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Torque & Statics

Unit: Rotational Dynamics

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Torque Worksheet
& Definition

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Under what conditions can a $24 \text{ kg}$ box balance a $12 \text{ kg}$ box?

Balanced boxes
Teacher CER Answer Key (Balancing)
Equation: $\Sigma \tau = 0 \implies m_1 g d_1 = m_2 g d_2$
Claim: The $12 \text{ kg}$ box must be twice as far from the pivot.
Reasoning: Torque is the product of weight and distance. To maintain an equal turning effect with half the mass, you must double the lever arm. $(24g) \cdot d = (12g) \cdot 2d$.

I. Torque ($\tau$ - Tau)

▶ Instructional Video 1 - p. 1 - 3

A) Rotating force. Which torque below is the greatest?

Comparing applications

B) depends on and to pivot point

Lever arm diagram

C) The distance between the Force and the pivot point is called the

________________ or ________________________

D) Maximum when Force is degrees to the lever arm

Angle dependence diagram

E) $\tau = d[F \sin \theta]$ units (mN)

F) Units -

Teacher CER Answer Key (Defining Torque)
Answers: Force, Distance, Lever Arm (or Moment Arm), 90°, mN (Meters-Newtons).
Reasoning: Rotation depends on the component of force acting perpendicularly to the radius. $\sin(90^\circ) = 1$ maximizes this component.

Quick Check

3 questions on 1. Balancing & Defining Torque — pick an answer to check yourself.

1. A 24 kg box sits 0.50 m from a seesaw's pivot. How far on the other side must a 12 kg box sit to balance it?

2. A 20 N force is applied at the end of a 0.50 m wrench, at 30° to the wrench handle. What torque does it produce?

3. To open a door with the least effort, where and how should you push?

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Torque Practice Problems

▶ Instructional Video 1 - p. 4 - 5

1. A uniform meter stick is balanced at its midpoint with several forces applied, as shown below. If the stick is in equilibrium, the magnitude of the force $X$ in newtons (N) is...

Stick balance

2. A uniform meter stick has a $45.0 \text{ g}$ mass placed at the $20 \text{ cm}$ mark as shown in the figure. If a pivot is placed at the $42.5 \text{ cm}$ mark and the meter stick remains horizontal in static equilibrium, what is the mass of the meter stick?

Finding mass

3. A massless rigid rod of length $3d$ is pivoted at a fixed point $W$, and two forces each of magnitude $F$ are applied vertically upward as shown. A third vertical force of magnitude $F$ may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude $F$ is applied at point:

Rod pivot

4. A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and perpendicular to the page. Four forces are exerted tangentially to the rims of the wheels, as shown. The magnitude of the net torque on the system about the axis is

Compound wheels

5. A meter stick of negligible mass is placed on a fulcrum at the $0.60 \text{ m}$ mark, with a $2.0 \text{ kg}$ mass hung at the $0 \text{ m}$ mark and a $1.0 \text{ kg}$ mass hung at the $1.0 \text{ m}$ mark. The meterstick is released from rest in a horizontal position. Immediately after release, the magnitude of the net torque on the meterstick about the fulcrum is most nearly

Teacher CER Answer Key
Answers: 1: (A), 2: (E), 3: (C), 4: (C), 5: (B).
Reasoning: For #2, stick's COM is at $50 \text{ cm}$. Lever arm for mass = $42.5 \text{ cm} - 20 \text{ cm} = 22.5 \text{ cm}$. Lever arm for stick = $50 \text{ cm} - 42.5 \text{ cm} = 7.5 \text{ cm}$. $45 \cdot 22.5 = M \cdot 7.5 \implies M = 135 \text{ g}$. For #4: $\Sigma \tau = (3FR + 3FR) - 4FR = 2FR$.
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Practice Problems

▶ Instructional Video p. 6 - 7
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Rotational Equilibrium

▶ Watch Support Forces Explanation

Ex 1) Mass Safe = $15,000. \text{ kg}$   Mass Beam = $1500. \text{ kg}$

Find $F_A$ (force of left leg on the table) and $F_B$ (force of right leg on the table)

Equilibrium scenario

$\Sigma \tau = 0$

  1. Draw all the FORCES on the picture above
  2. Find $F_A$ and $F_B$
Teacher CER Answer Key
Laws: $\Sigma F_y = 0$ and $\Sigma \tau = 0$
Claim: Support forces must sum to $165,000 \text{ N}$ ($150,000 + 15,000$).
Reasoning: By choosing A as the pivot, $F_A$ creates no torque. We find $F_B$ by balancing the safe's torque and the beam's COM torque against the upward $F_B$ at its position. Once $F_B$ is found, $F_A = Weight_{total} - F_B$.

Quick Check

3 questions on 4. Support Forces — pick an answer to check yourself.

1. A uniform 4.0 m, 100 N beam rests on supports at its ends. A 300 N box sits 1.0 m from the left end. Force from the right support?

2. A 150,000 N safe sits on a 15,000 N beam held by two legs. What must the two leg forces add up to?

3. When solving for support forces, why is it useful to put the pivot at one of the supports?

Conflicting ContentionsNo calculator
The Balanced Seesaw

A 30 kg child sits 2.0 m from the pivot of a seesaw. A 60 kg parent sits 1.0 m from the pivot on the other side. The seesaw is horizontal and balanced. Ignore the mass of the board.

30 kg 60 kg 2.0 m 1.0 m child parent pivot

Four students explain why the seesaw balances:

Ava: "The parent pushes down harder, so the parent's side should really go down. It only stays level because the board is stiff."

Ben: "It balances because the child's torque about the pivot and the parent's torque about the pivot are equal in size and opposite in direction. The forces don't have to be equal."

Cam: "It balances because the forces on the two sides are equal. The parent's weight is spread over less of the board, so it acts like the same force as the child's."

Dee: "The pivot doesn't exert any torque about itself, but it still pushes up on the board with a force of 900 N."

Which student(s), if any, do you agree with? Explain what is right or wrong about each statement.

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Ex 2) Compound Wheel

▶ Video Tutorial: Compound Wheels

Find Net Torque, Both Forces = $50. \text{ N}$, $r_1 = 30. \text{ cm}$, $r_2 = 50. \text{ cm}$

Compound disk torque

A $30. \text{ cm}$ disk and a $50. \text{ cm}$ disk are attached to an axle that passes through their centers. If a $50. \text{ N}$ force hangs from the $30. \text{ cm}$ disk and a $50. \text{ N}$ force is applied to the $50. \text{ cm}$ disk as shown above. What is the net torque on the compound wheel?

CW negative     CCW positive

Teacher CER Answer Key
Equation: $\tau_{net} = r_1 F_1 - r_2 F_2$
Answer: $10. \text{ m}\cdot\text{N}$ Counter-Clockwise.
Reasoning: Calculation: $(50. \text{ N} \cdot 0.50 \text{ m}) - (50. \text{ N} \cdot 0.30 \text{ m}) = 25 - 15 = 10 \text{ m}\cdot\text{N}$. The outer wheel provides a larger lever arm.

Quick Check

3 questions on 5. Compound Wheels — pick an answer to check yourself.

1. The same force is applied tangentially to the rim of a compound wheel, first on a disk of radius r, then on a disk of radius 2r. The torque…

2. On a compound wheel, a 50 N force on the 0.50 m disk turns it counterclockwise, while a 50 N force on the 0.30 m disk turns it clockwise. What is the net torque?

3. A 50 N force on the 0.50 m disk turns the wheel counterclockwise. What clockwise force on the 0.30 m disk would make the net torque zero?

Linked Multiple ChoiceNo calculator
Will the Meter Stick Turn?

A light meter stick is held at rest, horizontal, on a frictionless pivot at its center. Two forces are then applied as described in each row, and the stick is released. All "up" and "down" forces are vertical.

5040302010 0 1020304050 distance from pivot (cm) LEFTRIGHT

For each case, choose what the stick does right after release:

A. starts rotating clockwise    B. starts rotating counterclockwise    C. stays at rest (no rotation)    D. cannot be determined

CaseChoice
(a) 10 N down at 20 cm left; 5 N down at 40 cm right
(b) 10 N down at 30 cm left; 10 N up at 30 cm right
(c) 10 N down at 25 cm left; 20 N down at 10 cm right
(d) 8 N down at 50 cm right; 8 N pulling horizontally (along the stick) at 50 cm left
(e) 12 N down applied directly at the pivot; 6 N down at 20 cm right
(f) 10 N down somewhere on the left; 10 N down somewhere on the right

Explain the rule you used. Which cases surprised you?

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Ex 3) Beams &
Sign Suspension

A $2.20 \text{ m}$ uniform beam with a mass of $25.0 \text{ kg}$ has a $28.0 \text{ kg}$ sign attached to it. Find all the forces on the beam.

Sign from wall hinge

Find the tension on cable $T$ and the components of the wall's force on the beam. Find a) $F_{wall y}$, $F_{wall x}$ and b) Cable Tension.

$\Sigma \tau = 0$ (Use the wall as a pivot point)

b) Find $F_{wall y}$ & $F_{wall x}$

Teacher CER Answer Key
Equation: $\Sigma \tau_{hinge} = 0$
Claim: The vertical component of tension ($T\sin\theta$) must counteract the weights of the beam and the sign.
Reasoning: Beam weight acts at $1.1 \text{ m}$ (middle). Sign weight acts at $2.2 \text{ m}$. Setting torques about the hinge allows solving for $T$. Horizontal wall force equals $T\cos\theta$ to satisfy $\Sigma F_x = 0$.

Quick Check

3 questions on 6. Beams & Cables — pick an answer to check yourself.

1. When solving for the cable tension on a beam hinged to a wall, why choose the hinge as the pivot point?

2. A horizontal 2.0 m uniform 10 kg beam is hinged to a wall. A 20 kg sign hangs from its far end, and a cable at the far end pulls up at 30° above horizontal. Find the tension. (g = 10 m/s²)

3. For a horizontal beam held by a wall hinge and an angled cable, what does the wall's horizontal force balance?

Comparison TaskNo calculator
Two Ways to Hang a Sign

Two identical 120 N signs hang from the ends of identical light horizontal rods. Each rod is attached to a wall by a hinge and held up by a cable tied to the end of the rod. In Setup A the cable makes a 30° angle with the rod. In Setup B the cable makes a 60° angle with the rod.

Setup A Setup B 30° 60° 120 N 120 N hinge hinge cable cable

For each quantity, state whether it is greater in A, greater in B, or the same in both. If different, give the ratio.

QuantityAnswer
(a) Torque exerted by the sign's weight about the hinge
(b) Torque exerted by the cable about the hinge
(c) Tension in the cable
(d) Horizontal force the hinge exerts on the rod

Explain each answer. Then calculate the cable tension in each setup.

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Torque
- Ladder Against a Wall

▶ Instructional Video: Ladder Logic

$\Sigma F = 0$ & $\Sigma \tau = 0$

Note: all walls are frictionless ($\mu=0$), and all floors are rough ($\mu \neq 0$), unless otherwise indicated.

[1] A $10 \text{-meter-long}$ ladder leans against the wall, as shown. If the ladder weighs $100. \text{ N}$, what is $\mu_{min}$?

Ladder 1

[2] A $10 \text{-meter-long}$ ladder leans against the wall as shown. If the ladder weighs $200. \text{ N}$ and there is just enough frictional force to allow an $800. \text{ N}$ person to climb to the top safely, what is $\theta_{min}$? Note: $\mu_{Floor}=0.675$.

Ladder 2

3. A uniform $250.0 \text{ N}$ ladder that is $12.0 \text{ m}$ long rests against a frictionless wall at an angle of $58.0 \text{ degrees}$, the ladder just keeps from slipping.

Draw the ladder's FBD.

(a) What forces act on the bottom of the ladder?

(b) What is the coefficient of friction of the bottom of the ladder with the ground?

Teacher CER Answer Key
Answers: 1: $\mu \approx 0.312$. 2: $\theta \approx 35.8^\circ$.
Reasoning: Summing torques at the ground removes the Normal Force and Friction from the torque equation. $F_{wall} = \text{Friction}$. Normal force at the ground equals the total weight of the ladder + person.
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Translating a Torque

1. If the force is $10. \text{ N}$ and the box is $4.0 \text{ m}$ by $4.0 \text{ m}$, what is the torque?

Torque translation

A wooden square of side length $2.0 \text{ m}$ is on a horizontal tabletop and is free to rotate about its center axis. The square is subject to two forces and rotates.

Wooden square setup

4. Where should another $4.0 \text{ N}$ force be applied to place the block in an equilibrium state?

Equilibrium point

5. Where should another $4 \text{ N}$ force be applied to maximize its torque?

Max torque logic
Teacher CER Answer Key
Answers: 1: $20. \text{ m}\cdot\text{N}$. 4: Through the center of mass (to create zero torque). 5: At a corner, perpendicular to the diagonal radius.
Reasoning: For #1, the perpendicular distance to the line of action from the center is $2.0 \text{ m}$. $10 \cdot 2.0 = 20$. For #5, torque is maximized by increasing the lever arm (to the corner) and using a $90^\circ$ angle.

Check Your Understanding

5 quick questions — pick an answer to see if you’re right.

1. A 24 kg box balances a 12 kg box on a seesaw. Where must the 12 kg box be?

2. For a given force and lever arm, the torque is greatest when the force makes what angle with the lever arm?

3. Two 50. N forces act on a compound wheel: one at r = 0.50 m (counterclockwise), one at r = 0.30 m (clockwise). What is the net torque?

4. A massless meter stick pivots at 0.60 m with 2.0 kg hung at 0 m and 1.0 kg at 1.0 m. What is the net torque? (g = 10 m/s²)

5. You push hard on a wrench handle, but straight toward the bolt (along the handle). What torque do you produce?

More practice: Torque QuizMaster
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