Lecture 2 1D Newtons Law

Newton’s 3rd Law of Motion

  • Definition: When two bodies interact, the forces of interaction must be equal in magnitude but opposite in direction.

  • Example: A mass m at rest on a level surface under gravity.

    • Forces Acting: Let N = force of interaction between surface and block.

      • Condition at Rest: Total force downwards = 0

        • This leads to the equation:[ N + mg = 0 ][ N = -mg ]

      • Force Down on Surface: The force pressing down from the block onto the surface is mg.

      • Conclusion: The forces are exactly opposite, in accordance with Newton’s 3rd Law.

Tension in a Cable

  • Definition: Tension in a cable (rope, chain, etc.) refers to the forces acting in opposite directions within the system.

  • Characteristics: The cable is inextensible, meaning it cannot be stretched.

    • Force Application: Suppose a force F is applied at one end and the opposing force is -F at the other.

    • Analysis of Forces on Link:

      • Consider a small part of the cable (e.g., a link).

        • Force on Link 1 due to Link 2 = F.

        • Force on Link 2 due to Link 1 = -F.

  • Tension in the Cable:

    • The tension (T) in the cable is defined as T = F, where each part (link) experiences forces +F and -F.

    • In pulley systems, the direction of constant tension can change.

      • Forces on Pulley: There are other forces acting on the pulley (e.g., gravity, support reaction). In equilibrium, these forces cancel out.

Example: Car and Caravan System

  • Setup: A car with mass 1300 kg pulls a caravan of mass 700 kg with a pulling force of 2 kN.

    • Objective: Find the acceleration of the car and caravan system.

  • Variables:

    • Let ( a ) be the acceleration of both the car and caravan.

    • T = Force on caravan due to car.

      • From Newton’s 3rd Law: T = -Force on car due to caravan.

  • Equations to Solve:

    • Total force on car: [ 2000 - T = 1300a ]

    • Total force on caravan: [ T = 700a ]

  • Solving the Equations:

    • Rearranging gives:
      [ a = 1 , m/s^2 ]

    • Tension found to be: T = 700 N.

    • Additionally: Viewing the car and caravan as a single object of mass 2000 kg yields: [ 2000a = 2000 ] leading to the same acceleration result.

Example: Two Masses Connected by a String Over a Pulley

  • Setup: A mass m1 is attached to one end of an inextensible string, and mass m2 is harnessed to the other end.

    • The string runs over a smooth, fixed pulley.

  • Objective: Find the acceleration of the system under gravity.

    • Acceleration Definition: The acceleration of m1 downwards is equal to the acceleration of m2 upwards.

  • Forces Acting:

    • Total force downwards on m1: [ m1g - T ]

      • Which gives us: [ m1g - T = m1a ]

    • Total force downwards on m2: [ m2g - T ]

      • This yields: [ m2g - T = m2(-a) ]

  • Combining Equations:

  • Subtracting gives: [ (m1 - m2)g = (m1 + m2)a ]

    • Solving for ( a ): [ a = rac{(m1 - m2)}{(m1 + m2)}g ]

  • Finding Tension T:

  • Tension can be calculated using:[ T = m1(g - a) ]

    • Further simplified to:[ T = rac{2m1m2g}{(m1 + m2)} ]

Validating Results

  • Check Dimensionally:

    • [ a ] has units of LT^{-2} (Acceleration), validating correct.

    • [ T ] has units of MLT^{-2} (Force), validating correct.

  • Behavioral Analysis:

    • If ( m1 > m2 ), then ( a > 0 ).

    • If ( m1 < m2 ), then ( a < 0 ).

    • If ( m1 = m2 ), then both masses balance contributing to zero acceleration (T = mg).

    • Special Case: If ( m2 = 0 ), ( a = +g ) indicating free fall for m1, and T = 0.

    • Interchange Masses: If m1 and m2 are interchanged, a changes sign, but tension remains constant.