Lecture 1 1D Newtons Law

Newton’s Laws of Motion

1. Introduction to Dynamics

  • Dynamics is the study of forces and their effects on the behavior of objects in motion.

  • Newton's Laws of Motion form the foundation of dynamics.

2. Newton's First Law

  • A body will remain at rest or in uniform motion in a straight line unless acted upon by a net external force.

    • Remark: This law was first discovered by Galileo.

    • If the velocity is zero and constant, the body remains at rest.

    • This law follows from the implications of the Second Law.

3. Newton's Second Law

  • The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass.

    • Formula: F = ma (where F is the force, m is the mass, and a is the acceleration).

    • If no force is applied, the acceleration is zero, and thus the velocity is constant.

    • Proportionality constant m is known as the Inertial Mass, depending on the material composition of the object.

4. Units of Measurement

  • In the International System of Units (SI):

    • Mass (m) is measured in kilograms (kg).

    • Force (F) is measured with the dimension [F] = ML/T²; thus, force is measured in kg·m/s².

    • 1 Newton (N) = 1 kg·m/s².

    • Example: To accelerate a 1 kg mass by 1 m/s² requires 1 N of force.

5. Example Application of Newton's Second Law

5.1 Problem 1: Car Acceleration

  • Scenario: A car with a mass of 1100 kg accelerates to a speed of 30 km/h in 12 seconds.

  • Conversions:

    • 30 km/h = 8.33 m/s.

  • Calculation of acceleration:

    • Using the formula: a = (v - u)/t = (8.33 - 0)/12 ≈ 0.694 m/s².

  • Thus, the applied force is:

    • F = ma ≈ 764 N (or 0.764 kN).

5.2 Problem 2: Force and Distance

(a) For a force of 20 N applied to a 4 kg mass for 6 seconds (initial velocity = 15 m/s):

  • Calculation of acceleration:

    • a = F/m = 20/4 = 5 m/s².

  • Final velocity:

    • v = u + at = 15 + 5(6) = 45 m/s.

  • Distance covered:

    • s = ut + 1/2 at² = (15)(6) + 1/2(5)(6)² = 180 m.

(b) Applying a force to bring it to rest over a distance of 125 m:

  • Using the equation: v² - u² = 2as with u = 45, v = 0, and s = 125:

  • Solving for acceleration gives:

    • a = -8.1 m/s².

    • Force:

    • F = ma = -32.4 N (negative indicates deceleration).

6. Resultant Forces

  • If multiple forces act on a body, the acceleration is due to the net sum of all forces.

  • This net force is termed the resultant force (or total force).

7. Example Application: Jet Acceleration

7.1 Problem: Jet with Engine Force

  • Scenario: A jet with a mass of 8 tonnes has an engine producing 40 kN of force despite 4 kN air resistance.

  • Resultant force while engines are on:

    • F = 40 kN - 4 kN = 36 kN.

  • Acceleration while engines are on:

    • a = F/m = 36000 N/8000 kg = 4.5 m/s².

  • Speed when engines turn off:

    • v = u + at = 0 + (4.5)(12) = 54 m/s.

  • After the engine is off:

    • Air resistance only: F = -4 kN;

    • Acceleration during deceleration: a = -0.5 m/s²;

    • Distance till rest: s² = v² - 2as ⇒ 0 - (54)² = 2916 m.

8. Conclusion

  • The dynamics of motion are governed by Newton’s laws, allowing for calculations of acceleration, force, and mass in various scenarios.