooth: friction f_w = 0)
           |
         N_w <---\ (Contact force from wall)
           |      \ Ladder of length L, mass M
           |                  |        \  W = Mg (Acts at midpoint L/2)
           |         \ |
           |          v           |                       |             \ (Angle θ with floor)
       ----+--------------\--------------------> Ground (Rough: friction f_g)
           |               +---> f_g (Pushes right to prevent slip)
           |               |
           |              ^ N_g (Normal reaction from floor)
      

Equilibrium Balance Equations:
1. Vertical: N_g − M g = 0 ⇒ N_g = M g
2. Horizontal: f_g − N_w = 0 ⇒ f_g = N_w
3. Torque about base point: N_w (L sin θ) − M g (L/2 cos θ) = 0 ⇒ N_w = (M g / 2) cot θ
Slip condition: f_g ≤ μ N_g ⇒ (M g / 2) cot θ ≤ μ (M g) ⇒ tan θ ≥ 1 / (2μ).

Step-by-Step Solved Numerical Examples

Solved Example 1: Critical Angle of a Leaning Ladder

Problem: A uniform ladder of length 5 m and mass 20 kg leans against a smooth vertical wall with its base on a rough horizontal floor. The coefficient of static friction between the ladder and the floor is μ = 0.4. What is the minimum angle θ that the ladder can make with the floor without slipping?

Minimum Angle: θ = 51.34° (tan θ ≥ 1.25)

Step 1: Write Force Balance Equations:
Vertical equilibrium: N_floor = M g = (20)(9.8) = 196 N.
Horizontal equilibrium: f_friction = N_wall.

Step 2: Balance Torques About Base on Floor:
Torque due to wall reaction: τ_wall = N_wall (L sin θ) counter-clockwise.
Torque due to weight at midpoint: τ_weight = M g (L/2 cos θ) clockwise.
Equating torques: N_wall L sin θ = M g (L/2) cos θ ⇒ N_wall = (M g / 2) cot θ.

Step 3: Apply Friction Limit Condition:
To prevent slipping, the required horizontal friction cannot exceed maximum static friction:
f ≤ μ N_floor ⇒ (M g / 2) cot θ ≤ μ (M g)
cot θ ≤ 2μ ⇒ tan θ ≥ 1 / (2μ).
Given μ = 0.4: tan θ ≥ 1 / (2 × 0.4) = 1 / 0.8 = 1.25 ⇒ θ ≥ 51.34°.

Solved Example 2: Beam Supported on Two Knife-Edges

Problem: A uniform wooden rod of mass 6 kg and length 2 m rests horizontally on two knife-edges placed at distances of 0.2 m from each end. A mass of 4 kg is suspended at a distance of 0.5 m from the left end. Find the reaction forces R₁ and R₂ exerted by the knife-edges on the rod (take g = 10 m/s²).

R1 = 62.5 N · R2 = 37.5 N

Step 1: Establish Positions:
Let left end of rod be x = 0.
Left knife-edge 1: x = 0.2 m (Reaction R₁ upward).
Right knife-edge 2: x = 1.8 m (Reaction R₂ upward).
Distance between knife-edges: d = 1.8 − 0.2 = 1.6 m.
Suspended load: 4 kg at x = 0.5 m ⇒ distance from knife-edge 1 is (0.5 − 0.2) = 0.3 m.
Weight of uniform rod: W_rod = 6 × 10 = 60 N acts at center x = 1.0 m ⇒ distance from knife-edge 1 is (1.0 − 0.2) = 0.8 m.

Step 2: Translational Equilibrium:
R₁ + R₂ = W_load + W_rod = 40 + 60 = 100 N.

Step 3: Rotational Equilibrium About Knife-Edge 1:
R₂(1.6) − 40(0.3) − 60(0.8) = 0
1.6 R₂ = 12 + 48 = 60 ⇒ R₂ = 60 / 1.6 = 37.5 N.
Hence R₁ = 100 − 37.5 = 62.5 N.

5. Solved NEET & JEE Practice Questions

Q1. A couple produces:

A. Pure linear motion
B. Pure rotational motion
C. Both linear and rotational motion
D. No motion

Answer: B

Explanation: A couple consists of two equal and opposite forces whose lines of action do not coincide. Net force F_net = F − F = 0, so linear acceleration is zero. The net torque is non-zero, producing pure rotational motion.

Q2. For a uniform ladder leaning against a smooth vertical wall and resting on a rough floor with friction coefficient μ, the minimum inclination θ with the floor to prevent slipping is given by:

A. tan θ = μ
B. tan θ = 2μ
C. tan θ = 1 / (2μ)
D. tan θ = 1 / μ

Answer: C

Explanation: Balancing vertical forces (N = Mg), horizontal forces (f = N_w), and torques about the base (N_w L sin θ = Mg (L/2) cos θ) yields N_w = (Mg/2) cot θ. For no slipping, f ≤ μ N ⇒ cot θ ≤ 2μ ⇒ tan θ ≥ 1 / (2μ).

Q3. If the net external force on a rigid body is zero, does it guarantee that the net external torque is also zero?

A. Yes, always
B. No, a couple can exert net torque with zero net force
C. Only if the body is spherical
D. Only if the body is at rest

Answer: B

Explanation: A pair of equal and opposite forces acting along parallel displaced lines (a couple) has zero net force (∑ F = 0) but non-zero net torque (τ = F × d). Hence translational equilibrium does NOT imply rotational equilibrium.

Q4. A meter stick is balanced on a knife edge at its center. When two coins, each of mass 5 g, are put on top of the other at the 12 cm mark, the stick is found to be balanced at the 45 cm mark. What is the mass of the meter stick?

A. 56 g
B. 66 g
C. 76 g
D. 86 g

Answer: B

Explanation: Let mass of meter stick be M. Center of mass of the stick is at x = 50 cm.
Total mass of two coins: m = 5 + 5 = 10 g placed at x = 12 cm.
New fulcrum is at x = 45 cm.
Distance of coins from fulcrum: d_coins = 45 − 12 = 33 cm.
Distance of stick's center of mass from fulcrum: d_stick = 50 − 45 = 5 cm.
By Principle of Moments: m × d_coins = M × d_stick
10 × 33 = M × 5 ⇒ M = 330 / 5 = 66 g.

6. Frequently Asked Questions (FAQs)

Why does choosing a strategic pivot point simplify equilibrium problem-solving?

When a body is in translational equilibrium, the torque equation ∑ τ = 0 holds about any arbitrary point. The torque generated by any force passing directly through the chosen pivot is zero (lever arm r_perp = 0). By selecting a pivot where multiple unknown contact reactions act (such as a hinge or ground contact), those unknown forces drop out of the torque equation, leaving a single equation with one unknown.

Can a body be in rotational equilibrium while accelerating linearly?

Yes. A body in free fall has gravity acting through its CM, so it feels zero torque about its CM (rotational equilibrium) while it accelerates linearly at g.

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