Chapter 4- Forces and Motion

Think about the following questions:What is this object? Where is it? Why does it look like that?

IO is a moon of JupiterCompeting forces between Jupiter and the other Galilean moons cause the center of Io compress and melt. Consequently Iois the most volcanically active body in the solar system.

Erupting Volcano!!

Other examples of forces

What is a force?

IPC definition: A push or a pull exerted on some object

Better definition: Force represents the interaction of an object with its environment

The Unit for Force is a Newton

211s

kgmN

Two major types of forces

Contact Forces: Result from physical contact between two objects Examples: Pushing a cart, Pulling suitcase

Field Forces: Forces that do not involve physical contact Examples: Gravity, Electric/Magnetic Force

Force is a vector! (yay more vectors )

The effect of a force depends on magnitude and direction

Force Diagrams (p. 126)

Force Diagram: A diagram that shows all the forces acting in a situation

Free Body Diagrams p.127

Free Body Diagrams (FBDs) isolate an object and show only the forces acting on it

FBDs are essential! They are not optional! You need to draw them to get most problems correct!

How to draw a free body diagram

Situation: A tow truck is pulling a car(p. 127)We want to draw a FBD for the car

only.

Steps for drawing your FBD

Step 1: Draw a shape representing the car (keep it simple)

Step 2: Starting at the center of the object, Draw and label all the external forces acting on the object

Force of Tow Truck on Car=

5800 N

Add force of gravity

Gravitational force (Weight of car)=

14700 N

Force of Tow Truck on Car=

5800 N

Add force of the road on the car(Called the Normal Force)

Gravitational force (Weight of car)=

14700 N

Force of Tow Truck on Car=

5800 N

Normal Force =13690 N

Finally add the force of friction acting on the car

Gravitational force (Weight of car)=

14700 N

Force of Tow Truck on Car=

5800 N

Normal Force =13690 N

Force of Friction= 775 N

A Free Body Diagram of a Football Being Kicked

Fg

Fkick

A person is pushed forward with a force of 185 N. The weight of the person is 500 N, the floor exerts a force of 500 N up. The friction force is 30 N.

FN= 500 N

Fg= 500 N

Fapp= 185 NFf= 30 N

Forces you will need

Symbol of Force

Description

Fg Gravitational Force is the Weight of the Object (equal to mass x g= mg)

FN Normal Force= Force acting perpendicular to surface of contact

Ff Frictional Force- Opposes applied force; acts in direction opposite of motion

Fapp Applied Force

Sample Problem p. 128 #3

Draw a free body diagram of a football being kicked. Assume that the only forces acting on the ball are the force of gravity and the force exerted by the kicker.

Newton’s 1st Law of Motion

The Law of Inertia An object at rest remains at rest, and

an object in motion continues in motion with constant velocity (constant speed in straight line) unless the object experiences a net external force

The tendency of an object not to accelerate is called inertia

Acceleration

The net external force (Fnet) is the vector sum of all the forces acting on an object

If an object accelerates (changes speed or direction) then a net external force must be acting upon it

Equilibrium

If an object is at rest (v=0) or moving at constant velocity, then according to Newton’s First Law, Fnet =0

When Fnet =0, the object is said to be in equilibrium

How do we use this information?Sample Problem p. 133 #2

A crate is pulled to the right with a force of 82.0 N, to the left with a force of 115 N, upward with a force of 565 N and downward with a force of 236 N.

A. Find the net external force in the x direction B. Find the net external force in the y direction C. Find the magnitude and direction of the net

external force on the crate.

Step 1: Draw a FBDFup = 565 N

Fleft = 115 N

Fdown = 236 N

Fright = 82 N

Find the vector sum of forces

A. 82 N + (-115 N )= -33 N

B. 565 N + (-236 N) = 329 N

C. Find the resultant of the two vectors from part a and b.

33 N

329 NR = 331 N at 84.3 North of West

Newton’s 1st Law

Review Newton’s 1st Law: When Fnet=0, an object is in equilibrium

and will stay at rest or stay in motion

In other words, if the net external force acting on an object is zero, then the acceleration of that object is zero

Newton’s 2nd Law (p.137)

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

Fa net

Example p. 138 # 4

A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net external force acts on the otter along the incline?

Solving the problem

To calculate Fnet, we need m and a M=2.0 kg What is a? Vi= 0 m/s, t=0.50 s, displacement=85 cm=.85 m

Welcome back kinematic equations!

2

2

1attvx i

222

8.650.0

21

50.0085.0

21 s

mm

t

tvxa i

Ns

mkgmaFnet 148.62

2

Newtons’ 3rd Law

Forces always exist in pairs

For every action there is an equal and opposite reaction

Action- Reaction Pairs

Some action-reaction pairs:

Although the forces are the same, the accelerations will not be unless the objects have the same mass.

Everyday Forces

Weight= Fg = mg

Normal Force= FN= Is always perpendicular to the surface.

Friction Force= Ff

Opposes applied force There are two types of friction: static

and kinetic

Static Friction

Force of Static Friction (Fs) is a resistive force that keeps objects stationary

As long as an object is at rest: Fs = -Fapp

Kinetic Friction

Kinetic Friction (Fk) is the frictional force on an object in motion

Coefficients of Friction

The coefficient of friction (μ) is the ratio of the frictional force to the normal force

Coefficient of kinetic Friction

Coefficient of Static Friction

ForceNormal

ForceFrictionKinetic

F

F

N

kk

Friction Kinetic oft Coefficien

Force Normal

ForceFriction StaticFriction Static oft Coefficien

N

ss F

F

Sample Problem p. 145 #2

A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the char is in motion, a 327 N horizontal force keeps it moving at a constant velocity. A. Find coefficient of static friction B. Find coefficient of kinetic friction

Coefficient of Static Friction

In order to get the chair moving, it was necessary to apply 365 N of force to overcome static friction. Therefore Fs = 365 N.

The normal force is equal to the weight of the chair (9.81 x 25= 245 N)

5.1245

365

N

N

F

F

N

ss

Coefficient of Kinetic Friction

The problem states that the chair is moving with constant velocity, which means Fnet=0. Therefore, Fapp must equal -Fk.

Fapplied= 327 NFk= 327 N

Solve for Coefficient of Kinetic Friction

3.1245

327

N

N

F

F

N

kk

Forces at an angle

A woman is pulling a box to the right at an angle of 30 above the horizontal. The box is moving at a constant velocity. Draw a free body diagram for the situation.

FBD

Fg=Weight

FN= Normal ForceFapp= Applied Force

Ff= Friction Force F app,x

F app,y

What is Fnet?

Since the suitcase is moving with constant velocity, Fnet=0.

That means the forces in the x direction have to cancel out and the forces in y direction have to cancel out Fk = Fapp,x

FN + Fapp,y = Fg

NOTICE THAT NORMAL FORCE DOES NOT EQUAL WEIGHT IN THIS SITUATION

Let’s do an example. P. 154 #42

A 925 N crate is being pushed across a level floor by a force F of 325 N at an angle of 25 above the horizontal. The coefficient of kinetic friction is 0.25. Find the magnitude of the acceleration of the crate.

What do we need to know?

So we need mass and Fnet.

We have weight (925 N). So what is mass?

How to find Fnet? Find vector sum of forces acting on crate.

m

Fa net

kg 94.381.9

925

g

weight mass

2

smN

FBD

Fg=Weight=925 N

FN= Normal ForceFapp= 325 N

Ff= Friction Force F app,x

F app,y

Finding Fnet,y

Is box accelerating in y direction? No. Therefore Fnet in y direction is 0

So FN + Fapp,y = Fg

So FN = Fg- Fapp,y= 925 N- 325sin(25) FN= 787.65 N

Finding Fnet,x

Is box accelerating in x direction? Yes. Therefore Fnet,x is not 0

Fnet,x= Fapp,x – Ff

Fapp,x = Fappcos(25)=294.6 N

Use coefficient of friction to find Ff

Ff=μFN=(0.25)(787N)=197 N

Finish the Problem

Fnet,x = 294 N – 197 N= 97 N

So now we know that the Fnet on the box is 97 N since Fnet,y is 0

203.1

3.94

97

s

m

kg

N

m

Fa net

Another example. P. 154 #54 part a

A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35° below the horizontal.

If μk between the floor and the box is 0.57, how long does it take to move the box 4.00 m starting from rest?

DRAW FBD

FN

Fapp,y

Fapp= 485 NFg=319 N

Ff

Fapp,x

Find Fnet

Is box accelerating in y direction? No. Therefore Fnet in y direction is 0

So FN = Fapp,y + Fg

So FN = 485sin(35) + 319 N= 598 N

Fnet,x

Is box accelerating in x direction?Yes. Therefore Fnet,x is not 0

Fnet,x= Fapp,x – FfFapp,x = 485cos(35)=397.29 N

Use coefficient of friction to find FfFf=μFN=(0.57)(598)=341 N

Fnet, x = 397.29- 341= 57.29 N

So now we know that the Fnet on the box is

57.29 N since Fnet,y is 0

Weight of box is 319 N. Find mass by dividing by 9.81m= 32.52 kg

276.1

52.32

29.57

s

m

kg

N

m

Fa net

Finish the problem

We want to know how long it takes for the box to move 4.00 m.

Find vf so that you can solve for t

Solve for t

s

m1.76 a 0 v4m

2i s

mx

s

maxvv if 8.322

s

smsm

sm

a

vvt if 13.2

76.1

08.3

2

Forces on An Incline

A block slides down a ramp that is inclined at 30° to the horizontal. Write an expression for the normal force and the net force acting on the box.

Draw a Free Body Diagram

Fg

FNFF

θ

θ

Fg,x

Fg,y

Closer look at gravity triangle.

Solve for Fg,y and Fg,x

g

yg

F

F

h

a ,cos Fg

Fg,x

Fg,y

θ

g

xg

F

F

h

o ,)sin(

)cos(, mgF yg

)sin(, mgF xg

Coordinate system for inclined planes

X axis

Y axis

Fnet in the y direction

When a mass is sliding down an inclined plane, it is not moving in the y direction.

Therefore Fnet,y =0 and all the forces in the y direction cancel out.

Forces In the y-direction

So what are the forces acting in the y direction?

Look at your FBD

We have normal force and Fg,y

Since they have to cancel out…

FN= mgcos(θ)

Forces in the x direction

What is the force that makes the object slide down the inclined plane?

Gravity…but only in the x direction

Remember that Vectors can be moved parallel to themselves!!

Fg

FNFF

θ

Fg,x

Fg,y

θ

Forces in the x direction

So what are the forces acting in the x direction?

Friction Force (Ff) and Gravitational Force (Fg,x)

If the box is in equlibrium Fg,x = Ff

If the box is acceleratingFnet= Fg,x - Ff

What if there is an additional applied force?

Example: a box is being pushed up an inclined plane…

FN

θ

FfFg,y

Fapp

Fg,x

In that case…

FN= mgcosθ

Fnet = Fapp- Fg,x – Ff

If the object is in equilibrium then Fapp= Fg,x + Ff

An Example p. 153 #40

A 5.4 kg bag of groceries is in equilibrium on an incline of angle. Find the magnitude of the normal force on the bag.

Draw a FBD

FN

θ

Ff

Fg,y

Fg,x

Fg

Solve the Problem

The block is in equilibrium so… Fnet=0

Fg,y= FN=mgcosθ=(5.4kg)(9.81)cos(15)

FN=51 N

Additionally, what is the force of friction acting on the block?

Find Force of Friction

Fnet= 0

Fg,x= Ff= mgsinθ=5.4(9.81)sin(15)

Ff= 13.7N

Example p. 147 #3

A 75 kg box slides down a 25.0° ramp with an acceleration of 3.60 m/s2.

Find the μk between the box and the ramp

What acceleration would a 175 kg box have on this ramp?

FBD

FN

θ

Ff

Fg,y

Fg,x

Fg

What is Fnet?

They give mass and acceleration

So Fnet= ma= 75kg x 3.60 m/s2

Fnet= 270 N

FN= mgcosθ

Fnet= Fg,x – Ff=mgsinθ - Ff

Solve for Ff

Fnet= Fg,x – Ff=mgsinθ – Ff

Ff= mgsinθ – Fnet Ff = 75kg(9.8)sin(25) – 270 N

Ff = 40.62 N

Finish the Problem

We are trying to solve for μk

061.667

26.40

cos

26.40

N

N

mg

N

F

F

N

fk