A comprehensive unit on rotational mechanics, mapping the transition from linear motion to angular dynamics through technical analysis and real-world applications.
This force provides maximum torque due to its distance and perpendicular orientation.
P. 4 of 8
Equilibrium: The Hanging Sign
Static equilibrium requires balancing all linear forces and rotational torques.
Pivot
Mg
PHYSICS LAB
T
Observation Notes
Calculation Strategy
P. 5 of 8
Geometric Challenge: Square
Practice finding lever arm distances \( r \) and angles during the lecture.
Instructional Focus
Compare the rotational torque of a side-push vs. a corner-push relative to the central pivot.
P. 6 of 8
Geometric Challenge: Hexagon
Analyze torque on non-circular objects.
Determining Net Torque Sign
CCW Spin
POSITIVE (+)
CW Spin
NEGATIVE (-)
P. 7 of 8
The Law of Spin: Newton's Mirror
Angular acceleration (\(\alpha\)) results from the balance between Torque (the driver) and Moment of Inertia (the resistance).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem-Solving Workflow
1
Net Torque
Sum (+/-)
2
Identify
Inertia (I)
3
Calculate
Accel (\(\alpha\))
P. 8 of 8
Static Equilibrium Summary
Static equilibrium means an object remains at rest, requiring zero net force and zero net torque.
Translational Equilibrium
\[ \sum \vec{F} = 0 \]
Rotational Equilibrium
\[ \sum \vec{\tau} = 0 \]
Final Mastery Review
Identify pivot location and all lever arms.
Sum clockwise and counter-clockwise torques.
Ensure no net force in X or Y directions.
Dynamics Mastery Confirmed
F2
F3
F1 Analysis (Zero Torque)
Applied directly on the pivot point. Distance \( r = 0 \), so \( \tau = 0 \). No rotation.
F3 Analysis (Zero Torque)
This force is parallel to the rod. Push directed exactly away from pivot, angle is \( 0^\circ \). \( \sin(0^\circ) = 0 \), so \( \tau = 0 \).
F2 Analysis (Active Rotation)
This force rotates the rod. It has maximum distance and a 90° perpendicular angle.
P. 4 of 8
Equilibrium Case: The Hanging Sign
Static equilibrium requires balancing all linear pushes and all angular twists.
Pivot Axis
Beam weight
PHYSICS
LABORATORY
mg (Sign)
Tension (T) θ angle
Observation Notes
Calculation Strategy
P. 5 of 8
Geometric Challenge: Square
Practice finding lever arm distances \( r \) and angles.
Square Cross-Section Pivot Analysis
Instructional Focus
Compare the effect of a side-push vs. a corner-push on the square plate.
P. 6 of 8
Geometric Challenge: Hexagon
Practice calculating net rotational outcomes.
Determining Net Torque Sign
CCW Spin
POSITIVE (+)
&
CW Spin
NEGATIVE (-)
P. 7 of 8
The Law of Spin: Newton's Mirror
Dynamics concludes by linking torque to motion.
Linear
\[ \sum F = ma \]
Angular
\[ \sum \tau = I\alpha \]
Step-By-Step Workflow
1
Net Torque
2
Inertia (I)
3
Accel (α)
P. 8 of 8
Static Equilibrium Summary
Perfect rest requires no linear acceleration and no angular acceleration.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Final Mastery Review
Identify pivots and lever arms.
Analyze complex geometric components.
Dynamics Mastery Confirmed
P. 6 of 8
Geometric Challenge: Hexagon
Sign Conventions
CCW (+)
CW (-)
P. 7 of 8
Newton's Rotational Mirror
Linear
\[ \sum F = ma \]
Angular
\[ \sum \tau = I\alpha \]
Step-By-Step Workflow
1
Net Torque
2
Inertia (I)
3
Accel (α)
P. 8 of 8
Static Equilibrium Summary
Translation
\[ \sum \vec{F} = 0 \]
Rotation
\[ \sum \vec{\tau} = 0 \]
Identify pivots and lever arms.
Apply the conditions of equilibrium.
Dynamics Mastery Confirmed
Geometric Challenge: Square
Instructional Focus
Evaluate how lever arm distance changes based on application point. Compare side-center vs. corner pushes.
P. 6 of 8
Geometric Challenge: Hexagon
Sign Conventions
CCW (+)
Positive
CW (-)
Negative
P. 7 of 8
Newton's Rotational Mirror
Linear
\[ \sum F = ma \]
Angular
\[ \sum \tau = I\alpha \]
Step-By-Step Workflow
1
Net Torque
2
Inertia (I)
3
Accel (α)
P. 8 of 8
Static Equilibrium Summary
Translation
\[ \sum \vec{F} = 0 \]
Rotation
\[ \sum \vec{\tau} = 0 \]
Identify pivots and lever arms.
Apply the conditions of equilibrium.
Dynamics Mastery Confirmed
Zero Torque
Fixed Pivot
F1
F2
F3
Analysis: Force 1 (Applied on Pivot)
Torque = 0. Even though this force is perfectly perpendicular, it is applied directly on the pivot. This means the lever arm distance is zero (\(r = 0\)). Since \( \tau = (0)F \), this force produces Zero Torque and cannot rotate the rod.
Analysis: Force 3 (Parallel to Lever Arm)
Torque = 0. This force has a distance \( r \), but it points directly away from the axis. Since \( \sin(0^\circ) = 0 \), the twist is zero. You are just pulling the rod against the pivot.
Success: Force 2 (The Rotation Source)
This is the only force that will rotate the rod. It has a maximum lever arm (\(r = L\)) and is perfectly perpendicular (\(\theta = 90^\circ\)). It creates a Clockwise (negative) torque.
P. 4 of 8
Equilibrium: The Hanging Sign
Static equilibrium isn't just for seesaws. In engineering, beams supported by cables are used to hold weight over distance. To stay perfectly still, the torque from the cable must cancel the torque from the weight.
PIVOT
Tension (T) θ = 30°
PHYSICS LAB
ENTRANCE
Weight (W)
1. Downward Torque (CW)
The weight of the sign pulls down at a distance \( L \). Since it tries to spin the beam clockwise, its torque is Negative.
\( \tau_W = -L \cdot W \)
2. Upward Torque (CCW)
The cable pulls up and left. Only the vertical component (\( T \sin\theta \)) creates torque. It spins the beam counter-clockwise, so it's Positive.
\( \tau_T = +L \cdot T \sin\theta \)
Equilibrium Condition:
\( \tau_T + \tau_W = 0 \implies L \cdot T \sin\theta = L \cdot W \)
P. 5 of 8
Forces on Complex Geometry
In mechanical engineering, parts are rarely perfect circles. By analyzing force components on polygons, we can calculate precisely how much twist is being applied.
Cross-Sectional Analysis (Square Nut)
Force 1
Force 2
Force 3
Observation Challenge
Look at Force 2 in the square diagram above. Its line of action passes exactly through the top edge of the beam. Does it create rotation?
Because the force vector points directly along the lever arm path (parallel to the orientation of the top of the square), the angle \(\theta\) is zero. This results in Zero Torque.
P. 6 of 8
The Hexagon: Multi-Force Synthesis
When tightenining or loosening heavy hardware, we sum all applied torques. In this diagram, four different forces act on the same hexagonal body.
Force 1
Force 2
Force 3
Force 4
Sign Recognition Challenge
By observing the arrows above, can you tell which forces are working together to spin the nut in the same direction? (Answer: Forces 2 and 4 are both clockwise creators).
∑ τ = τ1 (+) + τ2 (-) + τ3 (0) + τ4 (-)
P. 7 of 8
The Law of Spin: Mirror Image
Newton's Second Law describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia). The result is Angular Acceleration.
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
"Mass resists being pushed. Moment of Inertia resists being twisted. To double the angular acceleration, you must either double the net torque or cut the moment of inertia in half."
P. 8 of 8
The Balancing Act: Static Equilibrium
Static Equilibrium is the physics of architecture and engineering. For a structure to remain at rest, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
No linear acceleration.
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
No angular acceleration.
Equilibrium Study: The Seesaw
r1
r2
To balance a heavy force with a lighter one, the lighter force must have a significantly longer lever arm to produce an equal and opposite torque.
The counterweight must be 2.75 meters to the right.
Synthesis Reference
Torque is the product of Force and Radius. If an object has a high resistance (high \( I \)), you need a high Net Torque to create acceleration. By using a long wrench, you increase \( r \), which mathematically multiplies your physical input force. This allows a human to generate the massive torque needed to overcome the static resistance of a stuck bolt.
Force 1
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Complex Support
Consider a uniform beam of mass M and length L. It is pivoted at a wall, supports a hanging sign of mass m, and is held up by a rope at an angle θ attached to the far end.
PIVOT
Mg (Beam)
EQUILIBRIUM
EXHIBIT
mg (Sign)
Tension (T) θ
Sign Conventions
Weight vectors (Sign & Beam): Both create clockwise rotation (Negative).
Tension vector: Only vertical component (\( T \sin\theta \)) creates counter-clockwise rotation (Positive).
Net Torque Balance:
\(\sum \tau = 0\)
\(\tau_{rope} - \tau_{beam} - \tau_{sign} = 0\)
\((T \cdot L \cdot \sin\theta) - (M g \cdot \frac{L}{2}) - (m g \cdot \frac{3}{4}L) = 0\)
P. 5 of 8
Geometric Observation
In mechanical engineering, we must identify the relationship between the force vector and the center of rotation purely by geometric observation.
Observational Case: Square Beam
FORCE A
FORCE B
FORCE C
Challenge: Which one twists?
Look at Force B. Its line of action passes exactly along the top edge of the square. It has no vertical component to pull the edge around the center. This results in Zero Torque.
P. 6 of 8
Geometric Synthesis: Hexagon
Observe the four forces acting on the hexagonal nut below.
F1
F2
F3
F4
Observation Task
Which force has the longest lever arm? Which one is parallel to its radius vector?
(Hint: F2 is at a vertex, further than F1 at a side midpoint).
P. 7 of 8
The Law of Spin
The final stage of dynamics is applying Newton's Second Law for Rotation.
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
Static Equilibrium: Dual Balance
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m at its far end.
AXIS
Mg (Beam)
PHYSICS
GALLERY
mg (Sign)
Tension (T) θ
The Static Equation
The sign and the beam's own mass create Clockwise (Negative) torques. The cable must create an equal Counter-Clockwise (Positive) torque to maintain balance.
Below is a square beam cross-section. Use the following descriptions to visualize where forces might be applied to create or prevent rotation.
Observational Case: Square Beam
Teaching Note: Use for live demonstration of force vectors.
Principles to demonstrate
Perpendicular Side Push: Push in the middle of a side. Lever arm is half the side length. Efficiency is 100%.
Parallel Edge Push: Push along the edge toward a corner. Since the angle is 0, torque is zero.
Vertex Push: Radius is larger (\(L\sqrt{2}/2\)), but angle matters immensely!
P. 6 of 8
The Hexagon: Multi-Force Synthesis
When tightenining or loosening heavy hardware, we sum all applied torques. Practice identifying where clockwise and counter-clockwise forces would be placed on this nut.
Demonstration Area: Plot Forces F1, F2, F3, F4
Sign Recognition Challenge
If a force pushes "Up" on the left side of the nut, is the torque positive or negative?
(Answer: Positive / CCW).
If a force pushes "Down" on the right side of the nut, is the torque positive or negative?
(Answer: Negative / CW).
P. 7 of 8
The Law of Spin
The final stage of dynamics is applying Newton's Second Law for Rotation.
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
Static Equilibrium: Dual Balance
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
r1
r2
Torque Mastery Confirmed
Identify pivots and lever arms.
Analyze geometric components.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls the rod but does not create a spin.
Force 2: The Driving Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle (\(\sin 90^\circ = 1\)).
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium involves balancing all linear pushes and all angular twists. Consider a uniform beam of mass M pivoted at a wall.
PIVOT
Mg (Beam)
MUSEUM
OF PHYSICS
mg (Sign)
Tension (T) θ
Equilibrium Rules
Forces: All downward weights must be canceled by upward tension component (\( T \sin\theta \)) and wall support.
Torques: Clockwise twists from Sign & Beam weight must equal the Counter-Clockwise twist from the rope.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Analysis Area
Principles of Square Torque
- Lever Arm: Distance from the center to any side midpoint is \( L/2 \).
- Vertex Radius: Distance to any corner is \( \frac{L}{\sqrt{2}} \).
- Inefficient Forces: Forces passing through the center pivot create zero rotation.
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Synthesis Area
Identifying the Twist
Draw forces at different locations to demonstrate how the sign (\(+/-\)) changes based on the side of application.
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
AXIS
Mg (Beam Weight)
MUSEUM OF
EQUILIBRIUM
mg (Sign Weight)
Tension (T) θ
The Static Balance
The sign and the beam's weight create Clockwise (Negative) torques. The cable must create an equal Counter-Clockwise (Positive) torque to maintain balance.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Analysis Area
Observation Logic
Identify how the torque changes based on distance from the center and the direction of the force vector.
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Synthesis Area
Torque Application Practice
Identify the sign (+/-) and relative strength of a force applied at different geometric points.
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Complex Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
AXIS
Mg (Beam)
MUSEUM OF
EQUILIBRIUM
mg (Sign Weight)
Tension (T) θ
The Static Balance
The sign and the beam's weight create Clockwise (Negative) torques. The cable must create an equal Counter-Clockwise (Positive) torque to maintain balance.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Analysis Area
Observation Logic
Identify how the torque changes based on distance from the center and the direction of the force vector.
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Synthesis Area
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
PIVOT
Mg (Beam)
PHYSICS
GALLERY
mg (Sign)
Tension (T) θ
The Static Balance
The weights create Clockwise (-) torques. The cable provides Counter-Clockwise (+) torque.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Analysis Area
Inquiry Points
Where would you push to get the most torque?
Where could you push and get zero rotation?
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Plotting Area
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
AXIS
Mg (Beam)
PHYSICS
MASTERCLASS
mg (Sign)
Tension (T) θ
The Static Balance
The weights create Clockwise (Negative) torques. The cable must create an equal Counter-Clockwise (Positive) torque to maintain balance.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Analysis Area
Inquiry Points
Where would you push to get the most torque?
Where could you push and get zero rotation?
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Synthesis Area
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
PIVOT AXIS
Mg (Beam)
MASTERCLASS
PHYSICS
mg (Sign)
Tension (T) θ
The Static Balance
Both the beam and sign pull the system Clockwise (-). The rope must pull Counter-Clockwise (+) to maintain rest.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Demonstration Area
Observation Logic
Draw forces at midpoints vs vertices.
Identify which forces pass through the center (Zero Torque) vs which have the largest lever arm.
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Synthesis Area
Torque Application Practice
Students: Use this area to plot multiple force vectors and calculate the Net Torque resulting from your choices.
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 1
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
PIVOT AXIS
Mg (Beam)
MUSEUM OF
EQUILIBRIUM
mg (Sign)
Tension (T) θ
The Static Balance
The sign and the beam's weight create Clockwise (Negative) torques. The cable must create an equal Counter-Clockwise (Positive) torque to maintain balance.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Analysis Area
Observation Logic
Identify how the torque changes based on distance from the center and the direction of the force vector.
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Synthesis Area
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 1
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
AXIS
Mg (Beam)
PHYSICS
MASTERCLASS
mg (Sign)
Tension (T) θ
The Static Balance
The sign and beam create Clockwise (-) torques. The cable creates a Counter-Clockwise (+) torque to maintain balance.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Demonstration Area
Observation Logic
Identify how the torque changes based on distance from the center and the direction of the force vector.
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Synthesis Area
Torque Application Practice
Identify the sign (+/-) and relative strength of a force applied at different geometric points.
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 1
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
AXIS
Mg (Beam)
PHYSICS
MASTERCLASS
mg (Sign)
Tension (T) θ
The Static Balance
The sign and beam weights create Clockwise (-) torques. The cable creates a Counter-Clockwise (+) torque to maintain balance.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Demonstration Area
Observation Logic
Identify how the torque changes based on distance from the center and the direction of the force vector.
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Plotting Area
Torque Application Practice
Students: Use this area to plot multiple force vectors and calculate the Net Torque resulting from your choices.
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Force 1
Force 2
Force 3
Force 1: Zero Torque (r = 0)
Even though this force is perpendicular, it is applied directly on the pivot. With no distance from the axis, there is no twist.
Force 3: Zero Torque (θ = 0°)
This force is at a distance, but it points directly away from the pivot. It pulls, but it does not spin.
Force 2: Active Torque
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium: Hanging Support
Static equilibrium requires balancing all pushes and all twists. Consider a beam of length L and mass M supporting a hanging sign of mass m.
AXIS
Mg (Beam)
PHYSICS
MASTERCLASS
mg (Sign)
Tension (T) θ
The Static Balance
The sign and beam weights create Clockwise (-) torques. The cable creates a Counter-Clockwise (+) torque to maintain balance.
In mechanical engineering, we must identify the relationship between a force vector and the center of rotation purely by geometric observation. Use the diagram below to demonstrate different force types.
Square Demonstration Area
Observation Logic
Analyze how force direction relative to the pivot determines the magnitude of torque.
P. 6 of 8
Geometric Synthesis: Hexagon
Geometry determines the available lever arms. Use this nut to practice identifying where clockwise and counter-clockwise forces would be placed.
Hexagon Plotting Area
Torque Application Practice
Identify the sign (+/-) and relative strength of a force applied at different geometric points.
P. 7 of 8
The Law of Spin
Newton's Second Law for Rotation describes the contest between the Driver (Torque) and the Resistor (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Problem Solving Workflow
1
Sum Net
Torque (∑τ)
2
Calculate
Inertia (I)
3
Solve for
Accel (α)
P. 8 of 8
The Balancing Act
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Balancing Act Analysis
Torque Mastery Confirmed
Identify pivots and lever arms.
Apply \(\sum \tau = I\alpha\).
Fixed Pivot
F1
F2
F3
F1 Analysis (Zero Torque)
Even though it is perpendicular, it is applied directly on the pivot point. Distance \( r = 0 \), so \( \tau = 0 \).
F3 Analysis (Zero Torque)
This force is parallel to the rod. Since the push is directed exactly away from the pivot, the angle is \( 0^\circ \). \( \sin(0^\circ) = 0 \), so \( \tau = 0 \).
F2 Analysis (Active Rotation)
This force has distance and a 90-degree angle. It creates a maximum twist!
P. 4 of 8
Equilibrium: Complex Support
Consider a uniform massy beam pivoted at a wall. To balance it, we attach a sign at 3/4 L and a rope at the far end at an angle θ.
Pivot Axis
Beam weight (Mg)
PHYSICS
EQUILIBRIUM
mg (Sign)
Rope Tension (T) θ angle
Equilibrium Rules
Sum of Forces: \(\sum F_y = 0\). All downward weights must be balanced by the vertical tension component.
Sum of Torques: \(\sum \tau = 0\). Clockwise twists (sign and beam) must be canceled by the counter-clockwise rope pull.
The Calculation Strategy:
\( \tau_{rope} - \tau_{beam} - \tau_{sign} = 0 \)
\( (T \cdot L \cdot \sin\theta) - (M g \cdot \frac{L}{2}) - (m g \cdot \frac{3}{4}L) = 0 \)
P. 5 of 8
Observation Challenge: Square
When teaching, add various force arrows to this diagram. Identify which orientation provides a spin and which results in Zero Torque.
Geometric Frame: Square Cross-Section
Teaching Checklist
A. Apply a perpendicular force to the side midpoint. Note the 90-degree angle to the center vector.
B. Apply a parallel force along the top edge. Observe how its path points through the axis, creating no twist.
C. Apply a force at the vertex (corner). Use this to discuss the larger radius \( r \) versus an angled \(\sin\theta\) loss.
P. 6 of 8
The Hexagon: Multi-Force Synthesis
Calculate the lever arm distance \( r \) for various application points on this hexagonal nut.
Determining Net Torque Sign
CCW Spin
POSITIVE (+)
VS
CW Spin
NEGATIVE (-)
P. 7 of 8
The Dynamic Law: Newton's Mirror
Dynamics concludes by linking torque to motion. The angular acceleration (\(\alpha\)) is the result of the battle between Drive (Torque) and Resistance (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Step-By-Step Workflow
1
Net Torque
Sum Sign (+/-)
2
Identify
Inertia (I)
3
Calculate
Accel (α)
P. 8 of 8
Perfect Balance: Static Equilibrium
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Final Mastery Review
Identify pivots and lever arms (\( r \)).
Explain why forces on the axis (\( r=0 \)) create no torque.
Understand mass distribution effects on Inertia (\( I \)).
Dynamics Mastery Confirmed
Fixed Pivot
F1
F2
F3
F1 Analysis (Zero Torque)
Even though it is perpendicular, it is applied directly on the pivot point. Distance \( r = 0 \), so \( \tau = 0 \).
F3 Analysis (Zero Torque)
This force is parallel to the rod. Since the push is directed exactly away from the pivot, the angle is \( 0^\circ \). \( \sin(0^\circ) = 0 \), so \( \tau = 0 \).
F2 Analysis (Active Rotation)
This force has distance and a 90-degree angle. It creates a maximum twist!
P. 4 of 8
Equilibrium: Complex Support
Static equilibrium requires balancing all linear pushes and all angular twists. Consider a uniform massy beam pivoted at a wall.
Pivot Axis
Beam weight (Mg)
PHYSICS
LABORATORY
mg (Sign)
Rope Tension (T) θ angle
Equilibrium Observation Notes
The Calculation Strategy
P. 5 of 8
Geometric Frame: Square
Apply forces during instruction to identify twists and zero-torque cases.
Square Cross-Section Pivot Analysis
Teaching Checklist
A. Apply a force to the side midpoint. Note the 90-degree angle to the center vector.
B. Apply a force along the top edge. Observe how its line of action passes through the pivot, creating no twist.
C. Apply a force at the corner vertex. Compare radius length vs angled force.
P. 6 of 8
The Hexagon: Multi-Force Synthesis
Calculate the lever arm distance \( r \) for various application points on this hexagonal hardware.
Determining Net Torque Sign
CCW Spin
POSITIVE (+)
&
CW Spin
NEGATIVE (-)
P. 7 of 8
The Dynamic Law: Newton's Mirror
Dynamics concludes by linking torque to motion. The angular acceleration (\(\alpha\)) is the result of the battle between Drive (Torque) and Resistance (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Step-By-Step Workflow
1
Sum Net
Torque (∑τ)
2
Identify
Inertia (I)
3
Calculate
Accel (α)
P. 8 of 8
Static Equilibrium Summary
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Final Mastery Review
Identify pivots and lever arms (\( r \)).
Explain why forces on the axis (\( r=0 \)) create no torque.
Understand mass distribution effects on Inertia (\( I \)).
Dynamics Mastery Confirmed
Fixed Pivot
F1
F2
F3
F1 Analysis (Zero Torque)
Even though it is perpendicular, it is applied directly on the pivot point. Distance \( r = 0 \), so \( \tau = 0 \).
F3 Analysis (Zero Torque)
This force is parallel to the rod. Since the push is directed exactly away from the pivot, the angle is \( 0^\circ \). \( \sin(0^\circ) = 0 \), so \( \tau = 0 \).
F2 Analysis (Active Rotation)
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium Case: The Hanging Sign
Consider a uniform beam of mass M pivoted at a wall. To achieve static equilibrium, the sum of all linear forces and all rotational torques must equal zero.
Pivot Axis
Mg (Beam)
PHYSICS
EQUILIBRIUM
mg (Sign)
Tension (T) θ angle
Observation Notes
Calculation Strategy
P. 5 of 8
Geometric Challenge: Square
"Geometry determines the lever arm." Use this clean diagram to practice drawing force vectors and identifying which will result in rotation.
Square Cross-Section Analysis
Dynamic Lesson Points
Demonstrate how a push along the top edge points toward the center (Zero Torque), whereas a perpendicular push at the side midpoint has a lever arm of \( \frac{1}{2}L \) and produces rotation.
P. 6 of 8
Geometric Challenge: Hexagon
Calculate the lever arm distance \( r \) for various application points on this hexagonal hardware.
Sign Recognition Task
CCW Spin
POSITIVE (+)
&
CW Spin
NEGATIVE (-)
P. 7 of 8
The Law of Spin: Newton's Mirror
Dynamics concludes by linking torque to motion. The angular acceleration (\(\alpha\)) is the result of the battle between Drive (Torque) and Resistance (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Step-By-Step Workflow
1
Net Torque
Sum Sign (+/-)
2
Identify
Inertia (I)
3
Calculate
Accel (α)
P. 8 of 8
Static Equilibrium Summary
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Final Mastery Review
Identify pivots and lever arms (\( r \)).
Explain why forces on the axis (\( r=0 \)) create no torque.
Understand mass distribution effects on Inertia (\( I \)).
Dynamics Mastery Confirmed
Fixed Pivot
F1
F2
F3
F1 Analysis (Zero Torque)
Even though it is perpendicular, it is applied directly on the pivot point. Distance \( r = 0 \), so \( \tau = 0 \).
F3 Analysis (Zero Torque)
This force is parallel to the rod. Since the push is directed exactly away from the pivot, the angle is \( 0^\circ \). \( \sin(0^\circ) = 0 \), so \( \tau = 0 \).
F2 Analysis (Active Rotation)
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium Case: The Hanging Sign
Static equilibrium requires balancing all linear pushes and all angular twists. Consider a uniform beam of mass M pivoted at a wall.
Pivot Axis
Mg (Beam)
PHYSICS
LABORATORY
mg (Sign)
Tension (T) θ angle
Observation Notes
Calculation Strategy
P. 5 of 8
Geometric Challenge: Square
"Geometry determines the lever arm." Use this clean diagram to practice drawing force vectors and identifying which will result in rotation.
Square Cross-Section Pivot Analysis
Dynamic Lesson Points
Demonstrate how a push along the top edge points toward the center (Zero Torque), whereas a perpendicular push at the side midpoint has a lever arm of \( \frac{1}{2}L \) and produces rotation.
P. 6 of 8
Geometric Challenge: Hexagon
Calculate the lever arm distance \( r \) for various application points on this hexagonal hardware.
Sign Recognition Task
CCW Spin
POSITIVE (+)
&
CW Spin
NEGATIVE (-)
P. 7 of 8
The Dynamic Law: Newton's Mirror
Dynamics concludes by linking torque to motion. The angular acceleration (\(\alpha\)) is the result of the battle between Drive (Torque) and Resistance (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Step-By-Step Workflow
1
Sum Net
Torque (∑τ)
2
Identify
Inertia (I)
3
Calculate
Accel (α)
P. 8 of 8
Perfect Balance: Static Equilibrium
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Final Mastery Review
Identify pivots and lever arms (\( r \)).
Explain why forces on the axis (\( r=0 \)) create no torque.
Understand mass distribution effects on Inertia (\( I \)).
Dynamics Mastery Confirmed
Parallel to the rod.
F2: Active Rotation
Has distance and angle.
P. 4 of 8
Equilibrium Case: The Hanging Sign
Static equilibrium requires balancing all linear pushes and all angular twists. Observe the connection points on this uniform massy beam.
Pivot Axis
Beam weight
Tension (T) θ angle
PHYSICS
LABORATORY
mg (Sign)
Observation Notes
Calculation Strategy
P. 5 of 8
Geometric Frame: Square
Identify twists and zero-torque cases by drawing your own vectors during the lecture.
Square Cross-Section Pivot Analysis
Demonstration Guide
Discuss how different orientations of push affect the resulting torque relative to the center pivot.
P. 6 of 8
Geometric Challenge: Hexagon
Practice finding lever arm distances \( r \) and the angle \( \theta \) for various force application points.
Sign Convention Key
CCW Spin
POSITIVE (+)
&
CW Spin
NEGATIVE (-)
P. 7 of 8
Synthesis: The Law of Spin
The contest between Torque (Driver) and Inertia (Resistor) results in Angular Acceleration.
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
P. 8 of 8
Static Equilibrium Summary
Perfect rest requires no linear acceleration and no angular acceleration.
Translational Balance
\[ \sum \vec{F} = 0 \]
Rotational Balance
\[ \sum \vec{\tau} = 0 \]
Final Mastery Review
Identify pivots and lever arms.
Analyze complex geometric components.
F1
F2
F3
F1 Analysis (Zero Torque)
Even though it is perpendicular, it is applied directly on the pivot point. Distance \( r = 0 \), so \( \tau = 0 \).
F3 Analysis (Zero Torque)
This force is parallel to the rod. Since the push is directed exactly away from the pivot, the angle is \( 0^\circ \). \( \sin(0^\circ) = 0 \), so \( \tau = 0 \).
F2 Analysis (Active Rotation)
This is the only force that will rotate the rod. It has both distance and a perpendicular angle.
P. 4 of 8
Equilibrium Case: The Hanging Sign
Static equilibrium requires balancing all linear pushes and all angular twists. Consider a uniform beam of mass M pivoted at a wall.
Pivot Axis
Beam weight
PHYSICS
LABORATORY
mg (Sign)
Tension (T) θ angle
Observation Notes
Calculation Strategy
P. 5 of 8
Geometric Challenge: Square
"Geometry determines the lever arm." Use this clean diagram to practice drawing force vectors and identifying which will result in rotation.
Square Cross-Section Pivot Analysis
Dynamic Lesson Points
Demonstrate how a push along the top edge points toward the center (Zero Torque), whereas a perpendicular push at the side midpoint has a lever arm of \( \frac{1}{2}L \) and produces rotation.
P. 6 of 8
Geometric Challenge: Hexagon
Calculate the lever arm distance \( r \) for various application points on this hexagonal hardware.
Sign Convention Key
CCW Spin
POSITIVE (+)
&
CW Spin
NEGATIVE (-)
P. 7 of 8
The Dynamic Law: Newton's Mirror
Dynamics concludes by linking torque to motion. The angular acceleration (\(\alpha\)) is the result of the battle between Drive (Torque) and Resistance (Moment of Inertia).
Linear (Translation)
\[ \sum F = ma \]
Angular (Rotation)
\[ \sum \tau = I\alpha \]
Step-By-Step Workflow
1
Sum Net
Torque (∑τ)
2
Identify
Inertia (I)
3
Calculate
Accel (α)
P. 8 of 8
Perfect Balance: Static Equilibrium
For an object to remain perfectly still, it must have a perfect balance of all pushes and all twists.
