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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 Jupiter Competing forces between

Jupiter and the other Galilean moons cause the

center of Io compress and melt. Consequently

Io is the most volcanically active body in the

solar system.

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

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

Add force of gravity

Add force of the road on the car (Called the

Normal Force)

Finally add the force of friction acting on the

car

A Free Body Diagram of a Football Being Kicked

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.

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.

Newtons 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 (v0) or moving at

constant velocity, then according to Newtons

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 FBD

Fup 565 N

Fright 82 N

Fleft 115 N

Fdown 236 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.

Newtons 1st Law

- Review Newtons 1st Law
- When Fnet0, 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

Newtons 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

objects mass

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
- M2.0 kg
- What is a?
- Vi 0 m/s, t0.50 s,
- displacement85 cm.85 m
- Welcome back kinematic equations! ?

(No Transcript)

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 - Fnet Fapp- Fk

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

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)

Coefficient of Kinetic Friction

- The problem states that the chair is moving with

constant velocity, which means Fnet0. Therefore,

Fapp must equal -Fk.

Solve for Coefficient of Kinetic Friction

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

What is Fnet?

- Since the suitcase is moving with constant

velocity, Fnet0. - 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

Lets 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.

FBD

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
- FfµFN(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

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,x

Ff

Fapp,y

Fg319 N

Fapp 485 N

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

Ff Fapp,x 485cos(35)397.29 N Use coefficient

of friction to find Ff FfµFN(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 9.81
- m 32.52 kg

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

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

?

Closer look at gravity triangle.

- Solve for Fg,y and Fg,x

Coordinate system for inclined planes

Y axis

X 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(?)

Forces in the x direction

- What is the force that makes the object slide

down the inclined plane? - Gravitybut only in the x direction

Remember that Vectors can be moved parallel to

themselves!!

?

Fg,y

?

Fg,x

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 accelerating
- Fnet Fg,x - Ff

What if there is an additional applied force?

- Example a box is being pushed up an inclined

plane

Fapp

Fg,x

Ff

Fg,y

?

In that case

- FN mgcos?
- 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

Ff

Fg,x

Fg,y

Fg

?

Solve the Problem

- The block is in equilibrium so
- Fnet0
- Fg,y FNmgcos?(5.4kg)(9.81)cos(15)
- FN51 N
- Additionally, what is the force of friction

acting on the block?

Find Force of Friction

- Fnet 0
- Fg,x Ff mgsin?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

Ff

Fg,x

Fg,y

Fg

?

What is Fnet?

- They give mass and acceleration
- So Fnet ma 75kg x 3.60 m/s2
- Fnet 270 N
- FN mgcos?
- Fnet Fg,x Ffmgsin? - Ff

Solve for Ff

- Fnet Fg,x Ffmgsin? Ff
- Ff mgsin? Fnet
- Ff 75kg(9.8)sin(25) 270 N
- Ff 40.62 N

Finish the Problem

- We are trying to solve for µk