FORCE

Newton’s Laws

Three Laws of MotionThree Laws of Motion

Aristotle’s Motion

Natural Motion is up or downNatural Motion is up or down Down for falling objectsDown for falling objects Up for smokeUp for smoke Circular for heavenly bodies since without endCircular for heavenly bodies since without end

Violent Motion Violent Motion Due to imposed forces such as wind pushing a Due to imposed forces such as wind pushing a

ship or someone pulling a cartship or someone pulling a cart Natural state of motion is restNatural state of motion is rest

A force is needed to keep something movingA force is needed to keep something moving

Aristotle’s Basic Error

Friction not understood as a forceFriction not understood as a force

Galileo’s Motion

Force is a push or a pullForce is a push or a pull Friction is a force that Friction is a force that

occurs when objects move occurs when objects move past each otherpast each other

Friction due to tiny Friction due to tiny irregularitiesirregularities

OnlyOnly when friction is when friction is present is a force required present is a force required to keep something movingto keep something moving

Galileo’s Inclined Planes

Ball rolling downhill Ball rolling downhill speeds upspeeds up

Ball rolling uphill Ball rolling uphill slows downslows down

He asked about ball on He asked about ball on smooth level surfacesmooth level surface

Concluded it would Concluded it would roll forever in absence roll forever in absence of frictionof friction

Inertia

Resistance to change in state of motionResistance to change in state of motion Galileo concluded all objects have inertiaGalileo concluded all objects have inertia Contradicted Aristotle’s theory of motionContradicted Aristotle’s theory of motion No force required to keep Earth in motion No force required to keep Earth in motion

around sun because no frictionaround sun because no friction

Newton

Born 1665Born 1665 Built on Galileo’s Built on Galileo’s

ideasideas Proposed three laws of Proposed three laws of

motion at age of 23motion at age of 23

Newton’s First Law

Every object continues in its state of rest, or of Every object continues in its state of rest, or of motion in a straight line at constant speed, unless motion in a straight line at constant speed, unless compelled to change that state by forces exerted compelled to change that state by forces exerted on it.on it.

Also called Law of Inertia: things move according Also called Law of Inertia: things move according to their own inertiato their own inertia

Things keep on doing what they are doingThings keep on doing what they are doing Examples: Hockey puck on ice, rolling ball, ball Examples: Hockey puck on ice, rolling ball, ball

in space, person sitting on couchin space, person sitting on couch

Ourtesy www.lakeheadu.ca/~alumni/ hockey.gif

Mass

Amount of inertia depends on amount of Amount of inertia depends on amount of mass…or amount of material (number and mass…or amount of material (number and kind of atoms)kind of atoms)

Measured in kilogramsMeasured in kilograms Question: Which has more mass, a Question: Which has more mass, a

kilogram of lead or a kilogram of feathers?kilogram of lead or a kilogram of feathers? Mass vs. Volume: volume is how much Mass vs. Volume: volume is how much

space something occupiesspace something occupies

Experiencing Inertia

Inertia is resistance to shakingInertia is resistance to shaking Which is easier to shake, a pen or a person?Which is easier to shake, a pen or a person? Why is it so hard to stop a heavy boat?Why is it so hard to stop a heavy boat?

Inertia in a Car Discuss three examples of inertia in a carDiscuss three examples of inertia in a car

•Car hitting a wall

•Car hit from behind by a truck

•Car going around a corner

Newton’s Second Law

Law of AccelerationLaw of Acceleration The acceleration produced by a net force on The acceleration produced by a net force on

an object is directly proportional to the an object is directly proportional to the magnitude of the net force, and is inversely magnitude of the net force, and is inversely proportional to the mass of the body.proportional to the mass of the body.

Acceleration = net force Acceleration = net force ÷mass÷mass F =maF =ma Acceleration is Acceleration is in direction of net forcein direction of net force

Units

F = maF = ma Unit of force is the Newton (N)Unit of force is the Newton (N) 1 N = 1 kg m/s1 N = 1 kg m/s22

Net Force

Net Force means sum of all forces actingNet Force means sum of all forces acting Sum is Vector sumSum is Vector sum

F1

F2

Resultant force

Understanding the Second Law

The cause of acceleration is…The cause of acceleration is… _________ resists acceleration_________ resists acceleration The greater the force, the ________ the The greater the force, the ________ the

____________________________ The greater the mass, the _________ the The greater the mass, the _________ the

acceleration.acceleration.

Force

Mass or inertia

acceleration

less

greater

F = ma is Three Equations

F F andand a a are vectorsare vectors So So FF = m = maa equation is really three equation is really three

FFxx = ma = max x FFyy = ma = may y FFzz = ma = mazz

Examples

What force is required to accelerate a 1000 What force is required to accelerate a 1000 kg car at 2.0 m/skg car at 2.0 m/s2 2 ??

Answer: F = ma = 1000 kg x 2.0 m/sAnswer: F = ma = 1000 kg x 2.0 m/s2 2 = = 2000 N.2000 N.

What is the acceleration of a 145 g baseball What is the acceleration of a 145 g baseball thrown with a force of 20.0 N?thrown with a force of 20.0 N?

a = F/m = 20.0 N/0.145kg = 138 m/sa = F/m = 20.0 N/0.145kg = 138 m/s22

F = ma Example; m unknown

An astronaut puts a 500.0 N force on an An astronaut puts a 500.0 N force on an object of unknown mass producing an object of unknown mass producing an accelerations of 0.462 m/saccelerations of 0.462 m/s2 2 . What was the . What was the mass?mass?

M = F/a = 500.0N/0.462 m/sM = F/a = 500.0N/0.462 m/s2 2 = 1082 Kg = = 1082 Kg = 1.08 x 101.08 x 103 3 KgKg

Net force exampleIf four teams are playing tug of war (imagine a rope If four teams are playing tug of war (imagine a rope

that looks like a cross, with the flag tied in the that looks like a cross, with the flag tied in the middle). Each team is 90middle). Each team is 90⁰ from each other. ⁰ from each other. Team A pulls with an overall force of 350 N to the Team A pulls with an overall force of 350 N to the North, Team B pulls with an overall force of 270 N North, Team B pulls with an overall force of 270 N to the South, Team C pulls with an overall force to the South, Team C pulls with an overall force of 150 N to the East and Team D pulls with an of 150 N to the East and Team D pulls with an overall force of 250 N to the West. If the flag in overall force of 250 N to the West. If the flag in the middle has a mass of .25 kg, what is the the middle has a mass of .25 kg, what is the magnitude and direction of its acceleration? magnitude and direction of its acceleration?

Putting it all together…….

Calculate the change in force of a car that has Calculate the change in force of a car that has a mass of 2500 kg if it goes from 45 m/s to a mass of 2500 kg if it goes from 45 m/s to rest in 7 seconds at a stop sign, then rest in 7 seconds at a stop sign, then accelerates up to 65 m/s in 5 seconds.accelerates up to 65 m/s in 5 seconds.

a= vf-vi/t or a = F/m

aa11 = 0-45/7 = -6.42 m/s = 0-45/7 = -6.42 m/s22

aa22 = 65-0/5 = 13 m/s = 65-0/5 = 13 m/s22

The difference between them is 19.42 m/sThe difference between them is 19.42 m/s22..

F = m x a = 2500 kg x 19.42 m/sF = m x a = 2500 kg x 19.42 m/s22

= 48550 N difference between = 48550 N difference between the the two accelerationstwo accelerations

Newton’s Third Law Forces always come in pairsForces always come in pairs Two forces on different objectsTwo forces on different objects Every action has an equal and opposite Every action has an equal and opposite

reactionreaction Whenever one object exerts a force on a Whenever one object exerts a force on a

second object, the second exerts an equal second object, the second exerts an equal and opposite force on the firstand opposite force on the first

Example: hammer hits nailExample: hammer hits nail

Example: pushing on wall

What are the forces when you push on a What are the forces when you push on a wall?wall?

You exert force on wallYou exert force on wall You accelerate in the opposite directionYou accelerate in the opposite direction Wall must have exerted a force on you in Wall must have exerted a force on you in

the direction you accelerated (by 2the direction you accelerated (by 2ndnd Law) Law)

Example: person walking

Foot exerts force Foot exerts force backward on groundbackward on ground

Ground exerts force Ground exerts force forward on footforward on foot

Example: Throwing ball

Pitcher exerts force on Pitcher exerts force on ballball

Ball exerts equal and Ball exerts equal and opposite force on opposite force on pitcherpitcher

Why doesn’t pitcher Why doesn’t pitcher move?move?

Example: Rocket

Rocket engine exerts Rocket engine exerts rearward force on gas rearward force on gas moleculesmolecules

Molecules exert Molecules exert forward force on forward force on rocket.rocket.

Book on Table

The mass of the book is one kg. What is The mass of the book is one kg. What is the force (magnitude and direction) the force (magnitude and direction) onon the the book?book?

9.8 N upward9.8 N upward

Really putting it all together……Calculate the Force necessary to launch a Calculate the Force necessary to launch a

cannonball with a mass of 15 kg if it is fired cannonball with a mass of 15 kg if it is fired at an angle of 43at an angle of 43⁰ if it hits a target 210 m ⁰ if it hits a target 210 m away in 6.3 seconds? away in 6.3 seconds?

What can we solve in this problem?What can we solve in this problem?What equations do we need to solve this What equations do we need to solve this

problem? problem?

What we need to solve the force

Vx = dx/t = 210/6.3 = 33.3 m/sVx = dx/t = 210/6.3 = 33.3 m/sVfVf22 = Vi = Vi2 2 + 2 a(d) Vi = 0 for this problem + 2 a(d) Vi = 0 for this problema = Vfa = Vf22/2d = 33.3/2d = 33.322 / 2(210) = 2.64 m/s / 2(210) = 2.64 m/s22

Force of the cannon: F = m(a)Force of the cannon: F = m(a)F = 15 kg (2.64 m/sF = 15 kg (2.64 m/s22) = 39.6 N) = 39.6 N

The Horse and the Cart Problem

If there is always an equal an opposite If there is always an equal an opposite reaction, how does anything move? For reaction, how does anything move? For example, if you have a horse and a cart, example, if you have a horse and a cart, how does the horse pull the cart?how does the horse pull the cart?

The Horse and Cart Problem.

A= - B B= - C C= -D A=B=C=D no acc!

These appear to be the equalizing forces.These appear to be the equalizing forces.

The Horse and Cart Problem.

Because it is accelerating, the force the horse exerts on the cart has increased. By Newton's third law, the force of the cart on the horse has increased by the same amount. But the horse is also accelerating, so the friction of the ground on its hooves must be larger than the force the cart exerts on the horse. The friction between hooves and ground is static (not sliding or rolling) friction, and can increase as necessary (up to a limit, when slipping might occur, as on a slippery mud surface or loose gravel). So, when accelerating, we still have B = -C, by Newton's third law, but D>C and B>A, so D>A.

More Examples

Can you think of some more examples of Can you think of some more examples of Newton’s Third Law in Action?Newton’s Third Law in Action?

Imagine an astronaut floating in deep space, Imagine an astronaut floating in deep space, with only his spacesuit. Is there any way for with only his spacesuit. Is there any way for him to move himself back to earth? him to move himself back to earth?

Mass vs. Weight

Mass is intrinsic property of any objectMass is intrinsic property of any object Weight measures gravitational force on an Weight measures gravitational force on an

object, usually due to a planetobject, usually due to a planet Weight depends on location of objectWeight depends on location of object Question 1: How does mass of a rock Question 1: How does mass of a rock

compare when on Earth and on moon?compare when on Earth and on moon? Question 2: How does its weight compare?Question 2: How does its weight compare?

Review Mass vs. Weight

What is mass?What is mass? Answer: quantity of Answer: quantity of

matter in something or matter in something or a measure of its inertiaa measure of its inertia

What is weight?What is weight? Answer: Force on a Answer: Force on a

body due to gravitybody due to gravity

Weight of 1 Kilogram

9.8 Newtons9.8 Newtons About 2.2 poundsAbout 2.2 pounds Compare the weight of 1 kg nails with 1 kg Compare the weight of 1 kg nails with 1 kg

styrofoamstyrofoam Answer: SameAnswer: Same

Weight Examples

What does a 70 kg person weigh?What does a 70 kg person weigh?

Weight = mass x g(acceleration due to gravity)Weight = mass x g(acceleration due to gravity)

W = mg = 70 kg x 9.80 N/mW = mg = 70 kg x 9.80 N/m2 =2 = 686 N 686 N An object weighs 9800 N on Earth. What is its An object weighs 9800 N on Earth. What is its

mass?mass? m = W/g = 9800 / 9.8 m/sm = W/g = 9800 / 9.8 m/s2 2 == 1000 kg1000 kg

Atwoods Lab You have 25 washers on your lab setup, if you have a unbalanced You have 25 washers on your lab setup, if you have a unbalanced

force, you will have acceleration. You will be using the stopwatch force, you will have acceleration. You will be using the stopwatch function of your data collector.function of your data collector.

Make a chart to record mass, time, acceleration and force.Make a chart to record mass, time, acceleration and force. Put all washers on one side, raise that side to the top, then release it Put all washers on one side, raise that side to the top, then release it

timing how long it takes to reach the bottom. Record this time. timing how long it takes to reach the bottom. Record this time. The mass of one washer is 16 g. It is the difference in mass that The mass of one washer is 16 g. It is the difference in mass that

causes the acceleration. Calculate the difference in mass and causes the acceleration. Calculate the difference in mass and record in table. 1record in table. 1stst mass is 25 x 16, 2 mass is 25 x 16, 2ndnd mass is 23 x 16, 3 mass is 23 x 16, 3rdrd mass is mass is 21 x 16 etc. 21 x 16 etc.

Calculate the Acceleration = 2d/tCalculate the Acceleration = 2d/t22 (d = 1 m for the fall) so (d = 1 m for the fall) so a = 2/ ta = 2/ t2 2

Calculate the Net force of the fall and record. (Calculate the Net force of the fall and record. (F= maF= ma)) Move one washer at a time over to the other side and repeat.Move one washer at a time over to the other side and repeat. Continue until the machine no longer turns (12 or 13 trials)Continue until the machine no longer turns (12 or 13 trials)

FRICTION

Sliding (motion) & Static (stationary)Sliding (motion) & Static (stationary)

Sliding Friction

Often called kinetic frictionOften called kinetic friction A force opposite to direction of motionA force opposite to direction of motion Due to bumps in surfaces and electric forcesDue to bumps in surfaces and electric forces

Ff

Surface under microscope

Kinetic Friction is…

Dependent on nature of the two surfacesDependent on nature of the two surfaces Directly proportional to the normal force Directly proportional to the normal force

between the surfacesbetween the surfaces Normal Force is perpendicular to the Normal Force is perpendicular to the

surface. If it is on a flat surface, it is equal surface. If it is on a flat surface, it is equal to the weight of the object.to the weight of the object.

Independent of velocityIndependent of velocity

Reducing Friction

In order to reduce friction we can:In order to reduce friction we can: A. Reduce surface areaA. Reduce surface area B. Reduce weight of objectB. Reduce weight of object C. Change type of frictionC. Change type of friction

- sliding(the greatest amount)- sliding(the greatest amount) - rolling (use wheels to ease friction)- rolling (use wheels to ease friction) - fluid ( Eliminate contact by using liquids or - fluid ( Eliminate contact by using liquids or

gases)gases)

Coefficient of friction k

Generally between zero and oneGenerally between zero and one Based on comparing Friction Force to Based on comparing Friction Force to

Normal ForceNormal Force Normal Force is always perpendicular to Normal Force is always perpendicular to

surfacesurface Calculate from FCalculate from Ff f / F/ FN = N = µµkk

Can be more thanCan be more than one for special rubberone for special rubber Very low for ice, Teflon, lubricated Very low for ice, Teflon, lubricated

surfaces, ball bearingssurfaces, ball bearings

Friction: Good or Bad

Mostly undesirable since reduces useful Mostly undesirable since reduces useful force and wastes energyforce and wastes energy

Friction produces heatFriction produces heat Necessary for walking!Necessary for walking! Necessary for braking Necessary for braking

Static Friction

Force to start something movingForce to start something moving Usually larger than kinetic friction for same Usually larger than kinetic friction for same

surfacessurfaces Requires force to be exerted Requires force to be exerted Before sliding begins, is equal and opposite Before sliding begins, is equal and opposite

to applied forceto applied force

Where are all the forces?

Block on an inclined planeBlock on an inclined plane

Free Body Diagram Example 1

If the box below accelerates to the right at 1 m/s2 Solve all of the following:

Solution 1 Fgrav = m x g = 5 x 9.8 = 49 NFgrav = m x g = 5 x 9.8 = 49 N Using the angle and the F applied, we can Using the angle and the F applied, we can

calculate the X and Y component of that force.calculate the X and Y component of that force. Fx= 15 sin 45 Fy = 15 cos 45Fx= 15 sin 45 Fy = 15 cos 45 Fx = 10.6 N Fy = 10.6 NFx = 10.6 N Fy = 10.6 N If the force of gravity is 49 N down and the If the force of gravity is 49 N down and the

applied force is 10.6 N up, then the normal applied force is 10.6 N up, then the normal force applied is the difference between the force applied is the difference between the two. F norm= 49-10.6 = 38.4 Ntwo. F norm= 49-10.6 = 38.4 N

Solution 1 cont. If the object has an a of 1 m/sIf the object has an a of 1 m/s22 and a mass and a mass

of 5 kg, then it has a net force of 5 N in the of 5 kg, then it has a net force of 5 N in the X direction.X direction.

If the applied force in the X is 10.6 and the If the applied force in the X is 10.6 and the net is 5, then the force of friction is the net is 5, then the force of friction is the difference between the two.difference between the two.

Ffric= 10.6-5 = 5.6 NFfric= 10.6-5 = 5.6 N To solve the coefficient of friction we use To solve the coefficient of friction we use

this equation: Fthis equation: Ff f = = kkFFN N

kk= F= Fff/F/FN N = 5.6/ 38.4 = .145= 5.6/ 38.4 = .145

Flat pull

If you pull a 2505 g box with a force of 15 If you pull a 2505 g box with a force of 15 N at an angle of 53N at an angle of 53⁰ to the horizon and the ⁰ to the horizon and the box accelerates at 2.0 m/sbox accelerates at 2.0 m/s22 to the right, to the right, calculate the following: calculate the following:

Fn, Fg, Ff, Fnet, Fapp, Fx, Fy and µFn, Fg, Ff, Fnet, Fapp, Fx, Fy and µ

Friction Lab Put a ramp flat in your lab space. Place two Put a ramp flat in your lab space. Place two

photogates relatively close together. photogates relatively close together. If the mass of the sled is .040 kg calculate the If the mass of the sled is .040 kg calculate the

Fnormal (Fn=Fg if on flat surface)Fnormal (Fn=Fg if on flat surface) Now, using your sled car (no wheels) launch the Now, using your sled car (no wheels) launch the

car with your rubber band. Make sure that it car with your rubber band. Make sure that it goes through both photo gates (you may have to goes through both photo gates (you may have to adjust photo gates). Use our acceleration adjust photo gates). Use our acceleration procedure from lab and calculate the rate of procedure from lab and calculate the rate of deceleration.deceleration.

Calculate Ffric= mass of sled x decelerationCalculate Ffric= mass of sled x deceleration Calculate µ = Ff/FnCalculate µ = Ff/Fn

Free body diagram example 2

Say a box is sitting on 30Say a box is sitting on 30⁰⁰ slope and is slope and is frictionless, so the only forces are the frictionless, so the only forces are the normal force and gravity. What is the normal force and gravity. What is the block's acceleration down the slope if the block's acceleration down the slope if the mass is 3.0 kg? What is the normal force?mass is 3.0 kg? What is the normal force?

Free Body Diagram example 3

A box is sitting on a 35A box is sitting on a 35⁰ inclined plane. It is ⁰ inclined plane. It is being pulled up the ramp by you with an being pulled up the ramp by you with an acceleration of 2.5 m/sacceleration of 2.5 m/s22. If the box has a . If the box has a mass of 25 kg and the force of friction is 3.5 mass of 25 kg and the force of friction is 3.5 N, solve all of the following: Fnet, Fnormal, N, solve all of the following: Fnet, Fnormal, Fgravity, Fapplied, and µ.Fgravity, Fapplied, and µ.

A 50-N applied force (30 degrees to the horizontal) A 50-N applied force (30 degrees to the horizontal) accelerates a box across a horizontal sheet of ice (see accelerates a box across a horizontal sheet of ice (see diagram). Glen Brook, Olive N. Glenveau, and diagram). Glen Brook, Olive N. Glenveau, and Warren Peace are discussing the problem. Glen Warren Peace are discussing the problem. Glen suggests that the normal force is 50 N; Olive suggests suggests that the normal force is 50 N; Olive suggests that the normal force in the diagram is 75 N; and that the normal force in the diagram is 75 N; and Warren suggests that the normal force is 100 N. Warren suggests that the normal force is 100 N. While all three answers may seem reasonable, only While all three answers may seem reasonable, only one is correct. Indicate which two answers are wrong one is correct. Indicate which two answers are wrong and explain why they are wrong.and explain why they are wrong.

Review: Newton’s Laws of Motion

Newton’s First Law:Newton’s First Law: Every object continues in its state of rest, or of motion in a Every object continues in its state of rest, or of motion in a

straight line at constant speed, unless compelled to change straight line at constant speed, unless compelled to change that state by forces exerted on it.that state by forces exerted on it.

Newton’s Second Law:Newton’s Second Law: The acceleration produced by a net force on an object is The acceleration produced by a net force on an object is

directly proportional to the magnitude of the net force, and directly proportional to the magnitude of the net force, and is inversely proportional to the mass of the body.is inversely proportional to the mass of the body.

Newton’s Third Law:Newton’s Third Law: Whenever one object exerts a force on a second object, the Whenever one object exerts a force on a second object, the

second exerts an equal & opposite force on the firstsecond exerts an equal & opposite force on the first

Action- Reaction Lab Adjust the smart track (or lab table) to be as level as Adjust the smart track (or lab table) to be as level as

possible(may have to put lab book under) put rubber possible(may have to put lab book under) put rubber band around one car.band around one car.

Squeeze two cars together and attach with the car link.Squeeze two cars together and attach with the car link. Position car in middle of track, making sure all wheels Position car in middle of track, making sure all wheels

are on track.are on track. With a quick upward motion, pull the link straight up With a quick upward motion, pull the link straight up

and out from the cars.and out from the cars. Describe how the cars move in a data table.Describe how the cars move in a data table. Start adding marbles to cars and repeat procedures Start adding marbles to cars and repeat procedures

aboveabove Make all these combinations of marbles in carsMake all these combinations of marbles in cars 0,0 0,1 0,2 0,3 1,1 1,2 1,3 2,2 2,3 3,3 0,0 0,1 0,2 0,3 1,1 1,2 1,3 2,2 2,3 3,3 Sum up the action reaction effect on cars and marbles.Sum up the action reaction effect on cars and marbles.

Draw the free body diagram

Draw the free body diagram, if a = .1 m/s2 and the force you push on the lawnmower is 25 N, solve for every force you know.

Say a box is sitting on 40Say a box is sitting on 40⁰⁰ slope ramp. If slope ramp. If the mass is 3.0 kg? What are all the forces the mass is 3.0 kg? What are all the forces acting on the box and what is µ?acting on the box and what is µ?

Renee is on Spring Break and pulling her Renee is on Spring Break and pulling her 21-kg suitcase through the airport at a 21-kg suitcase through the airport at a constant speed of 0.47 m/s. She pulls on the constant speed of 0.47 m/s. She pulls on the strap with 120 N of force at an angle of 38° strap with 120 N of force at an angle of 38° above the horizontal. Determine the normal above the horizontal. Determine the normal force and the total resistance force (friction force and the total resistance force (friction and air resistance) experienced by the and air resistance) experienced by the suitcase.suitcase.

For each collection of listed forces, For each collection of listed forces, determine the vector sum or the net force.determine the vector sum or the net force.

Set ASet A58 N, right58 N, right42 N, left42 N, left98 N, up98 N, up98 N, down98 N, down

Hector is walking his dog (Fido) around the Hector is walking his dog (Fido) around the neighborhood. Upon arriving at Fidella's neighborhood. Upon arriving at Fidella's house (a friend of Fido's), Fido turns part house (a friend of Fido's), Fido turns part mule and refuses to continue on the walk. mule and refuses to continue on the walk. Hector yanks on the chain with a 67.0 N Hector yanks on the chain with a 67.0 N force at an angle of 30.0° above the force at an angle of 30.0° above the horizontal. Determine the horizontal and horizontal. Determine the horizontal and vertical components of the tension force.vertical components of the tension force.

Helen is parasailing. She sits in a seat Helen is parasailing. She sits in a seat harness which is attached by a tow rope to a harness which is attached by a tow rope to a speedboat. The rope makes an angle of 51° speedboat. The rope makes an angle of 51° with the horizontal and has a tension of 350 with the horizontal and has a tension of 350 N. Determine the horizontal and vertical N. Determine the horizontal and vertical components of the tension force. components of the tension force.

Jerome and Michael, linebackers for South’s varsity Jerome and Michael, linebackers for South’s varsity football team, delivered a big hit to the halfback in last football team, delivered a big hit to the halfback in last weekend’s game. Striking the halfback simultaneously weekend’s game. Striking the halfback simultaneously from different directions with the following forces:from different directions with the following forces:

FFJeromeJerome = 1230 N at 53° = 1230 N at 53°

FFMichaelMichael = 1450 at 107° = 1450 at 107°

Determine the resultant force applied by Jerome and Determine the resultant force applied by Jerome and Michael to the halfback. (The directions of the two forces Michael to the halfback. (The directions of the two forces are stated as counter-clockwise angles of rotation with are stated as counter-clockwise angles of rotation with East.)East.)

2. A box is pulled at a constant speed of 2. A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform 0.40 m/s across a frictional surface. Perform an extensive analysis of the diagram below an extensive analysis of the diagram below to determine the values for the blanks.to determine the values for the blanks.

Use your understanding of force Use your understanding of force relationships and vector components to fill relationships and vector components to fill in the blanks in the following diagram in the blanks in the following diagram andand to determine the net force and acceleration to determine the net force and acceleration of the object. (Fof the object. (Fnetnet = m•a; F = m•a; Ffrictfrict = μ•F = μ•Fnormnorm; F; Fgravgrav

= m•g)= m•g)

Friday Problem 1

A 5-kg mass below is moving with a an A 5-kg mass below is moving with a an acceleration of 4 m/sacceleration of 4 m/s22 to the right. The to the right. The coefficient of friction for this surface is .2. coefficient of friction for this surface is .2. Use your understanding of force Use your understanding of force relationships and vector components to relationships and vector components to determine all your forces.determine all your forces.

  

The 5-kg mass below is moving with a constant speed of 4 m/s to the right. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (Fnet = m•a; Ffrict = μ•Fnorm; Fgrav = m•g)

 

Friday Problem 2

You are pushing a 200 kg block up a 20 You are pushing a 200 kg block up a 20 ⁰ ⁰ hill with a force of 200 N. If the box moves hill with a force of 200 N. If the box moves up the hill with a constant speed of 2 m/s, up the hill with a constant speed of 2 m/s, calculate all the forces involved and calculate all the forces involved and calculate µ.calculate µ.

Tuesday Problem 1

5. The following object is being pulled at a 5. The following object is being pulled at a constant speed of 2.5 m/s. Use your constant speed of 2.5 m/s. Use your understanding of force relationships and understanding of force relationships and vector components to fill in the blanks in vector components to fill in the blanks in the following diagram the following diagram andand to determine the to determine the net force and acceleration of the object. (Fnet force and acceleration of the object. (Fnetnet

= m•a; F= m•a; Ffrictfrict = μ•F = μ•Fnormnorm; F; Fgravgrav = m•g) = m•g)

At one moment during a walk around the block, there At one moment during a walk around the block, there are four forces exerted upon Fido - a 10.0 kg dog. The are four forces exerted upon Fido - a 10.0 kg dog. The forces are:forces are:

FFappapp = 67.0 N at 30.0° above the horizontal (rightward = 67.0 N at 30.0° above the horizontal (rightward

and upward)and upward)FFnormnorm = 64.5 N, up = 64.5 N, up

FFfrictfrict = 27.6 N, left = 27.6 N, left

FFgravgrav = 98 N, down = 98 N, down

Resolve the applied force (FResolve the applied force (Fappapp) into horizontal and ) into horizontal and

vertical components, then add the forces up as vectors to vertical components, then add the forces up as vectors to determine the net force and calculate the acceleration.determine the net force and calculate the acceleration.

A box is sliding down a ramp at an angle of A box is sliding down a ramp at an angle of 4747⁰ to the horizontal. If it is accelerating at ⁰ to the horizontal. If it is accelerating at 2.5 m/s2.5 m/s22 and has a mass of 150 kg, what is and has a mass of 150 kg, what is the Fnormal, Fnet, Fgravity, Ffric and µ?the Fnormal, Fnet, Fgravity, Ffric and µ?

Ramp Problem #1

Say a box is sitting on 30Say a box is sitting on 30⁰⁰ slope and is slope and is frictionless, so the only forces are the frictionless, so the only forces are the normal force and gravity. What is the normal force and gravity. What is the block's acceleration down the slope if block's acceleration down the slope if the mass is 3.0 kg? What is the normal the mass is 3.0 kg? What is the normal force?force?

Rotation & Centripetal Force

How to Keep it Straight Without How to Keep it Straight Without Getting DizzyGetting Dizzy

                        

Coutesy Space.com

Rotation

In addition to side to side (linear) motion, In addition to side to side (linear) motion, rotation plays an important role in physics, rotation plays an important role in physics, engineering, and life.engineering, and life.

Name some common phenomena or devices Name some common phenomena or devices that show rotationthat show rotation

Tops, planets, bicycle, car wheels, gears, pulleys, fans etc

Speed on a Wheel

Which horses on a Which horses on a carousel move the carousel move the fastest, inner or outer?fastest, inner or outer?

Outer

v = radius x angular speed

v = r

Mass at the End of a String

What force must theWhat force must the

string exert on the mass?string exert on the mass?

What is the direction of What is the direction of

this force?this force?

                   

A force toward the center of the circle

Centripetal Force

Any force directed toward the center of a circle is Any force directed toward the center of a circle is called called centripetal.centripetal.

Centripetal forces have clear causes such as Centripetal forces have clear causes such as tension in a string, gravity, friction etc.tension in a string, gravity, friction etc.

Some people call centripetal force a Some people call centripetal force a “pseudoforce.” (not real)“pseudoforce.” (not real)

They say “a real force such as friction They say “a real force such as friction providesprovides centripetal force.”centripetal force.”

How Big is Centripetal Force?

FFcc = mv = mv22/r/r

The faster the speed the more the forceThe faster the speed the more the force The tighter (smaller) the radius the more the The tighter (smaller) the radius the more the

forceforce vv22/r is called centripetal acceleration/r is called centripetal acceleration

Is a mass moving at steady speed in a circle accelerating? Yes. The direction is changingYes. The direction is changing What is the direction of this acceleration?What is the direction of this acceleration?

Toward the center of the circle

Car on a Curve

When auto rounds corner, sideways acting When auto rounds corner, sideways acting friction between tires and road provides friction between tires and road provides centripetal force that holds car on roadcentripetal force that holds car on road

             

Don’t Confuse Inertia With Force

Tub’s inner wall exerts Tub’s inner wall exerts centripetal force on centripetal force on clothes, forcing them into clothes, forcing them into circular pathcircular path

Water escapes throughWater escapes through

holes because it tends to holes because it tends to move by inertia in a move by inertia in a straight line pathstraight line path

Clothes WasherPhoto courtesy HowStuffWorks.com

How Can Water Stay In The Bucket? Bucket swung in aBucket swung in a

vertical circlevertical circle What force pushes on theWhat force pushes on the

water?water?

You have to swing the bucket fast enough for the bucket to fall as fast as the water

There must be a “normal” force exerted by the bottom of the bucket on the water, in addition to gravity

Weight and normal

force down

Centrifugal Force

The force ON THE The force ON THE PAIL is inward PAIL is inward (centripetal)(centripetal)

The force ON THE The force ON THE STRING is outward STRING is outward (centrifugal) (centrifugal)

If the string broke, If the string broke, which way would the which way would the can go?can go?

Tangent to the circle

Change Your Point of View

In rest frame of the In rest frame of the can there can there appears appears to to be a centrifugal force. be a centrifugal force. This pseudoforce(or This pseudoforce(or fictitious force) is a fictitious force) is a result of rotationresult of rotation

Unlike real forces, centrifugal force is not part of an interaction

Book on a Car Seat

When a car goes around a curve to the left, When a car goes around a curve to the left, a book slides a book slides

Which way does it slide?Which way does it slide? Why doesn’t it keep moving with the car?Why doesn’t it keep moving with the car?

There is not enough static friction force to keep it going in a circle. This friction must provide the necessary centripetal force.

The explanation in the rotating rest frame is different. How?

Roller Coaster Lab- Centripetal Force You are dropping the ball from 45You are dropping the ball from 45⁰, practice dropping the steel ball ⁰, practice dropping the steel ball

and the plastic ball to observe when it gets around the track.and the plastic ball to observe when it gets around the track. Attach the photogate and calculate the speed and centripetal force Attach the photogate and calculate the speed and centripetal force

of the marble at the top of the loop from various distances for both of the marble at the top of the loop from various distances for both marbles. Width of ball= .019 mmarbles. Width of ball= .019 m

Complete the table for both marbles. (as many trials as necessary) Complete the table for both marbles. (as many trials as necessary) steel = .028 kg plastic = .004 kg Fc= mvsteel = .028 kg plastic = .004 kg Fc= mv22/r radius of loop = .05 m /r radius of loop = .05 m

Draw a free body force diagram when the ball is at the top of the Draw a free body force diagram when the ball is at the top of the loop, label all forces. Do the following lab to solve for the minimum loop, label all forces. Do the following lab to solve for the minimum force needed to keep the ball (steel and plastic) on the loop. force needed to keep the ball (steel and plastic) on the loop.

Mass(kg) Weight(N) Photogate Time (sec)

Speed (m/s) Centripetal Force (N)

Did the marble stay on track?