20. Revision/fast recall

1. TITLE

Revision: 3 Cases of Pulley Constraints (Atwood Machine shortcuts) formula sheet + key points

2. QUICK FACTS BOX

Pulley Constraints — QUICK FACTS

  • String tension is constant throughout a massless, inextensible string.
  • Acceleration magnitude is identical for all masses connected by a tight string,.
  • Net external force on a system determines its overall acceleration ($F=ma$).
  • Internal forces (like tension) cancel out when considering the whole system.
  • Free-body diagrams are essential for analyzing individual connected masses.
  • Friction acts parallel to surfaces to oppose impending or actual relative motion.
  • Massless pulleys require no force to rotate, simplifying constraint equations.

3. FORMULA SHEET

FORMULAS:

  • Case 1: Vertical Atwood Machine

    • $a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g$ ($m/s^2$)
    • $T = \left( \frac{2m_1 m_2}{m_1 + m_2} \right) g$ ($N$)
      → Use when: Two masses ($m_2 > m_1$) hang vertically over a frictionless pulley.

  • Case 2: Horizontal Table Pulley (Block & Trolley)

    • $a = \frac{m_2 g - f_k}{m_1 + m_2}$ ($m/s^2$)
    • $T = m_1 a + f_k$ ($N$)
      → Use when: One mass ($m_2$) hangs vertically, pulling another ($m_1$) horizontally against kinetic friction ($f_k$),.

  • Case 3: Universal Shortcut Formula

    • $a = \frac{\text{Net Pulling Force}}{\text{Total Mass System}}$ ($m/s^2$)
      → Use when: Quickly finding system acceleration without solving simultaneous tension equations,.

4. MUST-REMEMBER VALUES/CONSTANTS

KEY VALUES:

  • Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$ (or $10 \text{ m/s}^2$ for simplicity),
  • Standard Force Unit = $1 \text{ N} = 1 \text{ kg m/s}^2$
  • Coefficient rule = $\mu_k < \mu_s$ (Kinetic friction is less than static friction),

5. COMPARISON SUMMARY

[Feature][Case 1: Vertical Atwood][Case 2: Horizontal Pulley]
[Driving Force][Difference in weights ($m_2g - m_1g$)][Weight of hanging mass ($m_2g$)]
[Opposing Force][Weight of lighter mass ($m_1g$)][Kinetic friction on horizontal mass ($f_k$)]
[System Mass][$m_1 + m_2$][$m_1 + m_2$]

6. COMMON MISTAKES — QUICK LIST

DON'T:

  • Forget to resolve the weight vector ($mg \sin\theta$ and $mg \cos\theta$) if a mass is on an inclined plane.
  • Assume tension is uniform if the string has significant mass or the pulley is not frictionless.
  • Include internal tension when applying Newton's Second Law to the entire system at once,.

DO:

  • Draw a separate free-body diagram (FBD) for each interacting body,.
  • Assume a uniform direction of acceleration across all connected parts to keep signs consistent,.

7. MNEMONICS & MEMORY TRICKS

REMEMBER:
* FBD: First Basic Duty — always draw a Free-Body Diagram before solving.
* Shortcut for $a$: "Pull minus Oppose over Total" ($a = \frac{F_{pull} - F_{oppose}}{M_{total}}$) to quickly find the acceleration of any connected system,.

8. HIGH-YIELD NEET POINTS

NEET LOVES TO ASK:

  • Calculating acceleration and tension when two masses are released over a vertical pulley (Standard Atwood).
  • Finding the tension in a string pulling a block across a rough horizontal surface,.
  • Identifying whether internal forces (like tension) contribute to the system's overall acceleration (They don't),.

9. ONE MCQ — SELF-TEST

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. What is the acceleration of the masses? (Take $g = 10 \text{ m/s}^2$)
(a) $10 \text{ m/s}^2$
(b) $5 \text{ m/s}^2$
(c) $2 \text{ m/s}^2$
(d) $0 \text{ m/s}^2$

Answer: (c) $2 \text{ m/s}^2$. Using the shortcut: $a = \frac{12g - 8g}{12+8} = \frac{40}{20} = 2 \text{ m/s}^2$.

10. NCERT PAGE REFERENCES

Revise from:
- Page 59: Tension and restoring forces in an inextensible string
- Page 61-62: Example 4.9 (Block and trolley system / Horizontal pulley case),
- Page 70: Exercise 4.16 (Vertical Atwood machine problem)

11. CTA

Detailed explanation → 3-Laws of Motion Guide
Quick test → [Revision Test: 3-Laws of Motion]