§12.1 — Equilibrium Conditions

A stationary object is satisfying. But when does an object stay put? When all external forces are balanced.

Definition of Equilibrium

An object is in static equilibrium if:

1. Translational equilibrium: the sum of all external forces is zero
2. Rotational equilibrium: the sum of all external torques is zero

Mathematically:

\[ \sum \vec F = 0 \]

\[ \sum \vec \tau = 0 \]

First Condition: Force Balance

If the sum of forces is not zero, the object accelerates (from Newton's second law):

\[ \sum \vec F = m\vec a \]

If we want \( \vec a = 0 \), then:

\[ \sum \vec F = 0 \]

Second Condition: Torque Balance

If the sum of torques is not zero, the object rotates:

\[ \sum \vec \tau = I\vec \alpha \]

If we want \( \vec \alpha = 0 \), then:

\[ \sum \vec \tau = 0 \]

Important: torque depends on the choice of axis. But if the sum of torques about one axis is zero, and all other forces are balanced, then the sum of torques about any other axis is also zero.

Simple Example: A Book on a Table

A book of weight W = mg = 10 N sits on a horizontal table.

Forces acting:

• Gravitational force: \( W \) (downward)
• Normal force from table: \( N \) (upward)

First condition (forces): \[ \sum F_y = N - W = 0 \] \[ N = W = 10 \text{ N} \]

Second condition (torques):

Both forces pass through the center of mass of the book (if the table presses uniformly). So their torques:

\[ \tau_W = 0 \quad (\text{passes through axis}) \] \[ \tau_N = 0 \]

Therefore the book is in equilibrium.

More Complex Example: Wires and Weight

Two wires hang from the ceiling. A weight of W = 100 N hangs from the wires. Both wires are vertical.

Forces acting (on the weight):

• Weight: \( W = 100 \) N (downward)
• Tension in left wire: \( T_L \) (upward)
• Tension in right wire: \( T_R \) (upward)

First condition: \[ T_L + T_R - W = 0 \] \[ T_L + T_R = 100 \text{ N} \]

If the wires are identical: \[ T_L = T_R = 50 \text{ N} \]

Second condition:

If we choose the center of the weight as our pivot, all three forces pass through it:

\[ \sum \tau = 0 \]

(Torque is zero!)

Independence of Axis Choice

Key principle: If \( \sum \vec F = 0 \) and you find \( \sum \vec \tau = 0 \) about one axis, this holds for all other axes too.

Proof: Suppose \( \sum \vec F = 0 \). About two different axes A and B:

\[ \sum \tau_A = \sum \vec r_A \times \vec F \] \[ \sum \tau_B = \sum \vec r_B \times \vec F \]

If \( \sum \vec F = 0 \): \[ \sum \tau_B = \sum (\vec r_B - \vec r_A + \vec r_A) \times \vec F \] \[ = \sum (\vec r_B - \vec r_A) \times \vec F + \sum \vec r_A \times \vec F \] \[ = (\vec r_B - \vec r_A) \times \sum \vec F + \sum \tau_A \] \[ = 0 + \sum \tau_A \]

Therefore: if \( \sum F = 0 \), then \( \sum \tau_A = 0 \) if and only if \( \sum \tau_B = 0 \) for any axis B.

Problem-Solving Strategy

To solve an equilibrium problem:

1. Identify forces: list all external forces
2. Components: resolve forces into x and y components
3. Equations: write \( \sum F_x = 0 \) and \( \sum F_y = 0 \)
4. Choose axis: usually pick the point of unknown force so its torque is zero
5. Torque equation: \( \sum \tau = 0 \)
6. Solve: use equations to find unknowns

Common Mistakes

• Forgetting the axis: when using \( \sum \tau = 0 \), specify the rotation axis
• Ignoring lever arms: measure all torque moment arms from a consistent reference point
• Sign confusion: be consistent: up is +, down is −; counterclockwise is +, clockwise is −

What You Should Know

• Two conditions of equilibrium: \( \sum \vec F = 0 \) and \( \sum \vec \tau = 0 \)
• Equilibrium is both translational and rotational
• Choice of axis is arbitrary (if \( \sum \vec F = 0 \) and one axis works, all work)
• Strategy: identify, resolve, write equations, solve

Preview of §12.2

Now we know when an object is in equilibrium. But where does this equilibrium act? The answer: at the center of mass (or center of gravity). Let's see where this special point is and why it matters.

📚 See also: Halliday Vol 1, Ch 12, §12.1 — Equilibrium conditions. 🔗 Reference: §4.2 (Newton's second law), §10.6 (Torque) — foundation for this section.

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