$\vec{\tau} = \vec{r} \times \vec{F}$

Magnitude: $\tau = rF\sin\theta$

where:

• $\tau$ = torque (N·m)
• $r$ = distance from axis of rotation to point where force is applied (m)
• $F$ = applied force (N)
• $\theta$ = angle between $\vec{r}$ and $\vec{F}$

💡 Key Idea: Torque depends not just on force magnitude, but also on WHERE and HOW the force is applied. Same force can produce different torques depending on position and angle.

Lever Arm (Moment Arm)

The lever arm (or moment arm) is the perpendicular distance from the axis of rotation to the line of action of the force:

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

Alternative torque formula: $\tau = r_{\perp} \cdot F$

Key Points:

• Maximum torque when $\theta = 90°$: $\tau_{max} = rF$

  • Force perpendicular to position vector

• Zero torque when $\theta = 0°$ or $180°$: $\tau = 0$

  • Force along the line to axis
  • No lever arm!

• Larger lever arm → more torque (easier to rotate)

  • Why door knobs are far from hinges
  • Why longer wrenches provide more turning ability

Direction of Torque

Sign Convention (for problems in a plane):

• Positive torque (+): tends to cause counterclockwise rotation
• Negative torque (−): tends to cause clockwise rotation

Right-Hand Rule (3D):

1. Curl fingers from $\vec{r}$ to $\vec{F}$
2. Thumb points in direction of $\vec{\tau}$

Net Torque

Net torque is the sum of all torques:

$\tau_{net} = \sum \tau_i = \tau_1 + \tau_2 + \tau_3 + ...$

Remember to include signs (clockwise vs. counterclockwise)!

Rotational Equilibrium

An object is in rotational equilibrium when:

$\tau_{net} = 0$

No angular acceleration - object either:

• Not rotating (at rest)
• Rotating at constant angular velocity

Complete Equilibrium

For complete static equilibrium, need BOTH:

1. Translational equilibrium: $\sum \vec{F} = 0$

   • No linear acceleration

2. Rotational equilibrium: $\sum \vec{\tau} = 0$

   • No angular acceleration

Conditions for Equilibrium

First Condition (Forces):

$\sum F_x = 0$ $\sum F_y = 0$

No net force in any direction.

Second Condition (Torques):

$\sum \tau = 0$

Net torque about ANY axis is zero.

Important: You can choose ANY point as the axis for calculating torques!

• Choose wisely to simplify calculations
• Often choose where unknown forces act (they contribute zero torque there)

Common Applications

Seesaws and Balance

• Balanced when torques are equal
• Heavier person sits closer to pivot
• $m_1 d_1 = m_2 d_2$ (for horizontal seesaw)

Door Hinges

• Hinge is axis of rotation
• Push far from hinge (large $r$) → easy to open
• Push close to hinge (small $r$) → hard to open
• Push perpendicular to door → maximum torque

Wrenches and Tools

• Longer wrench → larger lever arm → more torque
• Apply force perpendicular to handle
• Easier to loosen tight bolts

Levers

Three classes based on positions of fulcrum, effort, and load:

• Class 1: Fulcrum between effort and load (seesaw, crowbar)
• Class 2: Load between fulcrum and effort (wheelbarrow, nutcracker)
• Class 3: Effort between fulcrum and load (tweezers, biceps)

Center of Gravity

Center of gravity is the point where all the weight can be considered to act.

For uniform objects:

• Symmetric objects: at geometric center
• Irregular objects: found by suspension method or balancing

Torque due to weight: $\tau = mg \cdot d$

where $d$ is horizontal distance from pivot to center of gravity.

Problem-Solving Strategy

For Equilibrium Problems:

1. Draw a diagram showing all forces and their points of application
2. Choose coordinate system (which direction is +x, +y)
3. Choose axis of rotation (smart choice simplifies math!)
   • Often choose where unknown force acts
4. Apply $\sum F_x = 0$ and $\sum F_y = 0$
5. Apply $\sum \tau = 0$ (about chosen axis)
   • Assign + to counterclockwise, − to clockwise
   • Include ALL torques
6. Solve system of equations
7. Check: Do all answers make physical sense?

⚠️ Common Mistakes

Mistake 1: Forgetting the Angle

$\tau = rF\sin\theta$, NOT just $rF$!

• Only when $\theta = 90°$ does $\tau = rF$

Mistake 2: Using Wrong Distance

Must use perpendicular distance (lever arm), not just distance along the object!

Mistake 3: Sign Errors

Be consistent with sign convention:

• Counterclockwise: positive
• Clockwise: negative

Mistake 4: Axis Choice Confusion

Any axis works, but:

• Forces acting AT the axis contribute zero torque
• Choose axis through unknown force to eliminate it from torque equation

Mistake 5: Forgetting Weight

Weight acts at center of gravity - don't forget to include it!

Special Cases

Uniform Beam/Rod

• Weight acts at center (midpoint)
• $W = mg$ at $x = L/2$

Multiple Objects on Beam

• Each object contributes torque: $\tau_i = m_i g \cdot d_i$
• Sum all torques

Ladder Against Wall

• Weight at center
• Normal forces at two contact points
• Friction at ground prevents sliding
• Equilibrium: $\sum F = 0$ and $\sum \tau = 0$

Units

Torque: N·m (newton-meter)

Note: Same dimensions as energy (joules), but different physical meaning!

• Energy: scalar
• Torque: vector (direction matters)

DO NOT write torque in joules!

Torque vs. Force

Real-World Examples

Why Doors Have Handles Far from Hinges

• Maximize lever arm $r$
• Same force produces more torque
• Easier to open

Why Long Wrenches Are Better

• Longer wrench = larger $r$
• More torque for same force
• "Cheater bar" increases effective length

Balancing Objects

• Artist balancing mobile sculptures
• Equilibrium requires careful positioning
• Sum of clockwise torques = sum of counterclockwise torques

Steering Wheel

• Large diameter gives large torque
• Easier to turn wheels
• Race cars have smaller steering wheels (power steering helps)

Key Formulas Summary

Find: Position of adult for equilibrium

Step 1: Set up coordinate system

Choose the pivot as axis of rotation (torques about this point).

Distances from pivot:

• Left end: 2 m to the left of pivot
• Right end: 4 m to the right of pivot
• Center of beam: at 3 m from left = 1 m to right of pivot

Step 2: Identify all forces and distances

Child (at left end):

• Force: $F_c = m_c g = 30g$ (downward)
• Distance from pivot: $d_c = 2$ m (to left)
• Torque: $\tau_c = -30g(2) = -60g$ N·m (clockwise, negative)

Beam weight (at center):

• Force: $F_b = m_b g = 40g$ (downward)
• Distance from pivot: $d_b = 1$ m (to right)
• Torque: $\tau_b = 40g(1) = 40g$ N·m (counterclockwise, positive)

Adult (at unknown position):

• Force: $F_a = m_a g = 50g$ (downward)
• Distance from pivot: $d_a = ?$
• Torque: $\tau_a =$ depends on position

Step 3: Apply rotational equilibrium

$\sum \tau = 0$

Let $x$ = position of adult from left end.

Distance of adult from pivot = $(x - 2)$ m

Sign: If $x > 2$ (right of pivot): positive torque If $x < 2$ (left of pivot): negative torque

Step 4: Set up torque equation

Taking counterclockwise as positive:

$\tau_b + \tau_a + \tau_c = 0$

$40g(1) + 50g(x - 2) + (-60g)(2) = 0$

Divide by $g$:

$40(1) + 50(x - 2) - 60(2) = 0$

$40 + 50x - 100 - 120 = 0$

$50x - 180 = 0$

$50x = 180$

$x = 3.6 \text{ m from left end}$

Step 5: Verify the answer

Distance from pivot: $3.6 - 2 = 1.6$ m to the right

Check torques:

• Child: $-30(9.8)(2) = -588$ N·m (clockwise)
• Beam: $+40(9.8)(1) = +392$ N·m (counterclockwise)
• Adult: $+50(9.8)(1.6) = +784$ N·m (counterclockwise)

Sum: $-588 + 392 + 784 = +588$ N·m... wait, let me recalculate.

Actually: $-588 + 392 + 784 = 588$ N·m

Let me redo: $40 + 50(x-2) - 120 = 0$ $50(x-2) = 80$ $x - 2 = 1.6$ $x = 3.6$ m ✓

Check: $40(1) + 50(1.6) - 60(2) = 40 + 80 - 120 = 0$ ✓

Answer: The adult should sit 3.6 m from the left end (or 1.6 m to the right of the pivot).

Physical sense: Adult is heavier than child, so sits closer to pivot. Adult sits to right of pivot to counterbalance child on left. ✓

For no slipping: f ≤ μₛN_g 43.3 ≤ μₛ(150) μₛ ≥ 43.3/150 = 0.29

Minimum μₛ = 0.29

Find: Maximum distance $d$ before slipping

Step 1: Draw free body diagram and identify forces

At bottom of ladder (ground contact):

• Normal force from ground: $N_g$ (upward)
• Friction from ground: $f_s$ (horizontal, to the right)

At top of ladder (wall contact):

• Normal force from wall: $N_w$ (horizontal, to the left)
• No friction (wall is frictionless)

Weights:

• Ladder weight: $W_L = m_L g = 20g$ at center (2.5 m up ladder)
• Person weight: $W_p = m_p g = 70g$ at distance $d$ up ladder

Step 2: Apply force equilibrium

Horizontal forces ($\sum F_x = 0$): $N_w - f_s = 0$ $N_w = f_s$

Vertical forces ($\sum F_y = 0$): $N_g - W_L - W_p = 0$ $N_g = (20 + 70)g = 90g$

Step 3: Choose axis for torque calculation

Choose bottom of ladder as pivot (eliminates $N_g$ and $f_s$ from torque equation).

Step 4: Calculate perpendicular distances

Height of wall contact point: $h = L\sin\theta = 5\sin(60°) = 5(0.866) = 4.33$ m

Horizontal distance to wall: $b = L\cos\theta = 5\cos(60°) = 5(0.5) = 2.5$ m

Perpendicular distances from bottom of ladder:

For $N_w$ (acts horizontally at top):

• Lever arm = vertical distance = $L\sin\theta = 4.33$ m

For $W_L$ (acts at center, L/2 from bottom):

• Horizontal lever arm = $\frac{L}{2}\cos\theta = 2.5\cos(60°) = 1.25$ m

For $W_p$ (acts at distance $d$ from bottom):

• Horizontal lever arm = $d\cos\theta = d\cos(60°) = 0.5d$

Step 5: Apply rotational equilibrium ($\sum \tau = 0$)

Taking counterclockwise as positive, about bottom of ladder:

$N_w(L\sin\theta) - W_L\left(\frac{L}{2}\cos\theta\right) - W_p(d\cos\theta) = 0$

$N_w(4.33) - 20g(1.25) - 70g(0.5d) = 0$

$N_w(4.33) = 20g(1.25) + 70g(0.5d)$

$N_w(4.33) = 25g + 35gd$

$N_w = \frac{25g + 35gd}{4.33}$

Step 6: Apply friction condition

For no slipping: $f_s \leq \mu_s N_g$

At the verge of slipping: $f_s = \mu_s N_g$

Since $f_s = N_w$:

$N_w = \mu_s N_g = 0.4(90g) = 36g$

Step 7: Solve for d

$\frac{25g + 35gd}{4.33} = 36g$

$25g + 35gd = 36g(4.33)$

$25 + 35d = 155.88$

$35d = 130.88$

$d = 3.74 \text{ m}$

Answer: The person can climb up to 3.74 m (about 75% of the ladder's length) before the ladder starts to slip.

Check: Does this make sense?

• Person climbs 3.74 m out of 5 m ladder length ✓
• Person creates clockwise torque, wall force creates counterclockwise torque ✓
• Higher friction would allow climbing further ✓
• Steeper angle would allow climbing further (larger wall lever arm) ✓