Lesson 4-1 Changes in Motionimages.pcmac.org/SiSFiles/Schools/VA/TazewellSD/GrahamHigh/Uploads/Presentations/Holt...10/2/2017 9:03:55 1 AM Chapter 4 -- Changes in Motion A force is

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Chapter 4 -- Changes in Motion

A force is something that can change the

velocity of an object. (i.e. speed up, slow down,

or change direction).

In other words, a force can cause an object to accelerate.

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Changes in Motion

We often think of a force as a push or a pull.

We push a desk by exerting a force on it, and changing its velocity.

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Changes in Motion

If we push an object we exert a force for a period of time. If we stop pushing, the object will eventually stop. That is because other forces continue to act on it, changing its velocity.

Like?.... FRICTION

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Changes in Motion

Friction is only one example of a force. Forces can be divided into two major categories.

Contact forces are forces that exist between two objects that are in contact with each other.

Field forces are forces that exist between two objects that are separated by some distance.

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Newtons

Forces are measured in units called “Newtons” (N), Where a net force of 1.0 N will cause a mass of 1.0 kg to accelerate at a rate of 1.0 m/s2.

1 N = 1 kgm/s2

To convert a mass (in kg) to Newtons, you must multiply by acceleration due to gravity.

To convert Newtons to mass(in kg), you must divide by acceleration due to gravity.

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Changes in Motion

“Free-body diagrams” are used to focus on the forces acting on an object.

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Changes in Motion

How would we draw a free-body diagram for the situation shown below?

Fw = weight = mg = Newtons

FN = force of table on book (this is called the Normal force)

Holt Physics

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Changes in Motion

How would we draw a free-body diagram for a 5N football being kicked by a 67 N force at an angle of 56 degrees above horizontal?

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Changes in Motion

56

5 N

67 N

5 N

??

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Lesson 4-1 Changes in Motion


Ff

FN

Fw=mg

FA

Section Review pg. 128

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Newton’s Laws

1st Law - an object with no net force acting on it remains at rest or moves with a constant velocity in a straight line.

When there are forces, up or right is positive, and down or left is negative.

Inertia – the tendency of an object not to accelerate.

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Acceleration is determined by Net External Force

External force is a single force that acts on an object as a result of the interaction between the object and its environment.

Net external force is the sum of all forces acting on an object (the resultant force).

Ex: Tug of War…

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Ff

FN

FN is less than mg because of the upward pull

FA

Fw=mg

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FfFf

FN is more than mg because of the downward push

FN

FAFw=mg

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A man if pulling a crate with a force of 89 N at an angle of 15 degrees above horizontal. What are the x and y components of this force?

FAx = FAcos15 = 86N

FAy = FAsin15 = 23N

1589N

FAx

FAy

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A dog is pulled to the left with a force of 56 N and to the right with a force of 176 N, upward with a force of 780 N and downward with a force of 350 N. What is the net external force?

∑Fx = 176N-56N =120N

∑Fy =780N-350N= 430N

56N 176N

780N

350N120

430R

R2=1202+ 4302

R=446N

Tan-1(430/120)= 74 above horizontal

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Problems 4A

Problems on page 133

Section Review page 135 (1,2 and 3)

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Newton’s Second Law

Newton’s Second Law states – “The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the object’s mass.”

m

F a net

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Newton’s Second Law is often written with the net force isolated, as shown below;

ma Fnet Sometimes, the “net” is left out, because the “F” is understood to be the net force.

ma F

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Example 1... If the net external force on a desk with a mass of 35.0 kg is 45.0 N to the east, what acceleration (magnitude and direction) will the desk experience?

Fnet = ma

45.0 N = 35.0 kg (a)

a=45.0N/35.0kg=1.29 m/s2

east

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Example 2

A dachshund starts from rest at the top of a water slide that is 450.0 cm long and slides down to the bottom in 0.700 seconds. If the net external force on the dachshund is 25.8 N then what is the mass of the dachshund?

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A dachshund starts from rest at the top of a water slide that is 450.0cm long and slides down to the bottom in 0.700 seconds. If the net external force on the dachshund is 25.8 N then what is the mass of the dachshund?

Given: Fnet=25.8 N x = 450.0cm = 4.5 m

t = 0.700 sec

Formula: F = ma

Need: a =? And m=?

Old Formula: x = 1/2a t2

a = x/(1/2 t2) = 4.5 m/(1/2 (0.700)2)

a= 18.4 m/s2

F = ma, m=F/a = 25.8N/18.4m/s2 = 1.4 kg

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Questions to Consider

When a car hits a deer, why does the car take so much damage?

If you punch someone in the face, why does it hurt your hand?

When someone fires a gun, why do they feel a recoil?

When you blow up a balloon and release it, why does it fly around the room?

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Answer Newton’s Third Law!

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Newton’s Third Law

Newton’s Third LawIf two objects interact, the magnitude of theforce exerted on object 1 by object 2 is equalTo the magnitude of the force simultaneously

Exerted on object 2 by object 1, and theseForces are opposite in direction.

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What we call the “normal force” is a reaction force.

normalforceex.MOVnormalforceex.MOV

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Newton’s Third Law can also be used to explain propulsion.

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Newton’s Third Law can also be used to explain why your weight appears to change in an elevator.

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If every object that you pull pulls back with an equal and opposite force, how can you ever move anything?

../physlets/contents/mechanics/newtons_laws/illustration4_5.html../physlets/contents/mechanics/newtons_laws/illustration4_5.html

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Remember, when considering whether or not something will move, we want to find the net force acting on the individual object...the reaction force is on your hand!!!

Football and Newton’s Laws of Motion

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https://science360.gov/series/science-nfl-football/75c4deca-05f2-40f8-89ed-6756341d204f

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Problems 4B Due Thursday

Problems on page 138 (all)

Section Review pg.140 (2,3,5)

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Newton’s Laws Lab

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Equilibrium

Objects at rest or moving with a constant velocity are said to be in equilibrium.

– Therefore the net external force acting on a body in equilibrium must be zero.

– The vector sum of all the forces acting on it is zero.

Fx=0 and Fy=0

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Using Newton’s Laws

Friction - force that opposes the motion between two surfaces.

– The direction of the force of friction is parallel to the SURFACE and in a direction that opposes the slipping of the two surfaces.

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If the magnitude of the force applied is greater than the magnitude of static friction, movement occurs.

As long as the object does not move, the force of static friction (FS ) is always equal to and opposite in direction to the component of the applied force (FA ) that is parallel to the surface

FS = -FA WHEN IT DOES NOT MOVE

Also, FA is “just a hair” more than Fs at the start of motion (We’ll assume = for the most part.)

Static friction(FS ) - force that opposes the start of motion.

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Kinetic friction Fk - force between surfaces in relative motion

– Always LESS than static friction

– If moving at a constant velocity, Fk = -FA Variables Defined

FA = Applied Force

Ff = Force of friction

FN = Normal force

FW = Fg = Weight of the object

Fs = Force of static friction

Fk = Force of kinetic friction

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Equations

Fk= kFn s = coefficient of Fs,max= sFn static friction

Ff =Fn k = coefficient of kinetic friction

is a constant that depends on the two surfaces in contact.

(See page 144)

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Example

A 24 kg crate initially at rest on a horizontal floor requires a 75 Newton horizontal force to set it in motion to the left (this is FA and Fs,max). Find the coefficient of static friction (s) between the crate and the floor.

If it only requires 34 N to keep the crate moving at a constant velocity (this is Fk), what is the coefficient of kinetic friction

(k)?

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Fs = 75N

Fk = 34N

Fg=24kg X9.81m/s2

FN = 235 N

= 235 N

Fk = kFN

= s235N s= 75N / 235N = 0.32

= k235N k= 34N / 235N =0.14

Fs = 75N= sFN

FA

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Motion on a Horizontal

A desk with a mass of 75.0 kg sits on a floor, where the coefficient of static friction in between the two surfaces is 0.40. How much force must someone apply in a direction parallel to the surface of the floor to get the desk moving?

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Sketch the diagram:The net force in the y-

dimension is zero.

The box will move when the

applied force (FA) just

overcomes the maximum

force of static friction.

FN

s

s

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Given:FN

m = 75.0 kg g = 9.81 m/s2

0.40 s

Find: FN and then, Fs

Solution:

FN = -Fw = mg = -(75.0 kg)(-9.81 m/s2) = 736 N

Fs = sFN = (0.40)(736 N) = 290 N

so our applied force must be just greater than 290 N

s

s

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Problems 4C, page 145

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A boy exerts a horizontal force of 12.0 N on a 20.0 N box to slide it across a table where the coefficient of kinetic friction between the surfaces is 0.34. Sketch the free-body diagram and find the acceleration on the box.

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Sketch the diagram.

FNUse Fnet = ma. We will

need to find the net force

on the box and then divide

it by the mass, in order to

find the acceleration.k

k

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Sketch the diagram.

FN

The net force in the y-dimension

will be zero.

We will need to calculate the

force of friction (Fk), and

subtract it from the applied

force (Fa) to get the net force

(FNet).

k

k

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Given: FNFw = 20.0 N g = 9.81 m/s

2 Fa = 12.0 N

k = 0.34

Find: Fk, Fnet, m and then a

Solution:

Fk = kFN = kFw = (0.34)(20.0 N) = 6.8 N

Fnet = Fa - Fk = 12.0 N - 6.8 N = 5.2 N

2m/s 2.5

kg 2.04

N 5.2

m

F a net kg 2.04

m/s 9.81

N 20.0

g

F m

2

w

k

k

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Everyday ForcesMotion on a Horizontal

When the applied force (FA) is exerted at an angle above the horizontal, it is necessary to resolve the applied force into x (FAx) and y (FAy) components and subtract FAy from the Weight.

pullabovehoriz.MOVpullabovehoriz.MOV

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A girl is pulling a sled with a mass of 16.0 kg across a level patch of grass by exerting a force of 42.0 N along a rope that forms an angle of 33.0o above the ground. The coefficient of kinetic friction between the sled and the grass is 0.25. Sketch the free-body diagram and find the acceleration of the sled.

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Sketch the diagram.

Because a component of the

applied force is in the y-

dimension, the normal force is

going to be less than the weight

of the sled. The net force in the

y-dimension is still zero, and FN= (Fw - Fay).

Fk

k

Fn = (mg - Fay)

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Sketch the diagram.

The key to this problem is to

decrease the normal force (FN)

by the force applied in the y

(Fay), before you calculate the

force of friction (Fk).

k

Fk

Fn = (mg - Fay)

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Given: m = 16.0 kg Fa = 42.0 N = 33.0o k = 0.25

Find: Fax, Fay, FN, Fk, Fnet, a

Solution:

Fax = FaCos = (42.0 N)(Cos 33.0o) = 35.2 N

Fay = FaSin = (42.0 N)(Sin 33.0o) = 22.9 N

FN = (Fw - Fay) = (mg - Fay)

= (16.0 kg)(9.81 m/s2) - (22.9 N)) = 134 N

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Given: m = 16.0 kg Fa = 42.0 N = 33.0o k = 0.25

Find: Fax, Fay, FN, Ff, Fnet, a

Solution:

Fk = kFN = (0.25)(134 N) = 33.5 N

Fnet = Fax -Fk = 35.2 N -33.5 N = 1.7 N

2m/s 0.106

kg 16.0

N 1.7

m

F a net

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When the applied force (Fa) is exerted below the horizontal, the y component of the applied force (Fay) will increase the normal force (FN).

pushbelowhoriz.MOVpushbelowhoriz.MOV

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Example. A woman applies a force of 125.0 N on the handle of a 180 N lawnmower, at an angle of 35.00 below the horizontal. The coefficient of kinetic friction between the grass and lawnmower is 0.35. Draw a free-body diagram, and find the force of kinetic friction.

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Draw the free-body diagram and find the force of friction.

Fw = 180 N

k = 0.35

Fa = 25.0 N

Fax = FaCos = (125.0N)(Cos 35.0o) =

102.4 N

Fay = FaSin = (125.0N)(Sin 35.0o)

= 71.7 N

35.0o

FN = (Fw + Fay)= (180 N +71.7 N) = 251.7 N

Fk = kFN = (0.35) (251.7 N)

= 88 N

Fk

You can find a the same way as the previous problem

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Practice 4D, 1 and 4 on page 147

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Everyday ForcesMotion on an Inclined Plane

The weight (Fw) is directed straight down. It ignores the

incline.

Fw =Fg= mg

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The y component of the weight is (Fwy) is perpendicular to

the surface of the ramp.

Fwy = mgcos

Fw =Fg= mg

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The x component of the weight is (Fwx) is parallel to the surface

of the ramp.

Fw = mg

Fwy = mgcos

Fwx = mgsin

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We can redraw the x component of the weight (Fwx), to make it clear

that it helps to move the object down the ramp. We might also rename

it Fa If it is the only applied force working.

Fw = mg

Fwy = mgcos

Fwx = mgsin

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If there is friction between the ramp and the object, it will

oppose the motion of the object down the ramp.

Fw = mg

Fwy = mgcos

Fwx = mgsin

Ff

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If the object is in motion, the force of friction (Ff) can be calculated with

Fk = kFN. IF the object is being pulled up the ramp, then the force of

friction will be the other direction and Fg will also work against you.

Fw = mg

Fwy = mgcos

Fwx = mgsin

Fk = kFN

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The normal force (FN) will be equal and opposite to the y

component of the weight (unless you are pulling up or pushing

down at an angle).

Fw = mg

Fwy = mgcos

Fwx = mgsin

Ff = kFN

FN = -Fwy

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The two triangles in the picture are similar, so they share like

angles.

Fw = mg

Fwy = mgcos

Fwx = mgsin

Ff = kFN

FN = -Fwy

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Use the sheet provided….

A crate with a mass of 35.0 kg is sliding down an inclined plane with an angle of 43.0o above the horizontal. If the crate is accelerating at a rate of 3.45 m/s2, find the coefficient of kinetic friction between the ramp and crate. Start with a diagram.

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Fw=35.0kgX9.81m/s2 = 343N

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Given:m = 35.0 kg g = 9.81 m/s2 a = 3.45 m/s2

o43.0

Find: FN, Fwx, Fnet, Fk, k

Fwx = mgsin = (35.0 kg)(9.81 m/s2)(sin 43.0o) = 234 N

Solution:N 251 )43.0 )(cosm/s kg)(9.81 (35.0 cos mg cos F F F o2wwyN

Fnet = ma = (35.0 kg)(3.45 m/s2) = 121 N

Fk = Fwx - Fnet = 234 N – 121 N = 113 N

0.450 N 251

N 113

F

F

N

kk

Fnet=Fwx-Fk

Fk=Fwx-Fnet

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A 34.8kg BOX is moved up a ramp inclined at 26 degrees above horizontal. If the box starts from rest at the bottom of the ramp and is pulled at an angle of 35 degrees with respect to the incline and with a force of 372 N, what is the acceleration up the ramp? Assume that K is 0.25.

26

FA=372 NFN

34.8kg

26

Fwx=mgsin

Fwy=mgcos

35FK

X9.81=341N

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Find FK, FK = K FN = (0.25)(93N) = 23N

Find ax, ax = Fnet,x /m = = (FA,x- Fk - Fw sin26)/m

= (372cos35 – 23N – 341sin26)/34.8kg= 3.8 m/s2

Find FN, FN=(Fw,y-Fa,y )=(341N)cos26-372N(sin35)

= 93 N

4D 2 and 3 on page 147

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A 25kg crate is moved up a ramp inclined at 23 degrees above horizontal. If the box starts from rest at the bottom of the ramp and is pulled at an angle of 32 degrees with respect to the incline and with a force of 234 N, what is the acceleration up the ramp? Assume that K is 0.19.

23

FA=234 NFN

25kg

23

Fwx=mgsin

Fwy=mgcos

32FK

X9.81=245N

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Find FN, FN = (Fw,y - Fa,y )=

(25kg)(9.81m/s2)cos23 - 234N(sin32)= 102N

23

FN

245N

23

mgsin

mgcos

32FK

Fwx=mgsin

Fwy=mgcos

FA=234 N

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Find FK, FK = K FN =

(0.19)(102N) = 19.38N

23

234 NFN

245N

23

mgsin

mgcos

32FK

Fwy=mgcos

Fwx=mgsin

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Find ax, ax = Fnet,x /m =

= (FA,x- Fk - Fw sin23)/m

= (234cos32 – 19.38N – 245sin23)/m

= 83 kgm/s2/25kg = 3.3 m/s2

23

102 NFN

245N

23

mgsin

mgcos

32FK

234 N

WEIGHT and Fk are SUBTRACTED because they are working against the box going up the ramp.

Fwy=mgcos

Fwx=mgsin

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Using Newton’s Laws

– When an object is resting with no additional external forces on a horizontal surface, then FW = Fn

– When pushing down at an angle, force is added to FN(Fay+Weight), so FN increases to Weight+FAy.

– When pulling up at an angle, force is subtracted from the FN (weight-FAy), so FN decreases to Weight-FAy

– When moving at constant velocity Fk = FAx ,so Fk =Fn or FAx =Fn. This is also true of static friction until the maximum is reached.

– FAx means parallel to the surface

– When on an incline, Fgx makes the object go down and Fgy equals the normal force (unless pulled up at an angle, then Fgy is subtracted from the normal force.) Ff always opposes the motion.

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A block with a mass of 5.00 kg is placed on a frictionless inclined plane at an angle of 48.0o above the horizontal. Draw a diagram for the problem and find the acceleration of the block.

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Given:

m = 5.00 kg = 48.0o g = 9.81 m/s2

Find: Fw, Fwx, m and a

Solution:

N 49.1 )m/s kg)(9.81 (5.00 mg F Weight 2w

N 36.5 )48.0N)(Sin (49.1 SinF F Force Parallel owwx

2m/s 7.30

kg 5.00

N 36.5

m

F

m

F (a)on accelerati

pnet

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Ff Ff

FN

mgsin is the applied force!

Fw=mg

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Fw=mg

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Example 2What forces are at work on a 22N physics book on a 15 degree incline where the force of friction is 3N?What is the net external force?

15

3 N

Normal Force (NOT 22 N)

22N

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Now…can we figure out whether it will move? What else do we need?

15

3 N

FN

22N

15

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Now…sum all the x and y components!!!!!!!!!!!!!!!!!

15

3 N

FN

22N

15

Fwsin

Fwcos

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Assume x is in the direction of the inclined plane…

X components– (Fwsin15 – Ffriction)– 22sin15-3N = 5.7N-3N = +2.7N

Y components cancel– Fwcos15 - Normal Force (= Fwcos15) = 0

So….the book does move!– Newton’s Second Law can tell us how fast, but we’ll get to that later.

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Newton’s Third Law tells us that forces always come in pairs.

FSM05VD1.MOVFSM05VD1.MOV

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Newton’s Third Law tells us that forces always come in pairs.

FSA05VD1.MOVFSA05VD1.MOV

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These force pairs are called “action-reaction forces”, or “action-reaction pairs”.

Documents

Lesson 4-1 Changes in Motionimages.pcmac.org/SiSFiles/Schools/VA/TazewellSD/GrahamHigh/Uploads/Presentations/Holt...10/2/2017 9:03:55 1 AM Chapter 4 -- Changes in Motion A force is