🔧 Torque — The Rotational Cousin of Force

Part 1 of 7 — Torque: $\tau = rF\sin\theta$

You know that forces cause objects to accelerate in a straight line. But what causes objects to rotate? The answer is torque — a quantity that measures how effectively a force can cause rotation about a pivot point.

Think about opening a door: pushing near the hinge barely moves it, but pushing at the handle swings it easily. Same force, different torque.

Defining Torque

Torque ($\tau$, the Greek letter "tau") is defined as:

$\tau = rF\sin\theta$

Where:

• $r$ = distance from the axis of rotation (pivot) to where the force is applied
• $F$ = magnitude of the applied force
• $\theta$ = angle between the force vector and the position vector (from pivot to point of application)

Units

$[\tau] = \text{m} \cdot \text{N} = \text{N·m}$

Important: N·m for torque is NOT the same as a joule (J), even though $1\,\text{J} = 1\,\text{N·m}$. Torque and energy are fundamentally different quantities.

Maximum Torque

Torque is maximum when $\theta = 90°$ (force is perpendicular to the lever arm):

$\tau_{\text{max}} = rF$

Torque is zero when $\theta = 0°$ or $180°$ (force is parallel to the position vector — pushing directly toward or away from the pivot).

Sign Convention for Torque

By convention:

This follows the right-hand rule: curl the fingers of your right hand in the direction of rotation — your thumb points in the direction of the torque vector (out of the page for CCW, into the page for CW).

Example

A 50 N force is applied perpendicularly at a distance of 0.3 m from the pivot, causing clockwise rotation.

$\tau = -(0.3)(50)\sin 90° = -15 \text{ N·m}$

The negative sign indicates clockwise rotation.

Torque Concept Check 🎯

Torque Calculation Drill 🧮

1. A 40 N force is applied perpendicularly ($\theta = 90°$) at a distance of 0.25 m from the pivot. What is the magnitude of the torque? (in N·m)

2. A 100 N force is applied at $\theta = 30°$ at a distance of 0.5 m from the pivot. What is the magnitude of the torque? (in N·m)

3. You want to produce a torque of 60 N·m by applying a perpendicular force at a distance of 0.4 m from the pivot. What force is needed? (in N)

Torque Fundamentals Review 🔍

Exit Quiz — Torque Basics ✅

📏 Lever Arm and Moment Arm

Part 2 of 7 — The Lever Arm Approach

In Part 1 we computed torque as $\tau = rF\sin\theta$. There is an equivalent — and often more intuitive — way to think about torque using the concept of the lever arm (also called the moment arm).

What Is the Lever Arm?

The lever arm (moment arm) $r_{\perp}$ is the perpendicular distance from the axis of rotation to the line of action of the force.

$r_{\perp} = r\sin\theta$

So the torque formula becomes:

$\tau = r_{\perp} \cdot F = (r\sin\theta) \cdot F$

This is identical to $\tau = rF\sin\theta$ — just rearranged!

Line of Action

The line of action of a force is the infinite line drawn along the direction of the force through its point of application.

Why the Lever Arm Matters

The lever arm approach is especially useful when:

• Forces act at odd angles
• You can easily identify the perpendicular distance from a diagram
• Multiple forces act on the same object

Example

A horizontal beam of length 2 m is hinged at the wall. A cable attached to its end pulls upward at 60° above the beam with a force of 100 N.

• $r = 2$ m, $\theta = 60°$
• Lever arm: $r_{\perp} = 2\sin 60° = 1.73$ m
• Torque: $\tau = (1.73)(100) = 173$ N·m (CCW → positive)

The Component Method

Another equivalent approach: decompose the force into components parallel and perpendicular to the position vector.

• $F_{\parallel} = F\cos\theta$ → produces no torque (points toward/away from pivot)
• $F_{\perp} = F\sin\theta$ → produces all the torque

$\tau = r \cdot F_{\perp} = r \cdot F\sin\theta$

All Three Methods Give the Same Answer

Choose whichever is easiest for the given problem!

Lever Arm Concept Check 🎯

Lever Arm Calculations 🧮

1. A force of 80 N acts at the end of a 0.5 m beam at $30°$ to the beam. What is the lever arm? (in m, to 3 significant figures)

2. Using the lever arm from problem 1, what is the torque? (in N·m)

3. A horizontal force of 60 N is applied at the top of a vertical pole that is 3 m tall. The pivot is at the bottom. What is the torque magnitude? (in N·m)

Lever Arm Review 🔍

Exit Quiz — Lever Arms ✅

⚖️ Rotational Equilibrium

Part 3 of 7 — $\sum \tau = 0$

An object is in rotational equilibrium when it has no net torque acting on it. This means it is either not rotating at all, or rotating at a constant angular velocity (no angular acceleration).

The Condition for Rotational Equilibrium

$\sum \tau = 0$

This means the total clockwise torque equals the total counterclockwise torque:

$\sum \tau_{\text{CCW}} = \sum \tau_{\text{CW}}$

Choosing a Pivot Point

You can choose any point as your axis of rotation when applying $\sum \tau = 0$. The physics is the same regardless of your choice!

Pro tip: Choose a pivot point where an unknown force acts. That force will produce zero torque (since $r = 0$), simplifying your equation.

Example: Simple Balance

A 4 m uniform beam is balanced on a fulcrum at its center. A 20 N weight sits at the left end. Where should a 40 N weight be placed to balance the beam?

Taking torques about the fulcrum:

• 20 N weight: $\tau = +(20)(2) = +40$ N·m (CCW)
• 40 N weight at distance $d$ to the right: $\tau = -(40)(d)$ (CW)

Setting $\sum \tau = 0$: $40 - 40d = 0 \Rightarrow d = 1 \text{ m from center}$

Multiple Torques

When multiple forces act on an object, calculate the torque from each one about the same pivot, then add them with proper signs.

Strategy

1. Draw a clear diagram
2. Choose a convenient pivot point
3. Calculate each torque (with sign)
4. Set $\sum \tau = 0$ and solve

Weight of the Beam

For a uniform beam, the weight acts at the center of mass (geometric center). Don't forget to include the beam's own weight when it matters!

Rotational Equilibrium Quiz 🎯

Rotational Equilibrium Problems 🧮

1. A 6 m beam is balanced on a fulcrum at its center. A 50 N weight is 2 m to the left of the fulcrum. Where (in m from the fulcrum) should a 25 N weight be placed to the right for balance?

2. A see-saw has a fulcrum at its center. Child A (400 N) sits 2.5 m from the fulcrum. How far from the fulcrum (in m) must Child B (500 N) sit to balance?

3. A 10 m beam has a fulcrum 4 m from the left end. A 60 N force pushes down at the left end. What downward force (in N) at the right end balances the beam? (Ignore beam weight)

Equilibrium Concepts 🔍

Exit Quiz — Rotational Equilibrium ✅

🏗️ Static Equilibrium

Part 4 of 7 — $\sum F = 0$ AND $\sum \tau = 0$

An object in static equilibrium is completely at rest — not translating and not rotating. This requires TWO conditions to be satisfied simultaneously.

The Two Conditions for Static Equilibrium

Condition 1: Translational Equilibrium

$\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$

The net force in every direction is zero → no linear acceleration.

Condition 2: Rotational Equilibrium

$\sum \tau = 0$

The net torque about any axis is zero → no angular acceleration.

Both Are Required!

An object can satisfy one condition without the other:

• A spinning wheel at constant angular velocity has $\sum \tau = 0$ but may have $\sum F \neq 0$
• A sliding object may have $\sum F = 0$ but $\sum \tau \neq 0$ if forces create a couple

Problem-Solving Strategy

1. Draw a free-body diagram showing ALL forces with their points of application
2. Choose a coordinate system (usually x-horizontal, y-vertical)
3. Choose a pivot point (often where unknown forces act to eliminate them)
4. Apply the three equations:
   • $\sum F_x = 0$
   • $\sum F_y = 0$
   • $\sum \tau = 0$
5. Solve the system of equations

Example: Horizontal Beam

A uniform 4 m beam (weight 100 N) is attached to a wall by a hinge and supported by a cable at the far end making 30° with the beam. Find the cable tension.

Taking torques about the hinge: $\sum \tau = 0: \quad T\sin 30° \times 4 - 100 \times 2 = 0$ $2T = 200 \Rightarrow T = 100 \text{ N}$

Static Equilibrium Quiz 🎯

Static Equilibrium Problems 🧮

1. A 3 m uniform beam (weight 120 N) is supported at the left end (A) and the right end (B). A 200 N load sits 1 m from A. Find the reaction force at A. (in N, round to nearest whole number)

2. Find the reaction force at B for the beam in problem 1. (in N, round to nearest whole number)

3. A horizontal beam (negligible weight) is 5 m long, hinged at the wall. A cable at the free end makes 90° with the beam and supports a 400 N load at the free end. What is the cable tension? (in N)

Static Equilibrium Concepts 🔍

Exit Quiz — Static Equilibrium ✅

🎢 Balancing Beams and See-Saw Problems

Part 5 of 7 — Classic Equilibrium Scenarios

See-saws, balance beams, and supported structures are the bread and butter of torque problems on the AP exam. Let's master the patterns!

The See-Saw Principle

For a see-saw balanced on a fulcrum:

$m_1 g \cdot d_1 = m_2 g \cdot d_2$

Since $g$ cancels:

$m_1 d_1 = m_2 d_2$

Key Insight

The heavier person must sit closer to the fulcrum. The ratio of distances is inversely proportional to the ratio of weights:

$\frac{d_1}{d_2} = \frac{m_2}{m_1}$

Multiple Weights

With multiple weights on each side, sum the torques:

$\sum m_i g \cdot d_i \text{ (left)} = \sum m_j g \cdot d_j \text{ (right)}$

Including the Beam's Weight

For a uniform beam, its weight acts at the geometric center (center of mass).

Example

A 6 m uniform beam weighs 120 N and rests on a fulcrum 2 m from the left end. A 300 N weight hangs from the left end. Find the force needed at the right end to balance.

Taking torques about the fulcrum:

• 300 N weight: $\tau = +(300)(2) = +600$ N·m (CCW)
• Beam weight at center (3 m from left end = 1 m to right of fulcrum): $\tau = -(120)(1) = -120$ N·m (CW)
• Force $F$ at right end (4 m from fulcrum): $\tau = -(F)(4)$ (CW)

$600 - 120 - 4F = 0 \Rightarrow F = 120 \text{ N}$

Balancing Problems Quiz 🎯

Balance Beam Calculations 🧮

1. A see-saw is 8 m long with the fulcrum at the center. A 70 kg person sits at the left end. How far from the center (in m) should an 80 kg person sit on the right to balance?

2. A 10 m uniform beam (weight 200 N) has its fulcrum 3 m from the left end. A 500 N weight hangs at the left end. What upward force (in N) is needed at the right end to balance? (Round to nearest whole number)

3. Two children sit on a see-saw: 30 kg at 2 m left and 20 kg at 3 m right. What is the net torque about the fulcrum? (in N·m, use $g = 10$ m/s², positive = CCW)

See-Saw Physics 🔍

Exit Quiz — Beam and See-Saw Problems ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Torque & Equilibrium Practice

Time to put everything together! This workshop focuses on systematic problem-solving for torque and equilibrium scenarios.

Master Problem-Solving Strategy

Step-by-Step Approach

1. Identify the system — What object(s) are in equilibrium?
2. Draw a free-body diagram — Show ALL forces at their actual points of application
3. Choose a pivot — Pick a smart point (where unknown forces act)
4. Set up equations:
   • $\sum F_x = 0$
   • $\sum F_y = 0$
   • $\sum \tau = 0$
5. Solve — Usually you start with the torque equation

Common Mistakes to Avoid

• Forgetting the weight of the beam itself
• Incorrect lever arms (always measure perpendicular distance)
• Wrong sign convention (CCW = +, CW = −)
• Not choosing a strategic pivot point

Ladder Problem 🎯

A 5 m ladder (weight 200 N) leans against a smooth (frictionless) wall at 60° from the floor. The base rests on a rough floor.

Beam and Cable Problem 🧮

A uniform horizontal beam (length 4 m, weight 300 N) is attached to a wall by a hinge. A cable attached to the beam 3 m from the hinge connects to the wall above, making an angle of 30° with the beam. A 500 N load hangs from the free end.

1. What is the tension in the cable? (in N, round to nearest whole number)

2. What is the horizontal component of the hinge force? (in N, round to nearest whole number)

3. What is the vertical component of the hinge force? (in N, round to nearest whole number)

Strategy Check 🔍

Challenge Problem 🏆

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Torque and Rotational Equilibrium

Let's bring together everything from this topic and practice the types of questions you'll see on the AP Physics 1 exam.

Key Concepts Summary

Torque

$\tau = rF\sin\theta = r_{\perp}F = rF_{\perp}$

• Units: N·m
• CCW = positive, CW = negative

Rotational Equilibrium

$\sum \tau = 0$

Static Equilibrium (both conditions)

$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum \tau = 0$

Problem-Solving Tips

• Choose pivot at point of unknown force
• Don't forget beam weight (acts at center)
• $m_1 d_1 = m_2 d_2$ for see-saw balance
• Use component method for angled forces

AP-Style Questions — Set 1 🎯

AP Calculation Practice 🧮

1. A 1.5 m uniform beam (weight 80 N) is supported by a hinge at the left and a cable at the right end. The cable makes 90° with the beam. A 200 N weight hangs 1.0 m from the hinge. Find the cable tension. (in N, round to nearest whole number)

2. A wrench handle is 0.25 m long. What perpendicular force is needed to produce 50 N·m of torque? (in N)

3. A 60 kg person stands 2 m from the left end of a 5 m beam (mass 20 kg) supported at both ends. Find the left support force. (in N, use $g = 10$ m/s²)

Comprehensive Review 🔍

Final AP Review ✅