Chapter 4. Force and MotionChapter 4. Force and Motion

Chapter Goal: To establish a connection between force and motion.

Student Learning Objectives

• To recognize what does and does not constitute a force.

• To identify the specific forces acting on an object.• To draw an accurate free-body diagram of an object.• To begin the process of understanding the connection

between force and motion.

A force is an interaction between two objects

• A force is a push or a pull on an object.

• If I push a book across a table, the book pushes me back (inanimate objects can exert force!)

• A force is a vector. It has both a magnitude and a direction.

• The force (interaction) has the same magnitude for both me and the book. However the direction of the force on me is opposite to the direction of the force on the book.

A force is an interaction between two objects

•A force on an object requires an agent. The agent is another object.

•This is another way of confirming that a force is an interaction between objects. It takes two to tango!

• A force is either a contact force or a long-range (noncontact) force.

•Gravity is the only long-range force we will study this semester. It is an interaction between two objects, but they are not necessarily in contact.

•All other forces (this semester) only exist when the two objects are in contact.

A Short Catalog of Forces - Gravity• Gravity is a long-range attractive force between two objects.

•In this class, our emphasis is on the interaction between the Earth and objects on or near its surface (the weight force).

•Your text uses W for the weight force.

•Near the surface of the earth, W = mg, where m is the object mass in kilograms and g= 9.8 m/s2 is the acceleration due to gravity.

Earth

A Short Catalog of Forces: TensionRopes, strings, cables and hooks exert a tensional force. These agents always pull, they never push.

Although it seems strange, the sled exerts a tension force on the rope.

A Short Catalog of Forces - NormalThe normal force is a contact force between 2 surfaces that is perpendicular to the surfaces. Your book uses FN for the normal force

A Short Catalog of Forces - frictionThe frictional force is a contact force between 2 surfaces that is parallel to the surfaces

A Short Catalog of Forces

Other forces are usually shown as an with an identifying subscript.

pF

F

You’ve just kicked a rock, and it is now sliding across the ground about 2 meters in front of you. Which of these forces act on the rock at this instant? Choose all that apply

A. Gravity, acting downwardB. The normal force, acting upwardC. The force of the kick, acting in the

 direction of motionD. Friction, acting opposite the direction of

 motionE. All of the above

Drawing force vectors

Free-body diagrams• Identify all forces acting on the

object.

• Draw a coordinate system. If motion is along an incline, the coordinates system should be tilted relative to true horizontal and true vertical.

• Represent the object as a dot at the origin.

• Draw vectors representing each of the identified forces and label with the appropriate symbol.

truck not moving due to friction

Freebody diagrams

Draw a free body diagram for the truck.

There is no friction, but a cable attached to the wall keeps the truck from moving.

Freebody diagrams

Draw a free body diagram for the sled. Assume friction between the snow and the sled.

4.3 Newton’s Second Law of Motion

Newton’s Second LawAn object of mass m, subject to forces F1 , F2 , etc. will undergo an acceleration with a magnitude directly proportional to the net force and inversely proportional to the mass:

m

Fa

aF

m

The direction of the acceleration isthe same as the direction of the net force.

4.3 Newton’s Second Law of Motion

Newton’s First LawThis is a special case of the 2nd Law. If the net force equals 0, there is no acceleration and the velocity of the object will not change. If it is at rest, it will stay at rest.

0F

In this case the object is in equilibrium. •Static equilibrium – object is at rest for a finite period•Dynamic equilibrium –object is moving at constant velocity.

Graphical Interpretation of Newton’s Second Law

• Newton’s 2nd Law is the equation of a line with a 0 value for the y-intercept (in this case the a- intercept!).

mFa

a

F

A a

F

B a

FC

a

F

D a

F

The following graphs plot acceleration vs force for different objects. Which object has the greatest mass?

A. B. C. D.

Problem-Solving Strategy for Newton’s Law Problems

Problem-Solving Strategy: Equilibrium Problems

0yF

If the net force is equal to zero, must the object be at rest? The next example shows the answer to this:

0xF

Newton’s 1st Law: Towing a car up a hillA car with a weight (not mass) of 15,000 N is being

towed up a 20° frictionless slope at constant velocity. The tow rope is parallel to the slope surface. What is the tension on the tow rope?

Why is this a Newton’s 1st Law problem (ΣF = 0) and not a Newton’s 2nd Law problem (ΣF = ma)?

Draw a free body diagram for this problem.

Newton’s 1st Law: Towing a car up a hillA car with a weight (not

mass) of 15,000 N is being towed up a 20° frictionless slope at constant velocity. The tow rope is parallel to the slope surface. What is the tension on the tow rope?

What part of the story indicates that ΣF = 0?

Newton’s 1st Law: Towing a car up a hillOnce the pictorial representation is completed, use Newton’s 1st Law in component form:

T and n have only one non-zero component, but FG has non-zero components in both x and y directions. How do we write the components of FG in terms of θ ?

Newton’s 1st Law: Towing a car up a hill

Solving the 2nd equation tells us that n = FG cos θ, which is not necessary information for this problem.

Problem-Solving Strategy for Newton’s 2nd Law Problems

1. Use the problem-solving strategy outlined for Newton’s 1st Law problems to draw the free body diagram and determine known quantities.

2. Use Newton’s Law in component form to find the values for any individual forces and/or the acceleration.

3. If necessary, the object’s trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics.

4. Reverse # 2 and 3 if necessary.

Newton’s 2nd Law – Speed of a Towed Car

•Draw a free-body diagram. Find the acceleration using Newton’s second law in component form.

•Draw a pictorial representation and Find the velocity using one of the kinematic equations.

Speed of a towed car – pictorial representation of motion, motion diagram and free body diagram

Speed of a towed car

Use the free body diagram to write Newton’s 2nd Law in component form:

ΣFy = may = 0 (no change in speed up or down)ΣFy = n – FG , or n=FG (not pertinent in this problem)ΣFx = max T – f = max

Speed of a towed car

Speed of a towed car

Apparent Weight – What the scale says (even if there is no scale!)

• The value the scale reads when scale and object are accelerating or being acted upon by some force other than gravity (e.g the buoyant force in water)

• Not equal to the true weight (FG = mg)

• Apparent weight is either a normal force (step on scale) or tension force (hanging scale)

• An object inside an accelerating elevator has the same acceleration as the elevator.

• Acceleration is not a force so don’t include directly on free body diagram.

Typical FBD for an apparent weight problem

n

FG = mg, but n does not!

Apparent Weight – What the scale would say, even if there is no scale

It takes the elevator in a skyscraper 4.0 s to reach a constant speed of 10.0 m/s. A 60.0 kg passenger gets on at the ground floor. What is her apparent weight during those 4 seconds?

Use the problem-solving strategy (repeated in next slide).

Problem-Solving Strategy for Newton’s Law Problems

Problem-Solving Strategy for Newton’s 2nd Law Problems

1. Use Newton’s Law in component form to find the values for any individual forces and/or the acceleration.

2. If necessary, the object’s trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics.

3. Reverse # 1 and 2 if necessary (Hint: it is!)

Mass – a quantitative measure of inertia• An object’s mass (m) is a measure of the amount

of matter that it contains. • The SI unit of mass is the kilogram (kg). The

English unit of mass is the slug, which is virtually never used.

• The mass of an object does not change unless you add or subtract matter from it.

• The mass of an object is the same on any planet and in deep space.

• Scales do not directly measure mass, except for a balance scale, which compares one mass to another.

Weight• The weight of an object on or above the earth is the

gravitational force that the earth exerts on the object (FG or W).

• Since weight is a force, the SI unit of weight is the Newton (N); the English unit is the pound.

• The weight of an object changes on other planets since the gravitational force of other planets are different than that of the earth.

• A scale does not directly measure weight; it measures the normal force or (hanging scale) the tension force. In most situations, this force measured by the scale is numerically equal to the weight force.

Apparent weight or “what the scale says”

• In accelerating reference frames (e.g. the ever-popular elevator with a scale), the scale reading will differ from the true weight. This is called apparent weight.

• Metric scales assume measurement on earth and are calibrated to read in mass units (kg). American scales read in English force units (lb). It’s a culture thing.

4.7 The Gravitational Force

Newton’s Law of Universal Gravitation

• Every particle in the universe exerts an attractive force on every other particle. • A particle is a piece of matter, small enough in size to be regarded as having no volume. In practice we use the particle model even for larger bodies• The Law of Universal Gravitation is an example of an inverse square law. The force between the two bodies is inversely proportional to the square of the distance between them.

221

rmmFG

For two particles that have masses m1 and m2 and are separated by a distance r, the force has a magnitude given by:

221

1221 rmmGFF

2211 kgmN10673.6 G

F1-2 and F2-1 are equal in magnitude but point in opposite directions

If m1 has a mass of 12 kg and m2 has a mass of 25 kg and thetwo objects are 1.2 meters apart, what is the magnitude of the gravitational force between them?

2211 kgmN10673.6 G

221

1221 rmmGFF

Ratio Reasoning with Inverse Square Laws

Two masses, separated by a distance of r experience a gravitational attraction, F. Now the distance is increased by a factor of 3. By what factor does the magnitude of the force change (what is the ratio )?

a. 1/3 c. 9b. 3 d. 1/9

1

2

FF

A. much smaller and in the same direction.

B. much smaller and in the opposite direction.

C. the same size and same direction.D. the same size andopposite direction.

The figure shows the moon being attracted by the earth. The earth is also attracted by the moon. Compared to FEonM (the force shown), FMonE is:

moon

earth

FEonM

4.7 The Gravitational Force

Definition of Weight

The weight of an object on or above the earth is the gravitational force that the earth exerts on the object. The weight always acts downwards, toward the center of the earth.On or above another astronomical body, the weight is the gravitational force exerted on the object by that body.

SI Unit of Weight: newton (N)

Calculate g

2rmM

GW E

2E

E

RMGg

ME = 5.98 x 1024 kg

RE = 6.38 x 106 m (don’t forget to square!)

If the object is close (within 100 miles) to the earth, this becomes:

2E

E

RmM

GW Define the quantity g, for all the unchanging entities:

2211 kgmN10673.6 G

mgW

g = 9.8 m/s2 on Earth1 kg of mass x g = 9.8 N, where N = kg m/s2

The Gravitational Force

2E

E

RmM

GW

2E

E

RMGg

Ratio Reasoning

A space traveler weighs 540.0 N on earth.What will he weigh on another planet whose

radius is twice that of earth and whose mass is 3 times that of earth?

Friction

Kinetic Friction

Experiments show that the kinetic friction force is nearly constant and proportional to the magnitude of the normal force.

where the proportionality constant μk is called the coefficient of kinetic friction.

Static Friction

The box is in static equilibrium, so the static friction must exactly balance the pushing force:

Static friction• An object remains at rest as long

as fs < fs max

• The object slips when fs = fs max

• A static friction force fs > fs max is not physically   possible.

• fs max >fk for the same surfaces

where the proportionality constant μs is called the  coefficient of static friction.

A model of friction

• “ motion” indicates motion relative to the two surfaces• the max value static friction, fs max occurs at the very instant the

object begins to move (which often means 1 nanosecond before, for problem-solving purposes.

• for any given materials, μs > μk

Determine which of the frictional forces is the largest. The box and the floor are identical in all situations.

Which of these have the same frictional force?A. none C. c,d,e,B. a and d D. c,d

Example Problem w tilted axis

A 75-kg snowboarder starts down a 50-m high (not long!), 100 slope with μk = 0.06. Assume he has just started moving, but his starting velocity is essentially zero. What is his speed at the bottom?

• Draw the free body diagram (fbd) to determine the forces.• Use Newton’s Law in component form.• Determine the acceleration.• Use kinematics to solve for the final speed.

• Start with the fbd.

Example Dynamics Problem w tilted axisA 75-kg snowboarder starts

down a 50-m high, 100 slope, with μk = 0.06. What is his speed at the bottom?

Write Newton’s Law for y-axis values. Is this 2nd Law, or 1st Law (a=0)?

There is one force along the + y axis and one force with a negative y component.

n

FG

fk

Find: v1

uk = .006

Example Dynamics Problem w tilted axisA 75-kg snowboarder starts down

a 50-m high, 100 slope, with μk = 0.06. What is his speed at the bottom?

Write Newton’s Law for y-axis values. Is this 2nd Law, or 1st Law (a=0)?

ΣFy = n – mg cos θ = may

But no acceleration along y axis means the net force is 0 therefore

n = mg cos θ

n

FG

fk

Find: v1

uk = .006

Example Dynamics Problem w tilted axisA 75-kg snowboarder starts down

a 50-m high, 100 slope, with μk = 0.06. What is his speed at the bottom?

ΣFx = mg sin θ – fk = max

where fk = uk n

and we know that n = mg cos θ; therefore

mg sin θ – ukmg cos θ = max

You didn’t even need to know the mass; it cancels out!

plug n chug: a = 1.12 m/s2

n

FG

fk

Find: v1

uk = .006n = mg cos θ

Example Dynamics Problem

A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom? Couldn’t find a picture of a snowboarder…. but the idea is the same

Findv1

a, =1.12 m/s2

Example Dynamics ProblemA 75-kg snowboarder starts

down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom?

Time is not important.

Use the mantras of the Indian Princess to calculate the length (hypotenuse) of the slope. That’s our x1.

Then use the appropriate kinematic equation

v1 = 25.4 m/s

Findv1

a, =1.12 m/s2

Reference FramesA reference frame is a coordinate system that allows description of

time and position relative to some “observer”.

• The origin moves (or stays at rest) in the same way as the observer does.

• If a person is sitting still in a moving train, the description of the person's motion depends on the chosen frame of reference.

• If the frame of reference is the train (i.e origin on the train), the person is considered to be not moving IN THAT FRAME OF REFERENCE

• if the frame of reference is the Earth, (i.e. origin on the Earth), the person is considered to be moving IN THAT FRAME OF REFERENCE

• The “observer” doesn’t have to be a person

Inertial Reference Frames• An inertial reference frame is one

where the origin is either at rest (the Earth, for our purposes) or moving at constant velocity.

• Newton’s 1st Laws are valid in an inertial reference frame.

• Newton’s 1st Law appears to be broken if one is looking at motion from a non-inertial (i.e. accelerating) reference frame.

• The truck is accelerating to the right. Imagine 2 frames of reference; that of the truck driver and that of a pedestrian.

Reference frames, revisitedThe truck is accelerating to

the right. The truck driver thinks the

box in the back is not moving, because its not sliding toward the back of the truck.

The ground-based observer sees the box getting faster, just like the truck.

Which observer is in an inertial reference frame?

A. The truck driverB. The ground-based guyC. The box (if it could observe)D. None of the above.

Free body diagram of the boxIn a free-body diagram of

the box, there is a:a. static friction vector,

pointing rightb. static friction vector,

pointing leftc. kinetic friction vector,

pointing rightd. kinetic friction vector,

pointing lefte. no friction vector, the

box is stationary in the truck

Make sure the cargo doesn’t slide• A box is in the back of a flatbed

truck What is the maximum acceleration the truck can have without the box sliding? Coefficients of friction between truck and box are μs = 0.40 and μk = 0.20. To solve this problem:

• Is the object of interest the box, truck, or both?

• Do the forces on the object of interest include static friction, kinetic friction, both, or neither?

EOC 115 – Hanging a picture

A person is trying to judge whether a picture of mass 1.10 kg is properly positioned by pressing it against a wall. The pressing (normal) force is perpendicular to the wall. The coefficient of static friction between picture and wall is 0.660. What is the minimum amount of pressing force required?

Draw a freebody diagram. In which direction is the normal force in this problem? Does it have anything to do with the weight of the picture?

Push the crateA dockworker pushes a 50-kg crate with a

downward force of 300.0 N, at an angle of 30 degrees below horizontal. The coefficient of kinetic friction is 0.30.

A. What is the normal force?

B. What is the frictional force?

C. What is the acceleration?

• Draw a freebody diagram of the object of interest (will it have tilted axes?)

• Apply Newton’s Law in component form.

Push the crate - Freebody diagram

n

Fpush

Fg

fk

θ

Knownsm = 50 kgFpush = 300Nθ = 30˚μk = 0.30

Findn, fk ,a

Push the crate (problem 1 from Exploration 4)

A. ΣFy = may n - FG - Fp sin 30˚

= 0

this solves for n = 640NB. fk = μk |n| = 192NC.ΣFx = max

Fp cos 30˚ - μk |n| = max

This will give us ax which is a = 1.36 m/s/s

Keep it up! Problem 3 from Exploration 4

A man pushes with a 100 N force on a 10- kg crate, which is sliding down the wall (despite his efforts). The pushing force is directed 53 degrees above horizontal. The coefficient of kinetic friction is 0.60. Determine the magnitude of the friction force between wall and crate.

Draw a freebody diagram. In which direction is the normal force in this problem?

Use Newton’s law in the appropriate direction to determine the normal force.