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Motion of System of Particles and Rigid Body - Equilibrium of Rigid Bodies

Grade 11ICSEPhysics

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

🔑Concepts

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A rigid body is in Mechanical Equilibrium if its linear momentum and angular momentum remain constant with time. This requires both translational and rotational equilibrium.

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Translational Equilibrium: A body is in translational equilibrium if the vector sum of all external forces acting on it is zero, ∑F⃗ext=0\sum \vec{F}_{ext} = 0. This implies the center of mass moves with constant velocity or is at rest.

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Rotational Equilibrium: A body is in rotational equilibrium if the vector sum of all external torques acting on it about any arbitrary axis is zero, ∑τ⃗ext=0\sum \vec{\tau}_{ext} = 0. This implies the angular acceleration α\alpha is zero.

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Principle of Moments: For a body in rotational equilibrium about a fixed axis, the sum of the anti-clockwise moments of the forces is equal to the sum of the clockwise moments of the forces about the same axis.

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Couple: Two equal and opposite forces whose lines of action do not coincide form a couple. A couple produces only rotational motion without any translational motion. The moment of a couple is independent of the choice of the pivot point.

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Center of Gravity (CGCG): The point at which the total weight of the body acts. For a rigid body in a uniform gravitational field, the CGCG coincides with the Center of Mass (CMCM).

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Partial Equilibrium: A body may be in translational equilibrium but not rotational equilibrium (e.g., a couple), or vice versa (e.g., a rod rotating about its center while the center accelerates).

📐Formulae

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

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

τ⃗=r⃗×F⃗=rFsin⁡θn^\vec{\tau} = \vec{r} \times \vec{F} = r F \sin \theta \hat{n}

Moment of Couple=Force×Perpendicular distance between forces\text{Moment of Couple} = \text{Force} \times \text{Perpendicular distance between forces}

Total Torque=∑iτ⃗i=∑i(r⃗i×mig⃗)=r⃗cg×Mg⃗\text{Total Torque} = \sum_{i} \vec{\tau}_i = \sum_{i} (\vec{r}_i \times m_i \vec{g}) = \vec{r}_{cg} \times M \vec{g}

💡Examples

Problem 1:

A uniform rod of length L=1.0 mL = 1.0\text{ m} and mass M=2.0 kgM = 2.0\text{ kg} is supported on two knife-edges placed at 0.2 m0.2\text{ m} and 0.8 m0.8\text{ m} from the left end. A load of 4.0 kg4.0\text{ kg} is placed at 0.3 m0.3\text{ m} from the left end. Calculate the normal reactions R1R_1 and R2R_2 at the knife-edges. (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

  1. Translational Equilibrium: Sum of upward forces = Sum of downward forces. R1+R2=(Mrod+Mload)g=(2.0+4.0)×10=60 NR_1 + R_2 = (M_{rod} + M_{load})g = (2.0 + 4.0) \times 10 = 60\text{ N}.
  2. Rotational Equilibrium: Take moments about the first knife-edge (0.2 m0.2\text{ m} mark). Distance of load from R1=0.3−0.2=0.1 mR_1 = 0.3 - 0.2 = 0.1\text{ m}. Distance of CGCG of rod from R1=0.5−0.2=0.3 mR_1 = 0.5 - 0.2 = 0.3\text{ m}. Distance of R2R_2 from R1=0.8−0.2=0.6 mR_1 = 0.8 - 0.2 = 0.6\text{ m}. Sum of clockwise moments = Sum of anti-clockwise moments: (40 N×0.1 m)+(20 N×0.3 m)=R2×0.6 m(40\text{ N} \times 0.1\text{ m}) + (20\text{ N} \times 0.3\text{ m}) = R_2 \times 0.6\text{ m} 4+6=0.6R2  ⟹  10=0.6R2  ⟹  R2=100.6=16.67 N4 + 6 = 0.6 R_2 \implies 10 = 0.6 R_2 \implies R_2 = \frac{10}{0.6} = 16.67\text{ N}.
  3. Finding R1R_1: R1=60−16.67=43.33 NR_1 = 60 - 16.67 = 43.33\text{ N}.

Explanation:

We applied the conditions for both translational equilibrium (vertical forces must cancel) and rotational equilibrium (net torque about the first knife-edge must be zero).

Problem 2:

A uniform ladder of weight W=200 NW = 200\text{ N} leans against a smooth vertical wall, making an angle of 60∘60^\circ with the horizontal floor. Find the reaction forces from the wall and the floor.

Solution:

Let length of ladder be 2L2L. CGCG is at LL from bottom.

  1. Forces: Horizontal reaction from wall NwN_w, Vertical reaction from floor RvR_v, Horizontal friction from floor fsf_s, Weight WW acting at center.
  2. Translational Equilibrium: Horizontal: fs=Nwf_s = N_w Vertical: Rv=W=200 NR_v = W = 200\text{ N}
  3. Rotational Equilibrium (Torque about floor contact point AA): W×Lcos⁡60∘−Nw×2Lsin⁡60∘=0W \times L \cos 60^\circ - N_w \times 2L \sin 60^\circ = 0 200×L×0.5=Nw×2L×32200 \times L \times 0.5 = N_w \times 2L \times \frac{\sqrt{3}}{2} 100=Nw×3  ⟹  Nw=1003≈57.74 N100 = N_w \times \sqrt{3} \implies N_w = \frac{100}{\sqrt{3}} \approx 57.74\text{ N}. Reaction from floor Rfloor=Rv2+fs2=2002+57.742≈208.17 NR_{floor} = \sqrt{R_v^2 + f_s^2} = \sqrt{200^2 + 57.74^2} \approx 208.17\text{ N}.

Explanation:

Because the wall is smooth, it only exerts a normal reaction force. The floor provides both a normal reaction and a frictional force to maintain equilibrium.

Equilibrium of Rigid Bodies - Revision Notes & Key Formulas | ICSE Class 11 Physics