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.
• $F<em>x$ and $F</em>y$ 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:
  $\sum F = F<em>1 + F</em>2 + F<em>3 + … = F</em>R$

Superposition of Forces - Example

• Force vectors are most easily added using components:
  $R<em>x = F</em>{1x} + F<em>{2x} + F</em>{3x} + …$
  $R<em>y = F</em>{1y} + F<em>{2y} + F</em>{3y} + …$

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: $v<em>f^2 = v</em>i^2 + 2a(x<em>f - x</em>i)$
     $a = \frac{v<em>f^2 - v</em>i^2}{2(x<em>f - x</em>i)} = \frac{0 - (2.8 m/s)^2}{2 * 1 m} = -3.9 m/s^2$
  2. $\sum F = F<em>{net} = F</em>f = ma = 0.45 kg * (-3.9 m/s^2) = -1.8 N$

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 $v<em>f^2 = v</em>i^2 + 2gy$:
     $v_f = \sqrt{2gy} = \sqrt{2 * 9.8 m/s^2 * 10 m} = 14 m/s$ (downward)
  2. Find deceleration in the water:
     $a = \frac{v<em>f - v</em>i}{t} = \frac{0 - (-14 m/s)}{2 s} = 7 m/s^2$ (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 $m<em>1 = 10 kg$ and $m</em>2 = 7 kg$ 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.
    $a = \frac{F}{m<em>1 + m</em>2} = \frac{12 N}{10 kg + 7 kg} = 0.7 m/s^2$
  • b) Find the magnitude of the contact force between the two blocks.
    $P = m_2 a = 7 kg * 0.7 m/s^2 = 4.9 N$