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Torque & Statics

Balanced Beam Header

Unit: Rotational Dynamics

1. Balancing Logic 2. Defining Torque 3. MCQs 1-5 4. Reference Notes 5. Support Forces 6. Compound Wheels 7. Beams & Cables 8. Ladder Problems 9. Axis & Symmetry

Torque Worksheet

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

Balanced boxes
Teacher CER Answer Key
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. $(24g) \cdot d = (12g) \cdot 2d$.

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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
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.

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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: (B), 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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Detailed Reference Notes

▶ Instructional Video p. 6 - 7
Notes 1
Notes 2
Notes 3

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, $F_A = Weight_{total} - F_B$.

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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.

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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 $\Sigma F_x = 0$.

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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.

▶ Detailed Problem 3 Walkthrough

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.

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