Newton’s Laws .

Newton’s Laws

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Monday, September 15, 2008

Re-introduction to

Newton’s 3 Laws

Announcements

• Calendar has been updated for our new unit.

• Exam repair will occur by Thursday, end of day. Repair is to be done independently. Before school is the best time; but I will allow you to correct during the 2nd half of both lunch periods and after school.

What is Force?

• A force is a push or pull on an object.

• Unbalanced forces cause an object to accelerate…– To speed up– To slow down– To change direction

• Force is a vector!

Types of Forces

• Contact forces involve contact between bodies. – Normal, Friction

• Field forces act without necessity of contact.– Gravity, Electromagnetic, Strong, Weak

• Question: Is there really any such thing as a contact force?

Forces and Equilibrium

• If the net force (F) on a body is zero, it is in equilibrium.

• An object in equilibrium may be moving relative to us (dynamic equilibrium).

• An object in equilibrium may appear to be at rest (static equilibrium).

Galileo’s Thought Experiment

© The Physics Classroom, Tom Henderson 1996-2007

This thought experiment lead to Newton’s First Law.

Galileo’s Thought Experiment

© The Physics Classroom, Tom Henderson 1996-2007

Newton’s First Law

• Newton’s 1st Law is also called the Law of Inertia.

• A body in motion stays in motion in a straight line 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.

Newton’s Second Law

• Newton’s 2nd Law says that a body acted upon by a net external force will accelerate.

• The acceleration is proportional to the net force and inversely proportional to the mass. It is in the direction of the net force.

• F = ma

Newton’s Third Law

• Newton’s 3rd Law is commonly stated “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.

Commonly Confused Terms

• Inertia: or the resistance of an object to being accelerated

• Mass: the same thing as inertia (to a physicist).• Weight: gravitational attraction

inertia = mass weight

Sample Problem: Three forces act upon a 3.0 kg body moving at constant velocity. F1 = (4i – 6j + k) N and F2 = (i – 2j - 8k) N. Find F3.

• Sample Problem: Two forces, F1 = (4i – 6j + k) N and F2 = (i – 2j - 8k) N, act upon a body of mass 3.0 kg. No other forces act upon the body at this time. What do you know must be true?

• Sample problem: A tug-of-war team ties a rope to a tree and pulls hard horizontally to create a tension of 30,000 N in the rope. Suppose the team pulls equally hard when, instead of a tree, the other end of the rope is being pulled by another tug-of-war team such that no movement occurs. What is the tension in the rope in the second case?

A systematic approach for 1st or 2nd Law Problems

1. Identify the system to be analyzed. This may be only a part of a more complicated system. This is really the most important step.

2. Select a reference frame or coordinate system, stationary or moving, but not accelerating, within which to analyze your system.

3. Identify all forces acting on the system. Ignore the others. This is best done by drawing!

4. Set up ΣF = ma equations for each dimension. Solve for relevant unknowns.

5. Use kinematics or calculus where necessary to calculate more about the motion.

• Sample Problem: A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. The bullet is accelerated by expanding gases while it travels down the 0.820 m long barrel. Assuming constant acceleration and negligible friction, what is the force on the bullet?

• Sample Problem: A 3.00 kg mass undergoes an acceleration given by a = (2.50i + 4.10j) m/s2. Find the resultant force F and its magnitude.

Tuesday, September 16, 2008

The Normal Force

Tension

Announcements

• Exam repair due by Thursday PM

• Assignment from yesterday will be collected tomorrow.

Normal force

• The force that keeps one object from invading another object is called the normal force.

• “Normal” means “perpendicular”.

• Always determine the normal force by considering all forces that have components perpendicular to a surface.

Problem: determine the normal force acting on a 5.0 kg box sitting on a flat table.

Problem: Now determine the normal force acting on a 5.0 kg box sitting on a flat table.

F=16 N

40°

Problem: Now determine the normal force acting on a this 5.0 kg box sitting on a ramp at angle =30o.

Problem: Assume that the ramp is frictionless. What is the acceleration of the block down the ramp?

30o

F=20 N

20°

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.

Tension (static 2D)

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

15 kg

30o 45o

32 1

Problem: Determine the tension in all three ropes.

15 kg

30o 45o

32 1

Problem: What is the tension is the cable attached to a 5,000 kg elevator that starts on the ground floor at rest and accelerates upward, reaching a speed of 3.0 m/s in 2 seconds?

M

Wednesday, September 17, 2008

Magic Pulleys

Announcements

• Turn in HW Assignment #1 today

• Tomorrow: HW Assignment #2

Pulley problems

• Pulley’s simply bend the coordinate system

m1

m2

Sample problem: derive a formula for acceleration, assuming the table is frictionless.

m1

m2

m 1 m2

Sample problem: derive a formula for acceleration, assuming the table is frictionless.

m 1 m2

Sample problem: derive a formula for the tension T in the string.

Atwood machine

• A device for measuring g.

• If m1 and m2 are nearly the same, slows down freefall such that acceleration can be measured.

• Then, g can be measured.m1

m2

Problem: For the Atwood machine shown, derive an equation which can be used to find g, the gravitational acceleration, from a measured value of acceleration.

m1

m2

Atwood Machine Pre-lab

• Purpose: to determine gravitational acceleration using the Atwood machine.

• Hypothesis:?• Theoretical Background: See previous

slide.• Procedure: ?• Data: You are collecting data to determine

acceleration. Mass is important, too, You’ll need data tables ready to record data into.

Atwood Machine Lab

Thursday, September 18, 2008

Atwood Machine lab

• Write a full lab report in your lab notebook. The pre-lab you did yesterday has many of the relevant sections. Today, collect your data and analyze your results. Tabulate your data neatly.

• Calculate g from your mass and acceleration values. Do a standard deviation analysis on g, and do a percent error calculation on g as well.

• Write an analysis and conclusion.

Newton’s 2nd Law Workday

Friday, September 19, 2008

Friction


Announcements

• Turn in HW assignment:– Put problems 34, 36, 37 on top– Put problems 38,39, 56 on bottom– Make sure your name is on both pages.– Pass to the right.

• Lab books (Atwood lab) is due today.

Problem: How high up the frictionless ramp will the block slide?

5.0 kg20o

v = 12.0 m/s

Problem: Calculate acceleration of the 5 kg block. Table and pulley are magic and frictionless. Calculate Tension in the string.

20o

5.0 kg

1.0 kg

Friction• Friction opposes a sliding motion.• Static friction exists before sliding

occurs– (fs sN).

• Kinetic friction exists after sliding occurs–fk = kN

x

y

Draw a free body diagram for a braking car.

x

y

Draw a free body diagram for a car accelerating from rest.

Question:

• Why is it disadvantageous for cars to skid to a stop?

Friction mini-lab

• Use DataStudio to see if we can detect the difference in magnitude between static and kinetic friction.

Sliding down a ramp Sliding up a ramp

• Draw free body diagrams that include friction for a body which is

Sample problem: A 1.00 kg book is held against a wall by pressing it against the wall with a force of 50.00 N. What must be the minimum coefficient of friction between the book and the wall, such that the book does not slide down the wall?

F

Problem: Assume a coefficient of static friction of 1.0 between tires and road. What is the minimum length of time it would take to accelerate a car from 0 to 60 mph?

Friction – Day II

Tuesday, September 23, 2008

Announcements

• Lab books (Atwood lab) must be turned in today if they haven’t already been turned in.

• Tomorrow I will collect #44, 45, 50

• Tonight’s assignment is to work pages 8 and 9 in your classwork packet. I will check this for completion (effort) on Thursday.

Problem: Assume a coefficient of static friction of 1.0 between tires and road and a coefficient of kinetic friction of 0.80 between tires and road. How far would a car travel down a 15o incline after the driver applies the brakes if it skids to a stop? Assume the speed before brakes are applied is 26 m/s.

Review Time!

• Homework Problems --?

• Clicker Quiz – Newton’s Laws

• Ranking Tasks

• Video Surprise

Centripetal Force

Wednesday, September 24, 2008

Announcements

• Turn in Ch 5, problems #44, 45,50

Centripetal Force

• Inwardly directed force which causes a body to turn; perpendicular to velocity.

• Centripetal force always arises from other forces, and is not a unique kind of force.– Sources include gravity, friction, tension,

electromagnetic, normal.

• ΣF = ma• a = v2/r• ΣF = m v2/r

Highway Curves

R

r

z

Friction turns the vehicle

Normal force turns the vehicle

Problem: Find the minimum safe turning radius for a car traveling at 60 mph on a flat roadway, assuming a coefficient of static friction of 0.70.

Problem: Derive the expression for the period best banking angle of a roadway given the radius of curvature and the likely speed of the vehicles.

Centripetal Force Lab

Thursday, September 25, 2008

Announcements

• Open up class work packet to pages 8 and 9 for my stamp.

• Give yourself an AP score! Check your work against the scoring guide; please do not remove scoring guides from room.

• When you are done, begin work on a homework assignment (Ch 6: 9,11,13 or Ch 6: 14,15,21 or CW packet pages 10 and 11).

Conical Pendulum

What provides the centripetal force for a conical pendulum?

z

r

T

mg

L

Problem: Derive the expression for the period of a conical pendulum with respect to the string length and radius of rotation.

cos2

LT

g

Centripetal Force Lab

• This lab report will include only three parts:– Theoretical development section.– Data table from your class.– A graph the conical pendulum data from the

data such that the acceleration due to gravity can be obtained from the slope of a line.

2nd Period Conical Pendulum Data

# osc. Total time

(t, s)

Period

(T, s)

Radius

(R, m)

Length

(L,m)

10 14.9 1.49 0.07 0.70

5 14.96 2.95 0.50 2.21

5 5.97 1.19 0.09 0.57

3

2.70 0.31 1.86

1.31

10 11.02 0.06 0.39

7th Period Conical Pendulum Data

# osc. Total time

(t, s)

Period

(T, s)

Radius

(R, m)

Length

(L,m)

Non-uniform circular motion

Friday, September 26, 2008

Announcements

• Turn in Ch 6, problems #9,11,13

• Homework Questions?

Non-uniform Circular Motion

• Consider circular motion in which either speed of the rotating object is changing, or the forces on the rotating object are changing.

• If the speed changes, there is a tangential as well as a centripetal component to the force.

• In some cases, the magnitude of the centripetal force changes as the circular motion occurs.

Problem: You swing a 0.25-kg rock in a vertical circle on a 0.80 m long rope at 2.0 Hz. What is the tension in the rope a) at the top and b) at the bottom of your swing?

Problem: A 40.0 kg child sits in a swing supported by 3.00 m long chains. If the tension in each chain at the lowest point is 350 N, find a) the child’s speed at the lowest point and b) the force exerted by the seat on the child at the lowest point.

Problem: A 900-kg automobile is traveling along a hilly road. If it is to remain with its wheels on the road, what is the maximum speed it can have as it tops a hill with a radius of curvature of 20.0 m?

Look Around You

C:\Dont Archive\Science Videos\Look around you - series 1\look.around.you.module.0.-.calcium.dvd.extra.pilot.avi

Non-Constant Forces(Forces that are functions of time)


Announcements

• Turn in Ch 6, problems #9,11,13, if you didn’t turn them in Friday.

• Turn in lab book, if you didn’t already (technically it’s due today).

• Friday’s homework was Ch 6: #14,15,21

• Tonight’s homework is page 10-11 in your classwork packet.

Non-constant Forces

• Up until this time, we have mainly dealt with forces that are constant. These produce a uniform, constant acceleration. Kinematic equations can be used with these forces.

• However, not all forces are constant.• Forces can vary with time.• Forces can vary with velocity.• Forces can vary with position.

Calculus Concepts forForces That Vary With Time

• The Derivative• The derivative yields tangent lines and slopes.• Use the derivative to go from position -> velocity ->

acceleration

• The Integral• The integral yields the area under a curve.• Use the integral to go from acceleration -> velocity ->

position

• In these two approaches, F = ma, so methods for determining or using acceleration are extendable to force.

Tutorial on the Integral• Mathematically, the integral is used to calculate a sum

composed of many, many tiny parts.• In physics, the integral is used to find a measurable change

resulting from very small incremental changes. It’s useful to think of it as a fancy addition process or a multiplication process, depending on the situation.

• Here’s a specific example. Velocity times time gives displacement. If the velocity is changing with time, but a very tiny time change is used to calculate a very tiny displacement, we can nonetheless assume the velocity was constant during that tiny time change.

• If we calculate tiny displacements this way (recalculating our velocity for each time increment), then add the tiny displacements up to get a larger displacement, we have done “integration”.

Velocity as a Function of Time Displacement

• Velocity can be represented as follows:

• Rearrangement of this expression yields:

• What this means is that we can calculate a tiny displacement dx from the velocity v at a given time times a tiny time increment dt.

dxv

dt

dx vdt

Summing the Displacements

• When we sum up these tiny displacements, we use the following notation:

• This notation indicates that we are summing up all the little displacements dx starting at position xo at time to until we reach a final position and time, xf and tf. The velocity v may be a function of time, and may be slightly different for one time increment dt and the next time increment.

f f

o o

x t

x tdx vdt

Evaluating Integrals• We will evaluate polynomial integrals by reversing

the process we used in taking a polynomial derivative. This process is sometimes called “anti-differentiation”, or “doing an anti-derivative”.

• The general method for doing an anti-derivative is:

• Can you see how this is the reverse of taking a derivative? (Note: This “indefinite integral” requires us to add a constant C to compensate for constants that may have been lost during differentiation…but we will do “definite integrals” that do not require C.)

1

if then 1

nndx At

At x Cdt n

Evaluating Definite Integrals

• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)

2

2 2

4 1

23 2

41

2.03 2

1.0

Starting time: 1 s; ending time: 2 s.

Starting position:4 m

where 3 2 5

(3 2 5)

3 2 5

3 2 1

4 5

4 22 7 15

19

f

f

x

x

f

f

f

dxv v t t

dt

dx t t dt

t t tx

x t t t

x

x



2

2 2

4 1

23 2

4

1

2.03 2

1.0



where 3 2 5

(3 2 5)

3 2 5

3 2 1

4 5

4 22 7 15

19

f

f

x

x

f

f

f

dxv v t t

dt

dx t t dt

t t tx

x t t t

x

x



2

2 2

4 1

23 2

4

1

23 2

1



where 3 2 5

(3 2 5)

3 2 5

3 2 1

4 5

4 22 7 15

19

f

f

x

x

f

f

f

dxv v t t

dt

dx t t dt

t t tx

x t t t

x

x

• Sample problem: Consider a force that is a function of time:

F(t) = (3.0 t – 0.5 t2)N• If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the

resulting velocity and position of the particle?

• Sample problem: Consider a force that is a function of time:

F(t) = (16 t2 – 8 t + 4)N• If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the

resulting change in velocity of the particle?

Review: Graphical Integration

• When we are given a graph, we can perform graphical integration.

• This is equivalent of “multiplying the axes” or “determining the area under the curve”.

• Let’s take a look at an example.

Displacement

-1.0

t (s)

v (m/s)

1.0 2.0 3.0 4.0

1.0

Consider the graph of velocity versus time shown above. Now determine, first from an equation and then directly from the graph, the displacement during the first second (from 0 to 1.0 s).

Displacement

-1.0

t (s)

v (m/s)

1.0 2.0 3.0 4.0

1.0

The velocity follows the function shown above between zero and 1.0 s. An integral is performed on that function between zero and 1.0 s to determine the displacement during that time interval.

1.0

0 0

0.60

0.60

0.60

0.6f

o

x

v v at

v t

dxt

dtdx tdt

dx tdt

Displacement

-1.0

t (s)

v (m/s)

1.0 2.0 3.0 4.0

1.0

Evaluation of the integral yields a displacement of 0.30 m. This is exactly the same displacement you would get by taking the area of the colored triangle.

0

1.0

0 0

1.02

0

0.6

0.60

2

0.30

f

f

x

x

x

dx tdt

tx

x m

Non-Constant Forces(Forces that are functions of

velocity)Tuesday, September 30, 2008

Announcements

• Turn in Ch 6, problems #14,15,21

• Tomorrow I will stamp 10-11 in your class work packet.

• Tonight’s HW is Ch 6, problems #30,31,33

Drag Forces

• Drag forces are functions of velocity, rather than functions of time. They– slow an object down as it passes through a

fluid.– act in opposite direction to velocity.– vary with velocity or velocity squared.– depend upon the fluid and the projectile

characteristics.– will impose a terminal velocity if a falling

object reaches a high enough speed.

Drag as a Function of Velocity

• In general, a drag force acts like a velocity-dependent kinetic friction. The general expression is– fD = bv + cv2

• b and c depend upon– shape and size of object– properties of fluid

• b is more important at low velocity• c is more important at high velocity

Drag Force in Free Fall

mg mg

fD

mg

fD

mg

fD

when fD equals mg, terminal velocity has been reached

Drag Force in Free Fall

for fast moving objects

FD = c v2

c = 1/2 D AWhereD = drag coefficient = density of fluidA = cross-sectional area of projectile

FD = bv + c v2

for slow moving objects

FD = b v

mg

FD

Problem: For slow-moving objects, show that vT = mg/b, where vT is terminal velocity.

Problem: For slow-moving objects, show that prior to attaining terminal velocity, the velocity of an object subjected to the drag force varies as:

( ) 1bt

mmgv t e

b

Problem: For fast-moving objects, show that terminal velocity is

2T

mgv

D A

Problem: Derive an expression for the velocity of a falling fast-moving object as a function of time. The expression you derive should be valid prior to attainment of terminal velocity, as well as after terminal velocity is attained.

Workday – Drag Problems

Wednesday, October 1, 2008

Announcements

• Get out classwork packet for my stamp (pages 10-11).

• Check pages 10-11 against scoring guide.

• Check work for the problems we worked yesterday in class.

• Tomorrow Ch 6, problems #30,31,33

Coffee Filter Prelab

Thursday, October 2, 2008

Announcements

• Pass forward Ch 6, problems #30,31,33

• Tomorrow I will stamp CW Packet page 7 (drag problem).

• Tomorrow: coffee filter lab.

Newton’s Laws .

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