Lesson 1: Static vs. Dynamic Equilibrium

1. What is the actual difference between static and dynamic equilibrium if the net force is zero in both?

The difference is velocity.

• Static Equilibrium: The object is at rest (velocity = 0) and staying at rest.
• Dynamic Equilibrium: The object is moving at a constant velocity (velocity ≠ 0). In both cases, acceleration is zero, which means the Net Force is zero.

2. If an object is moving at a constant speed in a straight line, is it in dynamic equilibrium or not in equilibrium at all?

It is in Dynamic Equilibrium. Students often think "equilibrium" implies "stillness." It doesn't. It implies "balance." If you are sliding across ice at a perfectly constant 5 m/s, all forces on you (gravity, normal force, friction, air resistance) cancel out perfectly.

3. What are the two specific conditions required for a rigid body to be in total mechanical equilibrium?

Most students forget the second one:

• Net Force = 0 (Translational Equilibrium: It’s not accelerating in x or y).
• Net Torque = 0 (Rotational Equilibrium: It’s not starting to spin).

4. Can an object be in rotational equilibrium (net torque = 0) but NOT in translational equilibrium (net force = 0)?

Yes. Think of a spinning top that is falling off a table. It might be spinning at a constant rate (Torque = 0), but it is accelerating downward due to gravity (Force ≠ 0). Conversely, a block sliding down a ramp might have Net Force ≠ 0 (accelerating), but if it isn't tumbling, its Net Torque is 0.

5. Why do we treat the Center of Gravity as the single point where gravity acts when calculating torque?

Gravity actually pulls on every single atom of the object. However, mathematically, the average location of all those pulls is the Center of Gravity (CG). Treating the weight as a single force acting at the CG produces the exact same torque as summing up the torque on every individual atom. It is a valid simplification.

6. Why is static friction usually responsible for stability, and at what exact angle does it fail (Angle of Repose)?

Static friction locks objects in place. It adjusts its strength to match the pushing force up to a maximum limit (fs_max = μN).

The Limit: On a ramp, an object will begin to slide when the angle of the ramp theta satisfies tan(theta) = μ (where μ is the coefficient of static friction). This specific angle is the Angle of Repose.

Lesson 2: The "Ladder Problem": Solving Static Equilibrium with Torque

7. How do I decide where to put the "pivot point" (axis of rotation) when starting a ladder problem?

• Strategy: Place the pivot point at the location where the most unknown forces are acting.
• Usually, this is the base of the ladder (on the floor).

Why? The floor exerts both a Normal Force and a Friction Force. By putting the pivot there, the distance r for both forces is 0. This means their Torque (rF sin theta) is 0. You eliminate two unknowns in one step!

8. In a ladder problem, why isn't the Normal Force from the ground just equal to the weight of the ladder (N = mg)?

It often is, but not always.

• Case A (Smooth Wall): The wall only pushes horizontally. The floor pushes vertically. To balance vertical forces: N_floor = mg.
• Case B (Rough Wall): The wall has friction pushing up or down. Now, N_floor + Friction_wall = mg. If you assume N=mg here, you will get the wrong answer. Always sum the vertical forces to check.

9. If I pick a pivot point where a force is acting, does that force essentially "disappear" from the equation?

Yes. Torque is defined as Tau = Force * Lever Arm. If the force acts directly on the pivot, the lever arm is 0. Therefore, the torque is 0. The force still exists (it affects F=ma), but it does not cause rotation around that specific point.

10. When calculating Torque (Tau = rF sin theta), do I use the angle with the horizontal or the angle between the force and the lever arm?

Strictly, theta is the angle between the Force vector and the Position vector (r) (the line from the pivot to where the force pushes).

Shortcut: You can also use "Perpendicular Distance." Tau = Force * (perpendicular distance to pivot). This is often easier in ladder geometry.

11. How do I handle the signs (+ or -) for torque? Is clockwise always negative?

This is a convention, not a law, but standard physics classes use:

• Counter-Clockwise (CCW): Positive (+)
• Clockwise (CW): Negative (-)

Tip: Put your pencil on the paper at the pivot point. Push it in the direction of the force. Which way does the pencil rotate? That determines the sign.

12. In a ladder problem with friction on the floor and a smooth wall, how do I solve the system of equations without getting stuck?

Follow this exact order to avoid algebra blocks:

1. Sum of Torque = 0 (Pivot at base): This allows you to solve immediately for the Normal Force of the Wall.
2. Sum of Forces (Horizontal) = 0: The Wall Force equals the Friction on the floor.
3. Sum of Forces (Vertical) = 0: The Floor Normal Force equals gravity (mg).

13. "A person climbs a ladder..." — How do I calculate the maximum height they can reach before the ladder slips?

1. Set the friction force at the floor to its maximum possible value: f = μ * N_floor.
2. Set up the Sum of Torque equation with the pivot at the base.
3. The torque from the person (Gravity * distance x) will try to rotate the ladder down.
4. Solve for x (the distance up the ladder). This is your limit.

14. How do I solve for the tension in a cable holding up a beam (Boom/Strut problem) if the cable is at an angle?

You must decompose the Tension (T) into Tx and Ty components.

Torque: Usually, use the pivot at the hinge. The Tension provides a counter-clockwise torque. You can calculate this as T * L * sin(theta) (where theta is the angle between beam and cable).

15. If a ladder is leaning against a "rough" wall instead of a "smooth" wall, how does that change the free body diagram?

You must add a Friction vector at the top of the ladder.

Direction: Friction opposes motion. If the ladder wants to slide down the wall, friction points up.

This adds a new vertical force, changing your Sum of Fy equation.

Lesson 3: Elasticity in Physics

16. In simple terms, what does Young's Modulus actually measure—stiffness or stretchiness?

It measures Stiffness.

• High Young's Modulus = Very Stiff (e.g., Steel, Diamond). Hard to deform.
• Low Young's Modulus = Very Stretchy/Flexible (e.g., Rubber, Nylon). Easy to deform.

17. Why does Strain have no units while Stress has the same units as Pressure?

Stress is Force divided by Area (F/A). That is Newtons per square meter (N/m²), which is a Pascal (Pa)—the same as pressure.

Strain is a ratio of lengths: Change in Length divided by Original Length (Delta L / L). Since you are dividing meters by meters, the units cancel out. It is a pure number (often expressed as a %).

18. Why does steel have a higher Young's Modulus than rubber if rubber is easier to stretch?

Because Young's Modulus (Y) is the ratio Stress / Strain.

• To get a small strain (stretch) in steel, you need a HUGE stress (force). Big number / small number = Huge Y.
• To get a large strain in rubber, you only need a small stress. Small number / big number = Small Y.

19. Is the "Elastic Limit" the same thing as the "Breaking Point" (Fracture point) on a stress-strain graph?

No.

Elastic Limit: The point where, if you let go, the material won't snap back to its original shape. It is permanently stretched (plastic deformation), but it hasn't broken yet.

Breaking Point: The material physically snaps in two.

20. How do I rearrange the Young's Modulus formula (Y = FL / A delta L) to solve for the change in length?

This is the most common algebra error.

• Original: Y = (F * L) / (A * Delta L)
• Swap Y and Delta L:
• Delta L = (F * L) / (A * Y)

(Think: FLEA formula... FL / AY... wait, that spells FLAY. Just remember PL/AE is the standard engineering acronym: Play).

21. How do I calculate the cross-sectional area (A) for a wire if the problem only gives me the diameter in millimeters?

• Convert diameter to radius: r = diameter / 2.
• Convert radius to meters: Divide by 1000 (e.g., 2mm -> 0.002m).
• Use Area formula: A = pi * r^2.

Warning: Do not forget to square the radius! This is the #1 calculation error source.

22. Calculating the safety factor: How much force can a steel rod take before it hits the yield strength?

1. Find the Yield Strength (Stress limit) of steel in your textbook (e.g., 250 x 10⁶ Pa).
2. Use definition of Stress: Stress = F / A.
3. Solve for Force: F_max = Stress_yield * Area.

23. How do you calculate the elongation of a heavy hanging wire due to its own weight (not an external load)? This requires integration or a Center of Mass trick.

The weight is distributed along the wire. The top holds all the weight; the bottom holds no weight.

Trick: The average force on the wire is exactly half its total weight (mg / 2).

Formula: Delta L = (mg * L) / (2 * A * Y).

24. What is the physical difference between Stress/Strain and Force/Elongation? Why do we prefer the former?

• Force/Elongation depends on the specific object (a thick wire stretches less than a thin one).
• Stress/Strain depends on the material itself (Copper vs. Steel).
• Engineers use Stress/Strain because it allows them to compare materials regardless of how big or small the sample is.

25. What happens to the energy stored in a material when you stretch it past the elastic limit (Hysteresis)?

In the elastic region, energy is stored as potential energy (like a spring) and is fully recoverable. Once you pass the elastic limit, some energy is used to permanently rearrange the atoms (breaking bonds). This energy is lost as heat. If you stretch it and release it, the unloading curve follows a different path, and the area between the curves represents the energy lost to heat.