Newton’s Laws The Study of Dynamics

Isaac Newton Came up with 3 Laws of Motion to explain the observations and analyses of Galileo and Johannes Kepler. Invented Calculus. Published his Laws in 1687 in the book Mathematical Principles of Natural Philosophy.

What is Force? A force is a push or pull on an object. Forces cause an object to accelerate… To speed up To slow down To change direction

Newton’s First Law The Law of Inertia. A body in motion stays in motion at constant velocity and a body at rest stays at rest unless acted upon by an external force. This law is commonly applied to the horizontal component of velocity, which is assumed not to change during the flight of a projectile.

The First Law is Counterintuitive Aristotle firmly believed this. But Physics 1 students know better!

A force diagram illustrating no net force

A force diagram illustrating no net force

A force diagram illustrating no net force

A force diagram illustrating no net force

Another example illustrating no net force

Newton’s Second Law A body accelerates when acted upon by a net external force. The acceleration is proportional to the net force and is in the direction which the net force acts. This law is commonly applied to the vertical component of velocity.

Newton’s Second Law ∑F = ma where ∑F is the net force measured in Newtons (N) m is mass (kg) a is acceleration (m/s2)

Units of force Newton (SI system) 1 N = 1 kg m /s2 1 N is the force required to accelerate a 1 kg mass at a rate of 1 m/s2 Pound (British system) 1 lb = 1 slug ft /s2

The problem of weight Are weight and mass the same thing? No. Weight can be defined as the force due to gravitation attraction. W = mg

Newton’s Third Law For every action there exists an equal and opposite reaction. If A exerts a force F on B, then B exerts a force of -F on A.

Step 1: Draw the problem Working a Newton’s 2nd Law Problem Larry pushes a 20 kg block on a frictionless floor at a 45o angle below the horizontal with a force of 150 N while Moe pulls the same block horizontally with a force of 120 N. What is acceleration? 20 kg FL FM

Step 2: Diagram Working a Newton’s 2nd Law Problem Force diagram N FM FL FM FG N 20 kg FL FM FG N Force diagram Free Body diagram

Step 3: Set up equations F = ma Fx = max Fy = may Working a Newton’s 2nd Law Problem Step 3: Set up equations F = ma Fx = max Fy = may Always resolve two-dimensional problems into two one-dimensional problems.

Step 4: Substitute Make a list of givens from the word problem. Working a Newton’s 2nd Law Problem Step 4: Substitute Make a list of givens from the word problem. Substitute in what you know.

Step 5: Solve Plug-n-chug. Calculate your unknowns. Working a Newton’s 2nd Law Problem Step 5: Solve Plug-n-chug. Calculate your unknowns. Sometimes you’ll need to do kimematic calculations following the Newton’s 2nd law calculations.

Gravity as an accelerating force A very commonly used accelerating force is gravity. Here is gravity in action. The acceleration is g.

Gravity as an accelerating force In the absence of air resistance, gravity acts upon all objects by causing the same acceleration…g.

Gravity as an accelerating force The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g.

2-Dimensional problem Larry pushes a 20 kg block on a frictionless floor at a 45o angle below the horizontal with a force of 150 N while Moe pulls the same block horizontally with a force of 120 N. a) What is the acceleration? b) What is the normal force? FL FM FG N 20 kg

Flat surfaces – 1 D N = mg for objects resting on horizontal surfaces. N mg

Applied forces affect normal force. friction applied force weight normal N = applied force

Elevator Ride – going up! mg N Ground floor Normal feeling V = 0 A = 0 mg N Just starting up Heavy feeling V > 0 A > 0 mg N Between floors Normal feeling V > 0 A = 0 mg N Arriving at top floor Light feeling V > 0 A < 0 Elevator Ride – going up!

Elevator Ride – going down! mg N Top floor Normal feeling V = 0 A = 0 mg N Beginning descent Light feeling V < 0 A < 0 mg N Between floors Normal feeling V < 0 A = 0 mg N Arriving at Ground floor Heavy feeling V < 0 A > 0 Elevator Ride – going down!

Ramps – 2 D N mg N = mgcos   The normal force is perpendicular to angled ramps as well. It’s always equal to the component of weight perpendicular to the surface. N = mgcos N  mgcos mgsin mg 

Ramps – 2 D N mg N = mgcos   How long will it take a 1.0 kg block to slide down a frictionless 20 m long ramp that is at a 15o angle with the horizontal? N = mgcos N mg  mgcos mgsin 

Determination of the Coefficients of Friction Coefficient of Static Friction Set a block of one material on an incline plane made of the other material. Slowly increase angle of plane until the block just begins to move. Record this angle. Calculate s = tan.

Friction The force that opposes a sliding motion. Enables us to walk, drive a car, etc. Due to microscopic irregularities in even the smoothest of surfaces.

There are two types of friction Static friction exists before sliding occurs Kinetic friction exists after sliding occurs In general fk <= fs

Friction and the Normal Force The frictional force which exists between two surfaces is directly proportional to the normal force. That’s why friction on a sloping surface is less than friction on a flat surface.

Static Friction fs  sN fs : static frictional force (N) s: coefficient of static friction N: normal force (N) Static friction increases as the force trying to push an object increases… up to a point!

A force diagram illustrating Static Friction Normal Force Frictional Force Applied Force Gravity

A force diagram illustrating Static Friction Normal Force Bigger Applied Force Bigger Frictional Force Gravity

A force diagram illustrating Static Friction The forces on the book are now UNBALANCED! Normal Force Frictional Force Even Bigger Applied Force Gravity Static friction cannot get any larger, and can no longer completely oppose the applied force.

Kinetic Friction fk = kN fk : kinetic frictional force (N) k: coefficient of kinetic friction N: normal force (N) Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces.

Determination of the Coefficients of Friction Coefficient of Kinetic Friction Set a block of one material on an incline plane made of the other material. Slowly increase angle of plane until the block just begins to move at constant speed after giving it a slight tap. Record this angle. Calculate k = tan.

Magic Pulleys T mg N -x x m1 m2

Pulley problem Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find (a) the acceleration of each block and, (b) the tension in the connecting string. m1 m2

Pulley problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg) as shown. What must the minimum coefficient of static friction be to keep Mass 1 from slipping? m1 m2

Pulley problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If ms = 0.3 and mk = 0.2, what is a) the acceleration and b) the tension in the string? m1 m2

Tension A pulling force. Generally exists in a rope, string, or cable. Arises at the molecular level, when a rope, string, or cable resists being pulled apart.

Step 1: Identify the body to analyze. Working a Newton’s Law Problem Step 1: Identify the body to analyze. This may not be all that easy! It may be a knot, a nail, a hinge, a person, an “object” or a “particle”. It is the focus of your subsequent analysis.

Step 2: Select a reference frame. Working a Newton’s Law Problem Step 2: Select a reference frame. This should be an inertial reference frame which may be moving but not accelerating. Think of this as a coordinate system with a specific origin!

Step 3: Make a diagram of forces. Working a Newton’s Law Problem Step 3: Make a diagram of forces. Force diagram F1 F2 Free Body diagram F1 F2

Step 4: Set up force equations. Working a Newton’s Law Problem Step 4: Set up force equations. F = ma Fx = max Fy = may Always resolve two-dimensional problems into two one-dimensional problems.

Working a Newton’s Law Problem Step 5: Calculate! Substitute in what you know into the second law equations. Calculate unknown or unknowns.

Ramp (frictionless) N mg N = mgcos   The normal force is perpendicular to angled ramps as well. It’s usually equal to the component of weight perpendicular to the surface. N  mgcos mgsin mg 

Ramp (frictionless) N mg What will acceleration be in this situation? SF= ma mgsinq = ma gsinq = a N = mgcos N  mgcos mgsin mg 

Ramp (frictionless) N T mg How could you keep the block from accelerating? N = mgcos N T  mgcos mgsin mg 

T mg Tension (static 1D) SF = 0 T = mg The horizontal and vertical components of the tension are equal to zero if the system is not accelerating. T mg SF = 0 T = mg 15 kg

T1 T3 T2 T3 mg Tension (static 2D) SFx = 0 SFy = 0 The horizontal and vertical components of the tension are equal to zero if the system is not accelerating. SFx = 0 SFy = 0 30o 45o T1 T2 T3 T3 mg 15 kg

T M Mg Tension (elevator) When an elevator is still, the tension in the cable is equal to its weight. M

T M Mg Tension (elevator) What about when the elevator is just starting to head upward from the ground floor? M

T M Mg Tension (elevator) What about when the elevator is between floors? M

T M Mg Tension (elevator) What about when the elevator is slowing at the top floor? T Mg M

Tension (elevator) What about if the elevator cable breaks? M Mg

Pulley problems N T m1g m2g SF = ma m2g = (m1+m2)a -x x Magic pulleys simply bend the coordinate system. m1g N T m2g SF = ma m2g = (m1+m2)a -x x m1 m2

Pulley problems T N m1g m2g SF2 = m2a m2g - T = m2a SF1 = m1a Tension is determined by examining one block or the other SF2 = m2a m2g - T = m2a m1g N T m2g SF1 = m1a T-m1gsinq = m1a m1 m2 q