4-4 everyday forces and application of newton’s laws objectives 1.explain the difference between...

4-4 everyday forces and application of Newton’s Laws

Objectives

1.Explain the difference between mass and weight.

2.Find the direction and magnitude of the normal force.

3.Describe air resistance as a form of friction.

4.Use coefficients of friction to calculate frictional force.

• We are concerned with four types of forces:

• Gravity (weight)• Normal force• Friction force• Tension force• Applied force

Type of forces

• The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on earth is always equal to the weight of the object as found by the equation:

• Fgrav = m • g

• where g = 9.81 N/kg (on Earth) and m = mass (in kg)• Note: g is different at different locations

Gravity Force (Weight) Fgrav

Comparing Mass and Weight

Weight

• The force of gravity. • Vector, its direction is

downward. • W = mg • The weight of an object

(measured in Newton) will vary according to where in the universe the object is.

Mass • The mass of an object

refers to the amount of matter that is contained by the object;

• Scalar, has no direction

• The mass of an object (measured in kg) will be the same no matter where in the universe that object is located.

Fg Fg Fg

FgFg

Fg

The direction of gravity force is always downward

The slope of weight vs. mass

weight

mass

Slope is Gravitational acceleration g

Weight = mg

practice• Complete the following table showing the relationship between mass

and weight.

1.

2. Different masses are hung on a spring scale calibrated in Newtons.a) The force exerted by gravity on 5 kg = ______ N.

b) The force exerted by gravity on _______ kg = 98 N.

c) The force exerted by gravity on 70 kg = ________ N.

3. When a person diets, is their goal to lose mass or to lose weight? Explain.

Object Mass (kg) Weight (N)

melon 1

apple 0.98

Joe Samall 25

Fred 980

• The normal force is the support force exerted upon an object that is in contact with another stable object (usually a surface).

• The direction of the normal force is perpendicular to the surface, from the surface toward the object and on the object.

Normal Force (FN )

Fg Fg Fg

FgFg

Fg

FN FNFN

FN

FN

FN

Practice- indicate FN on each box with an arrow

What is the direction of normal force?

The magnitude of normal force

• The magnitude of normal force can be determined using Newton’s Laws.

• Example: a car is moving along a horizontal road.

mg

FNBecause the car is not moving up or down,

mgFN

Alert!

netN FF Normal Force

Net Force

• The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it.

• Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces.

• Friction force can be reduced by lubricant such as oil.

Friction Force (Ff)

The direction and magnitude of friction force:

– Friction depends upon the nature of the two surfaces (μ, dimensionless) and upon the degree to which they are pressed together(FN) .

Ff = μFN

– The friction force often opposes the motion of an object.

Kinetic versus Static Friction• kinetic friction results

when an object moves across a surface.

Ffrict = μk • Fnorm

• The symbol μ represents the coefficient of kinetic friction between the two surfaces. The coefficient value is dependent primarily upon the nature of the surfaces that are in contact with each other. It does not depends on area of contact, the angle of the area, or the temperature, etc.

• Static friction results when the surfaces of two objects are at rest relative to one another and a force exists on one of the objects to set it into motion relative to the other object.

• The static friction force balances the force that you exert on the box such that the stationary box remains at rest.

• Static friction can change.

Ffrict-static ≤ μs• Fnorm

Alert: which box has more friction?

A B

A and B have the friction.

Fg Fg Fg

FgFg

Fg

FN FNFN

FN

FN

FN

Ff

v

v

Ff

v

Ff

v

Ff

v

Ff Ff

v

Ff

Practice- indicate Ff on each box with an arrow

Example

• A 24 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor.

32.0s

Class work

• Text book page 145 Practice 4C #1-3

• Answers

1. 0.23

2. a. 1.5; b. 1.3

3. a. 870 N, 670 N; b. 110 N, 84 N; c. 1000 N, 500 N; d. 5N, 2 N

• The air resistance is a special type of frictional force that acts upon objects as they travel through the air. The force of air resistance is often observed to oppose the motion of an object. This force will frequently be neglected due to its negligible magnitude.

Air Resistance Force (Fair )

• The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.

Tension Force (FT )

• The spring force is the force exerted by a compressed or stretched spring upon any object that is attached to it. An object that compresses or stretches a spring is always acted upon by a force that restores the object to its rest or equilibrium position – directed toward equilibrium position.

Spring Force (Fspring )

kxFs

Fs is the force exerted by the spring on the object, k is spring constant and x is displacement from equilibrium position

Newton’s Laws Application

1. Drawing Free-Body Diagrams

2. Determining the Net Force

3. Apply Newton’s Laws

Drawing Free-Body Diagrams• Free-body diagrams are used to show the

relative magnitude and direction of all forces acting upon an object in a given situation.

• The size of the arrow in a free-body diagram reflects the magnitude of the force. The arrow shows the direction that the force is acting.

• Each force arrow in the diagram is labeled to indicate the exact type of force.

• It is generally customary to draw the force arrow from the center of the box outward in the direction that the force is acting.

A block of wood is sitting motionless on a table.What forces are acting on it?

FgWeight

FNNormal

Weight is the force of gravity

pulling an object toward the

CENTER OF THEEARTH

Normal Force is aREACTION

force that any object exerts

when pushed on

Ng FF

Class work

• Practice on free body diagrams

Finding the External Net Force and acceleration

• Fnet is the vector sum of all the individual forces. The three major equations that will be useful are

– Fnet = m•a,

– Fg = m•g,

– Ff = μ•FN

Example #1

• A man pushes a 50 kilogram crate across a frictionless surface with a constant force of 100 Newtons.

Draw a free-body diagram of the crate.What is the weight of the crate?What is the normal force that pushes on the crate?What is the net force on the crate?What is the crate’s acceleration?

FA

FN

Fg

Fg = mgFg = (50 kg)(9.81 m/s2)

Fg = 490.5 N

FN = Fg

FN = 490.5 N

Fnet will only bethe 100N horizontal

force

a = Fnet / ma = (100 N) / (50 kg)

a = 2 m/s2

Example #2• A horse pulls a 500 kilogram sled with a constant

force of 3,000 Newtons. The force of friction between the sled and the ground is 500 Newtons.

Draw a free-body diagram of the sled.What is the weight of the sled?What is the normal force that pushes on the sled?What is the net force on the sled?What is the sled’s acceleration?

FA

FN

Fg

Ff

Fg = mgFg = (500 kg)(9.81 m/s2)

Fg = 4905 N

FN = Fg

FN = 4905 N

Fnet = ΣFx

Fnet = 3000 N – 500 NFnet = 2500 N

a = Fnet / ma = (2500 N) / (500 kg)

a = 5 m/s2

the object is moving horizontally. Use the diagram to determine the normal force, the net force, the mass, and the acceleration of the object. (g =10 m/s2)

8 kg

80 N

40 N right

5 m/s2 right

Example #3

Example #4• Edwardo applies a 4.25-N rightward force to a 0.765-kg

book to accelerate it across a tabletop. The coefficient of friction between the book and the tabletop is 0.410. Determine the acceleration of the book.

Example #5• Lee Mealone is sledding with his friends when he

becomes disgruntled by one of his friend's comments. He exerts a rightward force of 9.13 N on his 4.68-kg sled to accelerate it across the snow. If the acceleration of the sled is 0.815 m/s/s, then what is the coefficient of friction between the sled and the snow?

Class work

• Application of Newton’s Laws – pages 1&2

Free Fall and Air ResistanceFree Fall

• Objects that are said to be undergoing free fall, are

• not encountering air resistance;

• falling under the sole influence of gravity. All objects will fall with the same rate of acceleration, regardless of their mass. This is due to that the acceleration is The ratio of force to mass (Fnet/m)

Falling with air resistance• As an object falls through air, it

usually encounters some degree of air resistance - the result of collisions of the object's leading surface with air molecules.

• The two most common factors that have a direct affect upon the amount of air resistance are

– the speed of the object: Increased speeds result in an increased amount of air resistance.

– the cross-sectional area of the object: Increased cross-sectional areas result in an increased amount of air resistance.

• As an object falls, it picks up speed. The increase in speed leads to an increase in the amount of air resistance. Eventually, the force of air resistance becomes large enough to balances the force of gravity. At this instant in time, the net force is 0 Newton; the object will stop accelerating. The object is said to have reached a terminal velocity.

Falling with air resistance – terminal velocity

• When forces acting at angles to the horizontal, Newton’s 2nd law still applies:

• Force is a vector quantity. Adding forces in 2 dimensions follows the rules for adding vectors.

∑F = ma

Net Force Problems in 2D

1. Resolve the vectors at an angle into x and y components.

2. Add all the x components together

3. Add all the y components together – usually the

4. Use Pythagorean Theorem to find the resultant (hypotenuse) Resultant2 = x2 + y2

5. Use trigonometric function to determine the direction: tanθ = opp / adj

Determine the Fnet mathematically

Example 1 - Pulling on an Angle

30˚

Fg

FN

This applied force (FA)can be broken into

COMPONENTS

FA

FAX

FAY

A block of 10 kg mass is pushed along a frictionless, horizontal

surface with a force of 100 N at an angle of 30° above horizontal.

The total vertical force mustbe 0, so

Ry = FN + FAY –Fg = 0FN = Fg – FAY

R = Rx = Fax

Acceleration depends only onFAX

X Y

FAX FAY

Fg

FN

Total = FAX Total = 0

FAx = 100cos(30o) = 87 N

FAy = 100sin(30o) = 50 N

Example 2 • A man pulls a 40 kilogram crate across a

smooth, frictionless floor with a force of 20 N that is 45˚ above horizontal.

What is the net force on the sled?

What is the crate’s acceleration?

Fnet = FA cos θFnet = (20 N)(cos 45°)

Fnet = 14.14 N

a = Fnet / ma = (14.14 N) / (40 kg)

a = 0.35 m/s2

How could the acceleration be increased?

Pushing at a smaller angle will make Fnet greater andtherefore increase acceleration.

Pushing on an Angle

-30˚

Fg

FN

FA

FAX

FAY

This applied force (FA)can be broken into

COMPONENTS

A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of

30° below horizontal.

The total vertical force mustbe 0, so

FN = Fg + FAY

Acceleration depends only onFAX

X Y

FAX FAY

Fg

FN

Total = FAX Total = 0

Example 3• A girl pushes a 30 kilogram lawnmower

with a force of 15 Newtons at an angle of 60˚ below horizontal.

Assuming there is no friction, what is the acceleration of the lawnmower?

What could she do to reduce her acceleration?

Fnet = FA cos θFnet = (15 N)(cos 60°)

Fnet = 7.5 N

a = Fnet / ma = (7.5 N) / (30 kg)

a = 0.25 m/s2

Push at an greater angle

Example 4 – determine Fnet and a

• The vertical forces are balanced (Fgrav, Fy, and Fnorm add up to 0 N),

• The horizontal forces add up to 29.3 N, right

• The net force is 29.3 N, right

• a = Fnet / m = 29.3 N / 10 kg = 2.93 m/s2, right

Sample problem 4D• A student moves a box of books by attaching a rope to

the box and pulling with a force of 90.0 N at an angle of 30.0 degrees. The box of books has a mass of 20.0 kg, and the coefficient of kinetic friction between the bottom of the box and the sidewalk is 0.50. find the acceleration of the box.

Class work

• Newton’s 2nd law practice – page 3 & 4

Equilibrium and Static• When all the forces that act upon an object are

balanced, then the object is said to be in a state of equilibrium.

• An object at equilibrium is either ...– at rest and staying at rest, or – in motion and continuing in motion with the same

speed and direction.

• "static equilibrium." refers to an object at rest

• A frame is shown with the given tension. Determine the weight of the frame.

Example

Rx = Ax + Bx + Cx = 0

-43 N + 43 N – Cx =0

Cx = 0

Ry = Ay + By + Cy = 0

25 N + 25 N + Cy = 0

Cy = -50 N

A B

C = ?

30o

C2 = Cx

2+ Cy2

R = 50. N

Ax = 50cos(150o) = -43 N

Ay = 50sin(150o) = 25 N

Bx = 50cos(30o) = 43 N

By = 50sin(30o) = 25 N

example• A sign is shown with the given mass of 5 kg.

Determine the tension of each cable. (g = 10 m/s2)

C = Fg

40o 40o

A = T B = T

Tsin40oTsin140o

Tcos40oTcos140o

Ry = Ay + By + Cy = 0

Tsin40o + Tsin140o – Fg = 0

T = 38 N= 50 N

An important principle• As the angle with the horizontal increases, the amount of

tensional force required to hold the sign at equilibrium decreases.

Fg = 10 N

Down the Slope• A tool used to move objects from one height to

another.

• Allows for the movement of an object without lifting it directly against gravity.

• The object accelerate downward due to the component gravity that is parallel to the plane.

Fg on Inclined Plane

Forces on an Incline Calculations

• Consider forces:– Perpendicular

• F┴ = Fg cos θ

• Cancel out Normal (FN )

– Parallel• F// = Fg sin θ

• Could be in the same or opposite of Friction (Ff )

Tilt you head method

θ

θ

Essential Knowledge• What happens to the component of weight that is

perpendicular to the plane as the angle is increased?Decreases – Fg perpendicular

• What happens to the component of weight that points ALONG the plane as the angle is increased?

Increases – Fg parallel• What happens to the normal force as the angle is

increased?Decreases – depends on Fg perpendicular

• What happens to the friction force as the angle is increased?

Decreases – depends on normal force

• The net force is the vector sum of all the forces. – All the perpendicular components (including

the normal force) add to 0 N. – All the parallel components (including the

friction force) add together to yield the net force. Which should directed along the incline.

Fnet = F//

mgsinθ = ma

a = gsinθ

In the absence of friction

Fnet = 0

With friction - Object is at equilibrium – at rest or moving with constant velocity

Ff

Horizontal:

F// = Ff

mgsinθ = μFN

mgsinθ = μ∙mgcosθ

tanθ = μ

Vertical:

F┴ = FN

mgcosθ = FN

Example

• What is the magnitude of the normal force?

FN = Fg perpendicular = Fg cos θ = 43.3 N

• If the box is sliding with a constant velocity, what is the magnitude of the friction force?

Ff = Fg parallel = Fg sin θ = 25 N

Fg = 50N

30°

Example 1• The free-body diagram shows the forces acting upon a 100-kg

crate that is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is 0.3. Determine the net force and acceleration of the crate.

In perpendicular direction:

Fnorm = F┴ = 850 N

In parallel direction:

Fnet = F// - Ff

Fnet = 500 N - µFnorm

Fnet = 235 Na = Fnet / m = 2.35 m/s2

F┴ = Fgrav∙cos30o = 850 NF// = Fgrav∙sin30o = 500 N

practice

Class work

Lab 10 – weight vs. mass• Purpose: determine the relationship between

mass and weight• Material: spring scale, known masses• Data table: include heading in data table• Data analysis: graph weight vs. mass• Conclusion:

– What is the relationship between weight and mass?– Determine the slope and indicate the meaning of the

slope in the graph.

Double Trouble (a.k.a., Two Body Problems)

• Two body-problems can typically be approached using one of two basic approaches. – One approach is the system analysis, the two

objects are considered to be a single object moving (or accelerating) together as a whole.

– Another approach is the individual object analysis, either one of the two objects is isolated and considered as a separate, independent object.

Example - system analysis• A 5.0-kg and a 10.0-kg box are touching each other. A

45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. Ignore friction forces and determine the acceleration of the boxes and the force acting between the boxes.

m = 15 kg

Fnet = 45 N

a = Fnet / m = 3 m/s2

Example - individual analysis

5a = 45 – 10a a = 3 m/s2

In vertical direction: FN = Fg = (5 kg) (9.81 m/s2) = 49 N

In horizontal direction: Fnet = Fapp - Fcontact

(5 kg)a = 45 N - Fcontact

In vertical direction: FN = Fg = (10 kg) (9.81 m/s2) = 98 N

In horizontal direction: Fnet = Fcontact

(10 kg)∙a = Fcontact

Example: system analysis • A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-

N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. The coefficient of kinetic friction is 0.200. Determine the acceleration and the contact force.

In vertical direction: FN = Fg = (15 kg) (9.81 m/s2) = 147 N

a = Fnet / m = (15.6 N/15.0 kg) = 1.04 m/s2

In horizontal direction: Fnet = Fapp - Ffrict = 45 N - μ•Fnorm

Fnet = 15.6 N

However, in order to find the contact force between the objects, we must make individual analysis.

In vertical direction: FN = Fg = (10 kg) (9.81 m/s2) = 98 N

In horizontal direction: Fnet = Fcontact - Ff

(10 kg)∙(1.04 m/s2) = Fcontact - μ•Fnorm

Example: individual analysis

10.4 = Fcontact – (0.2)(9.8)

Fcontact = 8.44 N

Lab 10 – weight vs. mass• Purpose: determine the relationship between

mass and weight• Material: spring scale, known masses• Data table: include heading in data table• Data analysis: graph weight vs. mass• Conclusion:

– What is the relationship between weight and mass?– What does the slope in the graph mean?