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System of Particles and Rotational Motion - Equilibrium of a Rigid Body

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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

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Translational Equilibrium: The vector sum of all external forces acting on the rigid body must be zero: ∑F⃗ext=0\sum \vec{F}_{ext} = 0. This implies a⃗cm=0\vec{a}_{cm} = 0.

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Rotational Equilibrium: The vector sum of all external torques acting on the rigid body about any point must be zero: ∑τ⃗ext=0\sum \vec{\tau}_{ext} = 0. This implies α=0\alpha = 0.

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A body may be in partial equilibrium: it can be in translational equilibrium but not rotational equilibrium (e.g., a 'couple' acting on a body), or vice versa.

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The Principle of Moments: For a body in rotational equilibrium, the sum of clockwise moments about any pivot is equal to the sum of anticlockwise moments about that same pivot.

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Center of Gravity (CGCG): It is the point where the total gravitational torque on the body is zero. For objects in a uniform gravitational field, the CGCG coincides with the Center of Mass (CMCM).

📐Formulae

∑F⃗i=0  ⟹  ∑Fx=0,∑Fy=0,∑Fz=0\sum \vec{F}_i = 0 \implies \sum F_x = 0, \sum F_y = 0, \sum F_z = 0

∑τ⃗i=0  ⟹  ∑r⃗i×F⃗i=0\sum \vec{\tau}_i = 0 \implies \sum \vec{r}_i \times \vec{F}_i = 0

τ=rFsin⁡θ\tau = r F \sin \theta

Mechanical Advantage (MA)=LoadEffort=l1l2\text{Mechanical Advantage (MA)} = \frac{\text{Load}}{\text{Effort}} = \frac{l_1}{l_2}

τ⃗net=τ⃗1+τ⃗2+...+τ⃗n=0\vec{\tau}_{net} = \vec{\tau}_1 + \vec{\tau}_2 + ... + \vec{\tau}_n = 0

💡Examples

Problem 1:

A uniform horizontal beam of length L=4 mL = 4\text{ m} and mass M=20 kgM = 20\text{ kg} is supported at its two ends by vertical pillars. A man of mass m=60 kgm = 60\text{ kg} stands at a distance of 1 m1\text{ m} from the left end. Find the reaction forces R1R_1 (left support) and R2R_2 (right support). (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

  1. For translational equilibrium: ∑Fy=0  ⟹  R1+R2−Mg−mg=0\sum F_y = 0 \implies R_1 + R_2 - Mg - mg = 0 R1+R2=(20+60)×10=800 NR_1 + R_2 = (20 + 60) \times 10 = 800\text{ N}.

  2. For rotational equilibrium, taking torque about the left end (pivot at R1R_1): ∑τ=0  ⟹  (mg×1 m)+(Mg×2 m)−(R2×4 m)=0\sum \tau = 0 \implies (mg \times 1\text{ m}) + (Mg \times 2\text{ m}) - (R_2 \times 4\text{ m}) = 0 (60×10×1)+(20×10×2)=4R2(60 \times 10 \times 1) + (20 \times 10 \times 2) = 4 R_2 600+400=4R2  ⟹  1000=4R2  ⟹  R2=250 N600 + 400 = 4 R_2 \implies 1000 = 4 R_2 \implies R_2 = 250\text{ N}.

  3. Substitute R2R_2 into the first equation: R1+250=800  ⟹  R1=550 NR_1 + 250 = 800 \implies R_1 = 550\text{ N}.

Explanation:

The problem uses both conditions of equilibrium. The weight of the uniform beam acts at its center (2 m2\text{ m} from ends). By setting the net torque about the left end to zero, we isolate and solve for the reaction force at the right end.

Equilibrium of a Rigid Body - Revision Notes & Key Formulas | CBSE Class 11 Physics