Newton's Laws of Motion

Force

  • A force is a push or a pull.
  • A force is an interaction between two objects or between an object and its environment.
  • A force is a vector quantity, with magnitude and direction.

Common Types of Forces

  • Normal Force: When an object pushes on a surface, the surface pushes back on the object perpendicular to the surface. This is a contact force.
  • Friction Force: This force occurs when a surface resists the sliding of an object and is parallel to the surface. Friction is a contact force.
  • Tension Force: A pulling force exerted on an object by a rope or cord. This is a contact force.
  • Weight: The pull of gravity on an object. This is a long-range force.

Magnitudes of Common Forces (Examples)

  • Sun's gravitational force on the earth: 3.5 × 10^{22} N
  • Thrust of a space shuttle during launch: 3.1 × 10^7 N
  • Weight of a large blue whale: 1.9 × 10^6 N
  • Maximum pulling force of a locomotive: 8.9 × 10^5 N
  • Weight of a 250-lb linebacker: 1.1 × 10^3 N
  • Weight of a medium apple: 1 N
  • Weight of smallest insect eggs: 2 × 10^{-6} N
  • Electric attraction between the proton and the electron in a hydrogen atom: 8.2 × 10^{-8} N
  • Weight of a very small bacterium: 1 × 10^{-18} N
  • Weight of a hydrogen atom: 1.6 × 10^{-26} N
  • Weight of an electron: 8.9 × 10^{-30} N
  • Gravitational attraction between the proton and the electron in a hydrogen atom: 3.6 × 10^{-47} N

Force Vectors

  • Use a vector arrow to indicate the magnitude and direction of the force.

Superposition of Forces

  • Several forces acting at a point on an object have the same effect as their vector sum acting at the same point.

Decomposing a Force into Component Vectors

  • Choose perpendicular x and y axes.
  • Fx and Fy are the components of a force along these axes.
  • Use trigonometry to find these force components.

Notation for Vector Sum

  • The vector sum of all the forces on an object is called the resultant of the forces or the net force:
    \sum F = F1 + F2 + F3 + … = FR

Superposition of Forces - Example

  • Force vectors are most easily added using components:
    Rx = F{1x} + F{2x} + F{3x} + …
    Ry = F{1y} + F{2y} + F{3y} + …

Newton’s First Law

  • An object at rest tends to stay at rest, and an object in motion tends to stay in uniform motion.
  • A body acted on by zero net force moves with constant velocity and zero acceleration.

Newton’s First Law (Continued)

  • If a net force acts, it causes acceleration.
  • If the net force is zero, there is no acceleration.

Newton’s Second Law

  • If the net force on an object is not zero, it causes the object to accelerate.

Uniform Circular Motion

  • An object in uniform circular motion is accelerated toward the center of the circle.
  • The net force on the object must point toward the center of the circle.

Force and Acceleration

  • The acceleration of an object is directly proportional to the net force \sum F on the object.
    a \propto \sum F

Mass and Acceleration

  • The acceleration of an object is inversely proportional to the object’s mass if the net force remains fixed.
    a \propto \frac{1}{m}

Newton’s Second Law of Motion

  • The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to the mass of the object.
    \sum F = ma
  • The SI unit for force is the newton (N).
    1 N = 1 kg \cdot m/s^2

Example Using Newton's Second Law

  • A worker applies a constant horizontal force of magnitude 20 N to a box with mass 40 kg resting on a level floor with negligible friction. To find the acceleration of the box:
    \sum F = F = 20 N = ma
    a = \frac{F}{m} = \frac{20 N}{40 kg} = 0.5 m/s^2

Example Using Newton's Second Law II

  • A waitress shoves a bottle with mass 0.45 kg to her right along a smooth, level counter. The bottle leaves her hand moving at 2.8 m/s, then slows down as it slides because of a constant friction force exerted on it by the countertop. It slides 1.0 m before coming to rest. To find the magnitude and direction of the friction force:
    1. Find acceleration from equation: vf^2 = vi^2 + 2a(xf - xi)
      a = \frac{vf^2 - vi^2}{2(xf - xi)} = \frac{0 - (2.8 m/s)^2}{2 * 1 m} = -3.9 m/s^2
    2. \sum F = F{net} = Ff = ma = 0.45 kg * (-3.9 m/s^2) = -1.8 N

Systems of Units

  • SI System: We primarily use this system.
  • British System: Force is measured in pounds, distance in feet, and mass in slugs.
  • cgs System: Mass is in grams, distance in centimeters, and force in dynes.

Mass and Weight

  • The weight of an object (on Earth) is the gravitational force that the Earth exerts on it.
  • The weight W of an object of mass m is:
    W = mg
  • The value of g depends on altitude.
  • On other planets, g will have an entirely different value than on Earth.

Example: High Diver

  • A high diver of mass 70 kg jumps off a board 10 m above the water. If her downward motion is stopped 2 s after she enters the water, what average upward force did the water exert on her?
    1. Find v just before the diver hits the water from the equation vf^2 = vi^2 + 2gy:
      v_f = \sqrt{2gy} = \sqrt{2 * 9.8 m/s^2 * 10 m} = 14 m/s (downward)
    2. Find deceleration in the water:
      a = \frac{vf - vi}{t} = \frac{0 - (-14 m/s)}{2 s} = 7 m/s^2 (upward)
    3. Find the force from \sum F = ma:
      F - mg = ma
      F = m(g + a) = 70 kg (9.8 m/s^2 + 7.0 m/s^2) = 1176 N

Newton’s Third Law

  • If you exert a force on a body, the body always exerts a force (the “reaction”) back upon you.
  • A force and its reaction force have the same magnitude but opposite directions.
  • These forces act on different bodies.

Applying Newton’s Third Law: Objects at Rest

  • An apple rests on a table. Identify the forces that act on it and the action-reaction pairs.

Applying Newton’s Third Law: Objects in Motion

  • A person pulls on a block across the floor. Identify the action-reaction pairs.

A Paradox?

  • If an object pulls back on you just as hard as you pull on it, how can it ever accelerate?

Free-Body Diagrams

  • A free-body diagram is a sketch showing all the forces acting on an object.

Multiple Objects Example

  • Two blocks of masses m1 = 10 kg and m2 = 7 kg are placed in contact with each other on a frictionless surface. A constant horizontal force F = 12 N is applied to m_1.
    • a) Find the acceleration of the system.
      a = \frac{F}{m1 + m2} = \frac{12 N}{10 kg + 7 kg} = 0.7 m/s^2
    • b) Find the magnitude of the contact force between the two blocks.
      P = m_2 a = 7 kg * 0.7 m/s^2 = 4.9 N