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Rotational Equilibrium
and Newton's First Law

Explore how torques balance to keep objects from spinning, and learn the rotational form of Newton's First Law: \(\sum \vec{\tau} = 0\).

AP Physics 1

What is Torque?

Torque is the rotational equivalent of force. While a force causes linear acceleration, a torque causes angular acceleration. Torque depends on three factors: the magnitude of the force, the distance from the pivot, and the angle at which the force is applied.

Visualizing Torque

Pivot \(r\) \(F\) \(\theta\) CW
\(\tau = rF\sin\theta\)

Sign Convention: CW vs CCW

+ Positive
Counterclockwise (CCW) torques are conventionally taken as positive. Think of turning a bolt to loosen it.
− Negative
Clockwise (CW) torques are conventionally taken as negative. Think of tightening a bolt.

Newton's First Law: Rotational Form

Just as an object at rest stays at rest (or in uniform motion) unless acted upon by a net force, a rotating object maintains its angular velocity unless acted upon by a net external torque.

The Equilibrium Condition

\(F_1\) \(r_1\) \(F_2\) \(r_2\)
\(\sum\tau = 0 \;\Rightarrow\; r_1 F_1 = r_2 F_2\)

Translational vs Rotational Equilibrium

Translational Rotational
Condition \(\sum \vec{F} = 0\) \(\sum \vec{\tau} = 0\)
Result No linear acceleration No angular acceleration
Can still move? Yes: constant velocity Yes: constant \(\omega\)
Quantity balanced Forces Torques
Static case Object at rest Object not rotating

Why Does This Matter?

Engineering
Bridges, cranes, and buildings must be in rotational equilibrium to avoid collapse.
Biology
Your joints act as pivot points: muscles provide torques to hold your arm steady.
Everyday Life
Seesaws, wrenches, doors: all rely on balancing or creating torques around a pivot.

How to Solve Equilibrium Problems

1
Draw a free-body diagram. Identify all forces and their points of application.
2
Choose a pivot point. Any point works, but choosing where an unknown force acts eliminates that variable.
3
Calculate each torque. Use \(\tau = rF\sin\theta\). Assign + for CCW, − for CW.
4
Set \(\sum\tau = 0\). Sum all torques and solve for the unknown.
5
Check with \(\sum F = 0\). Use force balance to verify or find remaining unknowns.

Example 1: Balanced Beam

A uniform beam of length \(4\,\text{m}\) and weight \(W = 200\,\text{N}\) rests on a fulcrum at its center. A \(100\,\text{N}\) weight is placed \(1.5\,\text{m}\) from the left end. If a downward force of \(50\,\text{N}\) is available, where must it be placed on the right side to balance the beam?

0 m 2 m 4 m \(100\,\text{N}\) 1.5 m \(50\,\text{N}\) \(d = ?\) W acts at center

Example 1: Solution

1
Pivot at the center. The beam's own weight creates no torque.
2
CCW Torque: The \(100\,\text{N}\) weight is \(0.5\,\text{m}\) left of the pivot, creating a counter-clockwise torque: \(\tau_{CCW} = 100\,\text{N} \times 0.5\,\text{m} = 50\,\text{N}\cdot\text{m}\).
3
CW Torque: The \(50\,\text{N}\) force is placed at an unknown distance \(d\) to the right. This creates a clockwise torque: \(\tau_{CW} = 50\,\text{N} \times d\).
4
Solve for \(d\): Set \(\sum\tau = 0 \Rightarrow \tau_{CCW} = \tau_{CW}\). So, \(50\,\text{N}\cdot\text{m} = 50\,\text{N} \times d\). This gives \(d = 1\,\text{m}\). The force must be placed \(1\,\text{m}\) to the right of the pivot.

Example 2: Sign on a Rod

A horizontal rod of length \(2\,\text{m}\) (negligible mass) is attached to a wall by a hinge. A wire at \(30^\circ\) above the rod supports the far end. A \(50\,\text{N}\) sign hangs from the end. Find the tension in the wire.

Hinge 2 m \(T\) \(30^\circ\) \(50\,\text{N}\)

Example 2: Solution

1
Choose pivot at hinge. This eliminates the unknown hinge force from the torque equation.
2
CW torque from sign: \(\tau_{\text{sign}} = rF = (2\,\text{m})(50\,\text{N}) = 100\,\text{N}\cdot\text{m}\).
3
CCW torque from tension: \(\tau_T = rF\sin\theta = (2\,\text{m})(T)(\sin 30^\circ) = T \cdot (1\,\text{m})\).
4
Set \(\sum\tau = 0\): \(\tau_T = \tau_{\text{sign}}\), so \(T \cdot (1\,\text{m}) = 100\,\text{N}\cdot\text{m}\). This gives \(T = \) \(100\,\text{N}\).

Example 3: Ladder Problem

A \(5\,\text{m}\) ladder of weight \(W = 120\,\text{N}\) leans against a frictionless wall at \(60^\circ\) above the ground. Find the normal force the wall exerts on the ladder.

5 m \(60^\circ\) \(W = 120\,\text{N}\) \(N_w\) \(f\) \(N_g\)

Ladder Solution: \(N_w\)

1
Pivot at the base (eliminates friction and ground normal from the torque equation).
2
CW torque from weight: acts at center of ladder (\(2.5\,\text{m}\) along). Horizontal distance from pivot: \(2.5\cos 60^\circ = 1.25\,\text{m}\). \(\tau_W = 120 \times 1.25 = 150\,\text{N}\cdot\text{m}\).
3
CCW torque from wall normal: acts at top. Vertical distance from pivot: \(5\sin 60^\circ \approx 4.33\,\text{m}\). \(\tau_{N_w} = N_w \times 4.33\).
4
Solve: \(N_w \times 4.33 = 150\), so \(N_w \approx\) \(34.6\,\text{N}\).

Common Mistakes to Avoid

✕ Forgetting the angle
Using \(\tau = rF\) instead of \(\tau = rF\sin\theta\). Only perpendicular components create torque.
✕ Wrong pivot choice
Any pivot works, but a poor choice leads to more unknowns. Choose the pivot where an unknown force acts.
✕ Sign errors
Mixing up CW (−) and CCW (+), or forgetting to assign consistent signs to all torques.

Question 1 of 5

A seesaw is balanced with a \(40\,\text{kg}\) child sitting \(2\,\text{m}\) from the pivot. Where must a \(60\,\text{kg}\) child sit to maintain rotational equilibrium?

Question 2 of 5

A force of \(20\,\text{N}\) is applied at the end of a \(0.5\,\text{m}\) wrench at an angle of \(90^\circ\) to the handle. What is the torque?

Question 3 of 5

A fan rotates at a constant angular velocity. Which statement is true about the net torque on the fan?

Question 4 of 5

A \(3\,\text{m}\) massless beam is supported by a pivot at its left end. A \(60\,\text{N}\) downward force acts at \(1\,\text{m}\) from the pivot, and an upward force \(F\) acts at the right end (\(3\,\text{m}\) from the pivot). What value of \(F\) keeps the beam in rotational equilibrium?

Question 5 of 5

When solving a torque problem, choosing the pivot at the point where an unknown force acts is advantageous because: