Equilibrium — Revision Notes

⚡ 30-Second Revision

Equilibrium Conditions: — \n * Translational: $\Sigma \vec{F} = 0$ (Net force is zero) \n * Rotational: $\Sigma \vec{\tau} = 0$ (Net torque is zero) \n- Types of Equilibrium: \n * Static: $v=0, \omega=0$ (at rest) \n * Dynamic: $v=\text{constant}, \omega=\text{constant}$ (moving without acceleration) \n- Stability (for Static): \n * Stable: CG at minimum, potential energy increases on displacement. \n * Unstable: CG at maximum, potential energy decreases on displacement. \n * Neutral: CG height unchanged, potential energy unchanged on displacement. \n- Torque Formula: $\tau = rF\sin\theta$

2-Minute Revision

Equilibrium is a state of balance where an object's motion (linear and rotational) remains constant. This means there's no acceleration. The two fundamental conditions for equilibrium are: (1) the net external force acting on the object must be zero (translational equilibrium), and (2) the net external torque acting on the object about any point must be zero (rotational equilibrium).

\n\nEquilibrium can be static, where the object is completely at rest, or dynamic, where it moves with a constant velocity (and/or rotates with constant angular velocity). For static equilibrium, we further classify stability: stable (returns to original position, CG at minimum potential energy), unstable (moves further away, CG at maximum potential energy), and neutral (finds new equilibrium, CG height unchanged).

Solving problems involves drawing accurate free-body diagrams, resolving forces, choosing a strategic pivot point for torque calculations, and setting up simultaneous equations from the equilibrium conditions.

5-Minute Revision

Equilibrium is a critical concept in mechanics, signifying a state where an object's linear and angular velocities are constant, implying zero linear and angular acceleration. This state is governed by two primary conditions: \n1.

Translational Equilibrium: The vector sum of all external forces acting on the object is zero ($\Sigma \vec{F} = 0$). This implies $\Sigma F_x = 0$, $\Sigma F_y = 0$, and $\Sigma F_z = 0$. \n2. Rotational Equilibrium: The vector sum of all external torques acting on the object about any arbitrary pivot point is zero ($\Sigma \vec{\tau} = 0$).

\n\nTypes of Equilibrium: \n* Static Equilibrium: The object is at rest ($v=0, \omega=0$). Example: A bridge. \n* Dynamic Equilibrium: The object is moving with constant velocity ($v \neq 0, \omega \neq 0$ or $v \neq 0, \omega = 0$ or $v = 0, \omega \neq 0$).

Example: A car at constant speed. \n\nStability of Static Equilibrium: \n* Stable: Object returns to original position after small displacement. Center of gravity (CG) is at a minimum height, so potential energy increases upon displacement.

\n* Unstable: Object moves further away after small displacement. CG is at a maximum height, so potential energy decreases upon displacement. \n* Neutral: Object remains in new position after small displacement.

CG height remains constant, so potential energy is unchanged. \n\nProblem-Solving Steps: \n1. Free-Body Diagram (FBD): Draw all forces (weight, normal, tension, friction) with correct directions and points of application.

\n2. Coordinate System & Resolution: Choose axes. Resolve forces into components. \n3. Translational Conditions: Set $\Sigma F_x = 0$ and $\Sigma F_y = 0$. \n4. Pivot Point & Rotational Condition: Choose a pivot (often where an unknown force acts).

Calculate torques ($\tau = rF\sin\theta$). Set $\Sigma \tau = 0$. \n5. Solve Equations: Solve the system of equations for unknowns. \n\nExample: A uniform rod of mass $M$ and length $L$ is supported at its ends.

A mass $m$ is placed at $L/4$ from the left end. Find reactions $R_1$ (left) and $R_2$ (right). \n* $\Sigma F_y = 0 \implies R_1 + R_2 - Mg - mg = 0$ \n* Pivot at left end: $\Sigma \tau = 0 \implies R_2 L - Mg(L/2) - mg(L/4) = 0$ \n* $R_2 = Mg/2 + mg/4$.

Then $R_1 = (M+m)g - R_2$.

Prelims Revision Notes

Equilibrium is defined by two conditions: zero net force (translational equilibrium, $\Sigma \vec{F} = 0$) and zero net torque (rotational equilibrium, $\Sigma \vec{\tau} = 0$). These conditions ensure that an object's linear and angular velocities remain constant, meaning zero linear and angular acceleration.

\n\nTranslational Equilibrium: \n* Implies $a=0$. \n* In 2D, this means $\Sigma F_x = 0$ and $\Sigma F_y = 0$. \n* Forces to consider: Weight ($mg$), Normal force ($N$), Tension ($T$), Friction ($f_s$ or $f_k$).

\n\nRotational Equilibrium: \n* Implies $\alpha=0$. \n* Torque ($\tau$) is the turning effect of a force: $\tau = rF\sin\theta$. \n* $r$ is the moment arm (perpendicular distance from pivot to line of action of force).

\n* Choose pivot strategically to eliminate unknown forces from torque equation. \n\nTypes of Equilibrium: \n* Static Equilibrium: Object is at rest ($v=0, \omega=0$). Example: Book on a table.

\n* Dynamic Equilibrium: Object is moving with constant velocity ($v=\text{constant}, \omega=\text{constant}$). Example: Satellite in orbit. \n\n**Stability of Static Equilibrium (related to potential energy $U_p$ and center of gravity CG):** \n* Stable: $U_p$ is minimum, CG is lowest.

Displacing it increases $U_p$. Object returns. (e.g., cone on base) \n* Unstable: $U_p$ is maximum, CG is highest. Displacing it decreases $U_p$. Object moves away. (e.g., cone on tip) \n* Neutral: $U_p$ is constant, CG height unchanged.

Displacing it finds new equilibrium. (e.g., sphere on flat surface) \n\nKey Problem-Solving Steps: \n1. Draw a clear Free-Body Diagram (FBD). \n2. Resolve forces into components along chosen axes.

\n3. Apply $\Sigma F_x = 0$ and $\Sigma F_y = 0$. \n4. Choose a pivot point and apply $\Sigma \tau = 0$. \n5. Solve the resulting system of equations.

Vyyuha Quick Recall

For Total Equilibrium, Forces Totally Equal Zero. (F for Force, T for Torque, E for Equilibrium, Z for Zero). \n\nOr, for types of stability: Stable = Sink (CG low), Unstable = Up (CG high), Neutral = No change (CG same height).