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.

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)

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

Units of force

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

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?

Working a Newton’s 2nd Law Problem

20 kgFL FM

Step 2: Diagram

Force diagram

Working a Newton’s 2nd Law Problem

Free Body diagram

20 kgFL FM

FG

N

FL

FM

FG

N

Step 3: Set up equationsF = maFx = max

Fy = may

Working a Newton’s 2nd Law Problem

Always resolve two-dimensional problems into two one-dimensional problems.

Step 4: Substitute Make a list of givens from the word problem.

Substitute in what you know.

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.

Working a Newton’s 2nd Law Problem

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.

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.

Gravity as an accelerating force

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?

20 kgFL FM

FG

N

Flat surfaces – 1 D

N = mg for objects resting on horizontal surfaces.

mg

N

Applied forces affect normal force.

applied forcefriction

weightnormal

N = applied force

Elevator Ride – going up!

mg

N

Groundfloor

Normal feeling

V = 0A = 0

mg

N

Juststarting up

Heavy feeling

V > 0A > 0

mg

N

Betweenfloors

Normal feeling

V > 0A = 0

mg

N

Arriving attop floor

Light feeling

V > 0A < 0

Elevator Ride – going down!

mg

N

Topfloor

Normal feeling

V = 0A = 0

mg

N

Arriving atGround floor

Heavy feeling

V < 0A > 0

mg

N

Betweenfloors

Normal feeling

V < 0A = 0

mg

N

Beginningdescent

Light feeling

V < 0A < 0

Ramps – 2 D

mg

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

mgsin

N = mgcos

Ramps – 2 D

mg

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

mgsin

N = mgcos

Determination of the Coefficients of FrictionCoefficient of Static Friction1) Set a block of one material on an

incline plane made of the other material.

2) Slowly increase angle of plane until the block just begins to move. Record this angle.

3) 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 frictionexists before sliding occurs

Kinetic frictionexists 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

Applied Force

Frictional Force

Normal Force

Gravity

A force diagram illustrating Static Friction

Bigger Frictional Force

Normal Force

Gravity

Bigger Applied Force

A force diagram illustrating Static Friction

Frictional Force

Normal Force

GravityEven Bigger Applied Force

The forces on the book are now UNBALANCED!

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 FrictionCoefficient of Kinetic Friction

1) Set a block of one material on an incline plane made of the other material.

2) Slowly increase angle of plane until the block just begins to move at constant speed after giving it a slight tap. Record this angle.

3) Calculate k = tan.

Magic Pulleys

m1

m2

T

mg

N

mgT

-x

x

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.

Pulley problem

m1

m2

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?

Pulley problem

m1

m2

Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If s = 0.3 and k = 0.2, what is a) the acceleration and b) the tension in the string?

Pulley problem

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

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

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. Force diagram

Working a Newton’s Law Problem

Free Body diagram

F1F2

F1

F2

Step 4: Set up force equations.

F = maFx = max

Fy = may

Working a Newton’s Law Problem

Always resolve two-dimensional problems into two one-dimensional problems.

Step 5: Calculate!

Substitute in what you know into the second law equations.

Calculate unknown or unknowns.

Working a Newton’s Law Problem

Ramp (frictionless)

mg

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

N = mgcos

Ramp (frictionless)

mg

What will acceleration be in this situation?F= mamgsin = magsin = a

N

mgcos

mgsin

N = mgcos

Ramp (frictionless)

mg

How could you keep the block from accelerating?

N

mgcos

mgsin

N = mgcos

T

Tension (static 1D)

The horizontal and vertical components of the tension are equal to zero if the system is not accelerating.

15 kg

T

mg

F = 0T = mg

Tension (static 2D)

The horizontal and vertical components of the tension are equal to zero if the system is not accelerating.

15 kg

30o 45o

T1T2

T3

Fx = 0Fy = 0T3

mg

Tension (elevator)

When an elevator is still, the tension in the cable is equal to its weight.M

T

Mg

Tension (elevator)

What about when the elevator is just starting to head upward from the ground floor?M

T

Mg

Tension (elevator)

What about when the elevator is between floors?

M

T

Mg

Tension (elevator)

What about when the elevator is slowing at the top floor?

M

T

Mg

Tension (elevator)

What about if the elevator cable breaks?

M Mg

-x

x

Magic pulleys simply bend the coordinate system.

Pulley problems

m1

m2

m1g

NT T

m2g

F = mam2g= (m1+m2)a

Tension is determined by examining one block or the other

Pulley problems

m 1 m2

m1g

NT

T

m2g

F2m2am2g - T = m2a

F1m1am1gsin = m1a