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Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

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Page 1: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Newton’s Laws

Applications

Page 2: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Tuesday, September 28, 2010

Introduction to Friction

Page 3: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Friction Friction is the force that opposes a sliding

motion. Friction is due to microscopic irregularities

in even the smoothest of surfaces. Friction is highly useful. It enables us to

walk and drive a car, among other things. Friction is also dissipative. That means it

causes mechanical energy to be converted to heat. We’ll learn more about that later.

Page 4: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Microscopic View

W

N

Friction may or may not exist between two surfaces. If it exists, it opposes the direction object “wants” to slide. It is parallel to the surface.

Fpushf

(friction)

Small view:Microscopic irregularities resist movement.

Big view:Surfaces look perfectly smooth.

Page 5: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Friction depends on the normal force.

The friction force that exists between two surfaces is directly proportional to the normal force.

Increasing the normal force increases friction; decreasing the normal force decreases friction.

This has several implications, such as… Friction on a sloping surface is less than friction on a flat

surface (since the normal force is less on a slope). Increasing weight of an object increases the friction

between the object and the surface it is resting on. Weighting down a car over the drive wheels increases

the friction between the drive wheels and the road (which increases the car’s ability to accelerate).

Page 6: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Static Friction This type of friction occurs

between two surfaces that are not slipping relative to each other.

fs sN fs : static frictional force (N) s: coefficient of static friction N: normal force (N)

Page 7: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

fs < sN is an inequality! The fact that the static friction equation is an

inequality has important implications. Static friction between two surfaces is zero unless

there is a force trying to make the surfaces slide on one another.

Static friction can increase as the force trying to push an object increases until it reaches its maximum allowed value as defined by s.

Once the maximum value of static friction has been exceeded by an applied force, the surfaces begin to slide and the friction is no longer static friction.

Page 8: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Static friction and applied horizontal force

Physics

N

W

Force Diagram

surface

There is no static friction since there is no applied horizontal force trying to slide the book on the surface.

Page 9: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Static friction and applied horizontal force

Physics

N

W

Force Diagram

surfaceFfs

Static friction is equal to the applied horizontal force, and there is no movement of the book since F = 0.

Page 10: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Static friction and applied horizontal force

Physics

N

W

Force Diagram

surfaceFfs

Static friction is at its maximum value! It is still equal to F, but if F increases any more, the book will slide.

Page 11: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Static friction and applied horizontal force

Physics

N

W

Force Diagram

surfaceFfk

Static friction cannot increase any more! The book accelerates to the right. Friction becomes kinetic friction, which is usually a smaller force.

Page 12: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Static friction on a ramp

Physics

N

surface

f s

At maximum angle before the book slides, let’s prove that s = tan

W = mg

Without friction, the book will slide down the ramp. If it stays in place, there is sufficient static friction holding it there.

Page 13: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Kinetic Friction This type of friction occurs between

surfaces that are slipping past each other.

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.

Page 14: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample ProblemA 10-kg box rests on a ramp that is laying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30.

a) What is the maximum horizontal force that can be applied to the box before it begins to slide?

b) What force is necessary to keep the box sliding at constant velocity?

Page 15: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Problem – Horizontal RampA 10-kg wooden box rests on a ramp that is lying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp if

a) no force horizontal force is applied to the box?

b) a 20 N horizontal force is applied to the box?

c) a 60 N horizontal force is applied to the box?

Page 16: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Problem – Ramp at an AngleA 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp if

a) the ramp is at a 25o angle?

b) the ramp is at a 45o angle?

c) what is the acceleration of the box when the ramp is at 45o?

Page 17: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Monday, October 4, 2010

Strings and TensionSprings

Page 18: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Announcements Homework

Page 19: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Tension Tension is a pulling force that arises when a

rope, string, or other long thin material resists being pulled apart without stretching significantly.

Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string.

Page 20: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

A physical picture of tension

Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string.

Page 21: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Tension examples

Note that the pulleys shown are magic! They do not affect the tension in any way, and serve only to bend the line of action of the force.

Page 22: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Tension Demo

Page 23: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample problem - TensionA. A 1,500 kg crate hangs motionless from a crane cable. What is

the tension in the cable? Ignore the mass of the cable.

B. Suppose the crane accelerates the crate upward at 1.2 m/s2. What is the tension in the cable now?

Page 24: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Springs (Hooke’s Law) The magnitude of the force exerted by a

spring is proportional to the amount it is stretched.

F = -kx F: force exerted by the spring (N) k: force constant of the spring (N/m or N/cm) x: displacement from equilibrium (unstretched

and uncompressed) position (m or cm) The direction of the force is back toward

the equilibrium (or unstretched) position.

Page 25: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample problem 1 - Spring• A 1.50 kg object hangs motionless from a spring with a

force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?

Page 26: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample Problem 2 - Spring• A 1.80 kg object is connected to a spring of force constant 120

N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60.

Page 27: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Connected ObjectsLot’s of Tension Today!

Page 28: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Announcements

Page 29: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample problem

A 5.0 kg object (m1) is connected to a 10.0 kg object (m2) by a string. If a pulling force F of 20 N is applied to the 5.0 kg object as shown,

A) what is the acceleration of the system?

B) what is the tension in the string connecting the objects?

(Assume a frictionless surface.)

Page 30: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Gravity

A very common accelerating force is gravity. Here is gravity in action. The acceleration is g.

Page 31: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

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.

Slowing gravity down

Page 32: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Magic pulleys on a flat table

Magic pulleys bend the line of action of the force without affecting tension.

Frictionless table

m1

m2

T

m2g

N

m1g

T

-x

x

F = mam2g + T – T = (m1 + m2)aa = m2g/(m1+m2)

Page 33: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

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 <Num 16>. (b)the tension in the connecting string <Num 17>.

Sample problem

m1

m2

Page 34: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary <NUM 18>.

Sample problem

m1

m2

Page 35: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Monday, October 8, 2007

More with Pulleys

Page 36: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Announcements HW #2 (FR#1) checked today. HW #3 (FR #2) checked tomorrow. HW #4 (FR #3) checked

Wednesday Clicker Questions: 19-22 Ranking Task Let’s go over HW #2

Page 37: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Tuesday, October 9, 2007

Pulleys and Ramps --Together at Last!

Page 38: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Announcements HW #3 (spring/pulley) checked

today. HW #4 (the dang monkey again!)

checked tomorrow. HW #5 (ramp and pulley) checked

Thursday.

Page 39: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If s = 0.30 and k = 0.20, what is (a)the acceleration of each block <NUM 23>? (b)the tension in the connecting string<NUM 24>?

Sample problem

m1

m2

Page 40: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

This week’s lab… Determine coefficient of static

friction for a block on a ramp. Determine coefficient of kinetic

friction for a block on a ramp.

Page 41: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Magic pulleys on a ramp It’s a little more complicated when a magic pulley is

installed on a ramp.

m 1 m2

F = mam2g -T + T – m1gsin = (m1+m2)am2g – m1gsin= (m1+m2)aa = (m2 – m1sin)g/(m1+m2)

m1g

N T

T

m2g

m1gsin

m1gcos

Page 42: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample problem Two blocks are connected by a string as

shown in the figure. What is the acceleration, assuming there is no friction?

10 kg5 kg

Page 43: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample problem - solution

10 kg5 kg

F = mam2g -T + T – m1gsin = (m1+m2)am2g – m1gsin= (m1+m2)aa = (m2 – m1sin)g/(m1+m2)a = [(5 – 10sin45o)(9.8)]/15a = - 1.35 m/s2

m1g

N T

T

m2g

m1gsin

m1gcos

Page 44: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample problem - solution

10 kg5 kg

F = mam2g -T + T – m1gsin = (m1+m2)am2g – m1gsin= (m1+m2)aa = (m2 – m1sin)g/(m1+m2)a = [(5 – 10sin45o)(9.8)]/15a = - 1.35 m/s2

m1g

N T

T

m2g

m1gsin

m1gcos

How would this change if there is friction on the ramp?

Page 45: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Ranking Tasks What do you know about

alignment of forces and magnitude of acceleration?

What do you remember about elevators?

Page 46: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Tuesday, October 17, 2006

Uniform Circular Motion

Page 47: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Uniform Circular Motion An object that moves at uniform speed

in a circle of constant radius is said to be in uniform circular motion.

Question: Why is uniform circular motion accelerated motion?

Answer: Although the speed is constant, the velocity is not constant since an object in uniform circular motion is continually changing direction.

Page 48: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Centrifugal Force Question: What is centrifugal force? Answer: That’s easy. Centrifugal force

is the force that flings an object in circular motion outward. Right?

Wrong! Centrifugal force is a myth! There is no outward directed force in

circular motion. To explain why this is the case, let’s review Newton’s 1st Law.

Page 49: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Newton’s 1st Law and cars•When a car accelerates forward suddenly, you as a passenger feel as if you are flung backward.

• You are in fact NOT flung backward. Your body’s inertia resists acceleration and wants to remain at rest as the car accelerates forward.

•When a car brakes suddenly, you as a passenger feel as if you are flung forward.

• You are NOT flung forward. Your body’s inertia resists acceleration and wants to remain at constant velocity as the car decelerates.

Page 50: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

You feel as if you are flung to the outside. You call this apparent, but nonexistent, force “centrifugal force”.

You are NOT flung to the outside. Your inertia resists the inward acceleration and your body simply wants to keep moving in straight line motion!

As with all other types of acceleration, your body feels as if it is being flung in the opposite direction of the actual acceleration. The force on your body, and the resulting acceleration, actually point inward.

When a car turns

Page 51: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Centripetal Acceleration Centripetal (or center-seeking)

acceleration points toward the center of the circle and keeps an object moving in circular motion.

This type of acceleration is at right angles to the velocity.

This type of acceleration doesn’t speed up an object, or slow it down, it just turns the object.

Page 52: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Centripetal Acceleration ac = v2/r

ac: centripetal acceleration in m/s2

v: tangential speed in m/s

r: radius in meters

v ac

Centripetal acceleration always points toward center of circle!

Page 53: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Centripetal Force

A force responsible for centripetal acceleration is referred to as a centripetal force.

Centripetal force is simply mass times centripetal acceleration.

Fc = m ac

Fc = m v2 / r Fc: centripetal force in N v: tangential speed in m/s r: radius in meters

Fc

Always toward center of circle!

Page 54: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Any force can be centripetal The name “centripetal” can be applied

to any force in situations when that force is causing an object to move in a circle.

You can identify the real force or combination of forces which are causing the centripetal acceleration.

Any kind of force can act as a centripetal force.

Page 55: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Static friction

As a car makes a turn on a flat road, what is the real identity of the centripetal force?

Page 56: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Tension

As a weight is tied to a string and spun in a circle, what is the real identity of the centripetal force?

Page 57: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Gravity

As the moon orbits the Earth, what is the real identity of the centripetal force?

Page 58: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Normal force with help from static friction

As a racecar turns on a banked curve on a racing track, what is the real identity of the centripetal force?

Page 59: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Tension,with some help from

gravity

As you swing a mace in a vertical circle, what is the true identity of the centripetal force?

Page 60: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Gravity, with some help from

the normal force

When you are riding the Tennessee

Tornado at Dollywood, what is the real identity of

the centripetal force when you are on a vertical loop?

Page 61: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample problem• A 1200-kg car rounds a corner of radius r = 45 m. If the

coefficient of static friction between tires and the road is 0.93 and the coefficient of kinetic friction between tires and the road is 0.75, what is the maximum velocity the car can have without skidding?

Page 62: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Thursday, October 19, 2006

Class workday

Page 63: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Friday, October 20, 2006

Inquiry Based Lab

Page 64: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Announcements HW due TODAY:

Newton’s #3-#6. Full lab report from Monday’s lab.

My room will be unavailable at lunch today, as I will be writing letters of recommendation. I will be available every morning but Thursday.

Page 65: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Centripetal Force and Friction Lab• Using only the hand strobe, a penny, stopwatch, chalk, and ruler,

determine the coefficient of friction between a penny and the hand strobe.

• Rules, hints, and tips:– You may write on the hand strobe with the chalk.– You must use centripetal force in your analysis.– Can you spin the hand strobe at a gradually increasing rate?– Can you spin the hand strobe at a constant rate?– The lowest rotational speed necessary to make the penny fly off the hand

strobe is an important number! How will you measure this speed with high accuracy?

• One report per group (handwritten) is due at the end of class. It should include a procedure, a good free body diagram, all data you collect, and a clear Newton’s 2nd Law analysis. Names of group members must be on the report. You will be graded on how well you develop a procedure to do this analysis correctly, your application of Newton’s 2nd Law, and your results.

Page 66: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Monday, October 30, 2006

Begin Exam Review forNewton’s Laws II

Page 67: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Announcements Federal Forms – does anyone have

them. Free response exam Wednesday. Lunch bunch resumes Wednesday. Today, we will review Newton’s Laws by

starting the packet. This MUST be completed by tomorrow when you come to class if you are going to be allowed to correct the free response exam given on Wednesday.

Page 68: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Sample problemYou whirl a 2.0 kg stone in a horizontal circle about your head. The rope attached to the stone is 1.5 m long.

a) What is the tension in the rope? (The rope makes a 10o angle with the horizontal).

b) How fast is the stone moving?

Page 69: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Monday, October 31, 2006

Exam Review

Page 70: Newtons Laws Applications. Tuesday, September 28, 2010 Introduction to Friction.

Announcements

• Federal Forms – does anyone have them?• Free response exam tomorrow.• Get out your free response packet for my

inspection.