Dynamics Laws of Motion More About...

DynamicsLaws of Motion

More About Forces

Lana Sheridan

De Anza College

Oct 10, 2017

Overview

• Newton’s first and second laws

Warm Up: Newton’s Second Law Implications

Question. If an object is not accelerating, can there be forcesacting on it?

A Yes.

B No.


Question. If an object is not accelerating, can there be forcesacting on it?

A Yes. ←B No.


Quick Quiz 5.3.1 You push an object, initially at rest, across africtionless floor with a constant force for a time interval ∆t,resulting in a final speed of v for the object. You then repeat theexperiment, but with a force that is twice as large. What timeinterval is now required to reach the same final speed v?

A 4∆t

B 2∆t

C ∆t2

D ∆t4

4S&J page 116.


Quick Quiz 5.3.1 You push an object, initially at rest, across africtionless floor with a constant force for a time interval ∆t,resulting in a final speed of v for the object. You then repeat theexperiment, but with a force that is twice as large. What timeinterval is now required to reach the same final speed v?

A 4∆t

B 2∆t

C ∆t2 ←

D ∆t4

4S&J page 116.

Overview

• mass and weight

• fundamental forces

• fields

• Newton’s 3rd Law

• action-reaction pairs of forces

• some types of mechanical forces

Mass

What is mass?

Very loosely speaking, it is a measure of the amount of “stuff” inan object.

As Newton’s second law implies, an object with greater mass ismore resistant to forces changing its velocity.

We call this resistance to changes in velocity “inertia”.

The mass that appears in Newton’s second law is sometimes calledinertial mass.

It happens to be equal to gravitational mass, because the strengthof gravitational interactions depends on mass. (More on this in aminute...)

The Difference between Mass and Weight

What is the difference between mass and weight?

Mass is a measure of inertia. Weight is a force an objectexperience due to a gravitational interaction.

The Difference between Mass and Weight

mass

A measure of the amount of matter in an object. Also, a measureof the inertia of an object, that is, its resistance to changes in itsmotion.

weight

The force due to gravity on an object.

Objects in free-fall can be said to be weightless, but they still havemass.

Weight Fg ,Fg = mg

Units: Newtons.

Forces at a Fundamental Level

Previously, we talked about two kinds of forces: contact forces andfield forces (ie. forces that act at a distance).

In mechanics problems, usually gravity is the only field force thatwe need to consider. The rest are all contact forces.

However, at a fundamental level, all forces that we know of arefield forces.

Forces at a Fundamental Level

Contact forces are a result of electrostatic repulsion at very smallscales.

The fundamental forces (interactions):

Force ∼ Rel. strength Range (m) Attract/Repel Carrier

Gravitational 10−38 ∞ attractive gravitonElectromagnetic 10−2 ∞ attr. & rep. photonWeak Nuclear 10−13 < 10−18 attr. & rep. W+,W−,Z 0

Strong Nuclear 1 < 10−15 attr. & rep. gluons

Gravity is actually quite a weak force, but it is the only one that(typically) matters on large scales.

Fields

field

A field is any kind of physical quantity that has values specified atevery point in space and time.

Fields were first introduced as a calculation tool. A force-field canbe used to identify the force a particular particle will feel at acertain point in space and time based on the other objects in itsenvironment that it will interact with.

We do not need a description of the sources of the field to describewhat their effect is on our particle.

FieldsTo be clear: When we adopt a field model of force interactions weseparate two interacting objects, placing one in the system and theother in the environment.

System

The Earth

Fg

FieldEnvironment

Examples of Fields

Gravity and the electrostatic force have associated fields.

g = −9.8 j N/kg, the gravitational field strength.

FG = mg FE = qE

We can also think of g as an acceleration. (m/s2 = N/kg)

Can we think of E as an acceleration (due to the electrostaticforce)? No. q 6= m

Examples of Fields



FG = mg FE = qE


Can we think of E as an acceleration (due to the electrostaticforce)?

No. q 6= m

Examples of Fields



FG = mg FE = qE


Can we think of E as an acceleration (due to the electrostaticforce)? No. q 6= m

Representing Fields

Fields are drawn with lines showing the direction of force that atest particle will feel at that point. The density of the lines at thatpoint in the diagram indicates the approximate magnitude of theforce at that point.

Gravitation: Electrostatic:

25.4 Obtaining the Value of the Electric Field from the Electric Potential 755

25.4 Obtaining the Value of the Electric Field from the Electric Potential

The electric field ES

and the electric potential V are related as shown in Equation 25.3, which tells us how to find DV if the electric field E

S is known. What if the situ-

ation is reversed? How do we calculate the value of the electric field if the electric potential is known in a certain region? From Equation 25.3, the potential difference dV between two points a distance ds apart can be expressed as

dV 5 2 ES

? d sS (25.15)

If the electric field has only one component Ex, then ES

? d sS 5 Ex dx . Therefore, Equation 25.15 becomes dV 5 2Ex dx, or

Ex 5 2dVdx

(25.16)

That is, the x component of the electric field is equal to the negative of the deriv-ative of the electric potential with respect to x. Similar statements can be made about the y and z components. Equation 25.16 is the mathematical statement of the electric field being a measure of the rate of change with position of the electric potential as mentioned in Section 25.1. Experimentally, electric potential and position can be measured easily with a voltmeter (a device for measuring potential difference) and a meterstick. Conse-quently, an electric field can be determined by measuring the electric potential at several positions in the field and making a graph of the results. According to Equa-tion 25.16, the slope of a graph of V versus x at a given point provides the magnitude of the electric field at that point. Imagine starting at a point and then moving through a displacement d sS along an equipotential surface. For this motion, dV 5 0 because the potential is constant along an equipotential surface. From Equation 25.15, we see that dV 5 2 E

S? d sS 5 0;

therefore, because the dot product is zero, ES

must be perpendicular to the displace-ment along the equipotential surface. This result shows that the equipotential sur-faces must always be perpendicular to the electric field lines passing through them. As mentioned at the end of Section 25.2, the equipotential surfaces associated with a uniform electric field consist of a family of planes perpendicular to the field lines. Figure 25.11a shows some representative equipotential surfaces for this situation.

Figure 25.11 Equipotential surfaces (the dashed blue lines are intersections of these surfaces with the page) and elec-tric field lines. In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point.

q

!

A uniform electric field produced by an infinite sheet of charge

A spherically symmetric electric field produced by a point charge

An electric field produced by an electric dipole

a b c

ES

Examples of Fields

The gravitational field caused by the Sun-Earth system can berepresented as:

1Figure from http://www.launc.tased.edu.au

Examples of Fields

The electrostatic field caused by an electric dipole system can berepresented as:

25.4 Obtaining the Value of the Electric Field from the Electric Potential 755

25.4 Obtaining the Value of the Electric Field from the Electric Potential

The electric field ES

and the electric potential V are related as shown in Equation 25.3, which tells us how to find DV if the electric field E

S is known. What if the situ-

ation is reversed? How do we calculate the value of the electric field if the electric potential is known in a certain region? From Equation 25.3, the potential difference dV between two points a distance ds apart can be expressed as

dV 5 2 ES

? d sS (25.15)

If the electric field has only one component Ex, then ES

? d sS 5 Ex dx . Therefore, Equation 25.15 becomes dV 5 2Ex dx, or

Ex 5 2dVdx

(25.16)

That is, the x component of the electric field is equal to the negative of the deriv-ative of the electric potential with respect to x. Similar statements can be made about the y and z components. Equation 25.16 is the mathematical statement of the electric field being a measure of the rate of change with position of the electric potential as mentioned in Section 25.1. Experimentally, electric potential and position can be measured easily with a voltmeter (a device for measuring potential difference) and a meterstick. Conse-quently, an electric field can be determined by measuring the electric potential at several positions in the field and making a graph of the results. According to Equa-tion 25.16, the slope of a graph of V versus x at a given point provides the magnitude of the electric field at that point. Imagine starting at a point and then moving through a displacement d sS along an equipotential surface. For this motion, dV 5 0 because the potential is constant along an equipotential surface. From Equation 25.15, we see that dV 5 2 E

S? d sS 5 0;

therefore, because the dot product is zero, ES

must be perpendicular to the displace-ment along the equipotential surface. This result shows that the equipotential sur-faces must always be perpendicular to the electric field lines passing through them. As mentioned at the end of Section 25.2, the equipotential surfaces associated with a uniform electric field consist of a family of planes perpendicular to the field lines. Figure 25.11a shows some representative equipotential surfaces for this situation.

Figure 25.11 Equipotential surfaces (the dashed blue lines are intersections of these surfaces with the page) and elec-tric field lines. In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point.

q

!

A uniform electric field produced by an infinite sheet of charge

A spherically symmetric electric field produced by a point charge

An electric field produced by an electric dipole

a b c

ES

1Figure from Serway & Jewett

Newton’s Third Law

Newton III

If two objects (1 and 2) interact the force that object 1 exerts onobject 2 is equal in magnitude and opposite in direction to theforce that object 2 exerts on object 1.

F1→2 = −F2→1

Or, as commonly stated: “every action has an equal and oppositereaction.”

Newton’s Third Law: Action Reaction Pairs

5.6 Newton’s Third Law 119

When it is important to designate forces as interactions between two objects, we will use this subscript notation, where F

Sab means “the force exerted by a on b.” The

third law is illustrated in Figure 5.5. The force that object 1 exerts on object 2 is popularly called the action force, and the force of object 2 on object 1 is called the reaction force. These italicized terms are not scientific terms; furthermore, either force can be labeled the action or reaction force. We will use these terms for conve-nience. In all cases, the action and reaction forces act on different objects and must be of the same type (gravitational, electrical, etc.). For example, the force acting on a freely falling projectile is the gravitational force exerted by the Earth on the projectile F

Sg 5 F

SEp (E 5 Earth, p 5 projectile), and the magnitude of this

force is mg. The reaction to this force is the gravitational force exerted by the pro-jectile on the Earth F

SpE 5 2 F

SEp. The reaction force F

SpE must accelerate the Earth

toward the projectile just as the action force FS

Ep accelerates the projectile toward the Earth. Because the Earth has such a large mass, however, its acceleration due to this reaction force is negligibly small. Consider a computer monitor at rest on a table as in Figure 5.6a. The gravita-tional force on the monitor is F

Sg 5 F

SEm. The reaction to this force is the force

FS

mE 5 2 FS

Em exerted by the monitor on the Earth. The monitor does not acceler-ate because it is held up by the table. The table exerts on the monitor an upward force nS 5 F

Stm, called the normal force. (Normal in this context means perpendicu-

lar.) In general, whenever an object is in contact with a surface, the surface exerts a normal force on the object. The normal force on the monitor can have any value needed, up to the point of breaking the table. Because the monitor has zero accel-eration, Newton’s second law applied to the monitor gives us g F

S5 nS 1 mgS 5 0,

so n j 2 mg j 5 0, or n 5 mg. The normal force balances the gravitational force on the monitor, so the net force on the monitor is zero. The reaction force to nS is the force exerted by the monitor downward on the table, F

Smt 5 2 F

Stm 5 2nS.

Notice that the forces acting on the monitor are FS

g and nS as shown in Figure 5.6b. The two forces F

SmE and F

Smt are exerted on objects other than the monitor.

Figure 5.6 illustrates an extremely important step in solving problems involv-ing forces. Figure 5.6a shows many of the forces in the situation: those acting on the monitor, one acting on the table, and one acting on the Earth. Figure 5.6b, by contrast, shows only the forces acting on one object, the monitor, and is called a force diagram or a diagram showing the forces on the object. The important picto-rial representation in Figure 5.6c is called a free-body diagram. In a free-body diagram, the particle model is used by representing the object as a dot and show-ing the forces that act on the object as being applied to the dot. When analyz-ing an object subject to forces, we are interested in the net force acting on one object, which we will model as a particle. Therefore, a free-body diagram helps us isolate only those forces on the object and eliminate the other forces from our analysis.

2

1

F12S

F12 !S

F21S

"F21S

Figure 5.5 Newton’s third law. The force F

S12 exerted by object 1

on object 2 is equal in magnitude and opposite in direction to the force F


on object 1.

Pitfall Prevention 5.6n Does Not Always Equal mg In the situation shown in Figure 5.6 and in many others, we find that n 5 mg (the normal force has the same magnitude as the gravita-tional force). This result, however, is not generally true. If an object is on an incline, if there are applied forces with vertical components, or if there is a vertical acceleration of the system, then n ? mg. Always apply Newton’s second law to find the relationship between n and mg.

Pitfall Prevention 5.7Newton’s Third Law Remember that Newton’s third-law action and reaction forces act on different objects. For example, in Figure 5.6, nS 5 F

Stm 5 2mgS 5 2 F

SEm. The

forces nS and mgS are equal in magnitude and opposite in direc-tion, but they do not represent an action–reaction pair because both forces act on the same object, the monitor.

Pitfall Prevention 5.8Free-Body Diagrams The most important step in solving a problem using Newton’s laws is to draw a proper sketch, the free-body dia-gram. Be sure to draw only those forces that act on the object you are isolating. Be sure to draw all forces acting on the object, includ-ing any field forces, such as the gravitational force.

! FtmS

FmtS

FmES

nS ! FtmS

nS

! FtmS

nS

! FgS S

! FgS

FEmS

! FgS

FEmS

a b c

FEm Figure 5.6 (a) When a computer monitor is at rest on a table, the forces acting on the monitor are the normal force nS and the gravitational force F

Sg. The reaction to nS is the force F

Smt

exerted by the monitor on the table. The reaction to FS

g is the force F

SmE exerted by the monitor on the Earth. (b) A force

diagram shows the forces on the monitor. (c) A free-body diagram shows the monitor as a black dot with the forces acting on it.





Sg 5 F



SpE 5 2 F





Sg 5 F


FS

mE 5 2 FS




S5 nS 1 mgS 5 0,


Smt 5 2 F

Stm 5 2nS.



SmE and F



2

1

F12S

F12 !S

F21S

"F21S





on object 1.



Stm 5 2mgS 5 2 F

SEm. The



! FtmS

FmtS

FmES

nS ! FtmS

nS

! FtmS

nS

! FgS S

! FgS

FEmS

! FgS

FEmS

a b c



Smt


g is the force F



Force Diagrams

Question. Do the two forces shown in the diagram that act on themonitor form an action-reaction pair under Newton’s third law?





Sg 5 F



SpE 5 2 F





Sg 5 F


FS

mE 5 2 FS




S5 nS 1 mgS 5 0,


Smt 5 2 F

Stm 5 2nS.



SmE and F



2

1

F12S

F12 !S

F21S

"F21S





on object 1.



Stm 5 2mgS 5 2 F

SEm. The



! FtmS

FmtS

FmES

nS ! FtmS

nS

! FtmS

nS

! FgS S

! FgS

FEmS

! FgS

FEmS

a b c



Smt


g is the force F



(A) Yes.

(B) No.

Force Diagrams

Question. Do the two forces shown in the diagram that act on themonitor form an action-reaction pair under Newton’s third law?





Sg 5 F



SpE 5 2 F





Sg 5 F


FS

mE 5 2 FS




S5 nS 1 mgS 5 0,


Smt 5 2 F

Stm 5 2nS.



SmE and F



2

1

F12S

F12 !S

F21S

"F21S





on object 1.



Stm 5 2mgS 5 2 F

SEm. The



! FtmS

FmtS

FmES

nS ! FtmS

nS

! FtmS

nS

! FgS S

! FgS

FEmS

! FgS

FEmS

a b c



Smt


g is the force F



(A) Yes.

(B) No. ←

Some types of forces

Gravitation

The force that massive objects exert on one another.

Newton’s Law of Universal Gravitation

FG =Gm1m2

r2

for two objects, masses m1 and m2 at a distance r .

G = 6.67× 10−11 Nm2 kg−2.(Challenge: check the units of G .)


Gravitation cont’d

For the moment, we will care about this force in that it givesobjects weight, w .

Fg = w = mg

and

g =GMEarth

R2Earth

The force Fg or w, acts downwards towards the center of theEarth.

Some types of forcesThe Normal Force

The reactive force of a surface, as implied by Newton’s Third Law.It acts in a direction perpendicular to the surface.

For an object of weight Fg sitting on a level surface, the normalforce, FN (or n in the textbook) is

FN = −Fg

Be careful! There are many cases in which the aboveequation is not true!

1Figure from www.sparknotes.com


Friction (will not be on 1st test)

The force a surface exerts on an object to resist its motion.

There are two kinds of friction, static friction, Fsf , and kineticfriction, Fkf .

The amount of friction depends on properties of the object and thesurface. More about this to follow!

The friction force always acts to oppose motion. That means thekinetic friction Fkf always points opposite to the velocity vector.


Tension

The force exerted by a rope or chain to suspend or pull an objectwith mass.

Problems involving tensions often require solving systems of vectorequations.

1Figure from www.sparknotes.com


Elastic Forces

Springs exert forces as they are being compressed or extended.They have a natural length, at which they remain if there are noexternal forces acting.

Hooke’s Law givesFspring = −kx

where k is a constant. x is the amount of displacement of one endof a spring from it’s natural length. (The amount of compressionor extension.

1Figure from CCRMA Stanford Univ.

Springs

Question. A mass is attached to the end of a spring which obeysHooke’s Law. The spring is compressed and then released. Is theacceleration of the mass constant?

(A) Yes.

(B) No.

Springs

Question. A mass is attached to the end of a spring which obeysHooke’s Law. The spring is compressed and then released. Is theacceleration of the mass constant?

(A) Yes.

(B) No. ←

Summary

• mass and weight

• forces and fields

• Newton’s 3rd law

• types of forces

First Test Friday (Oct 13).

(Uncollected) Homework Serway & Jewett,

• Ch 5, onward from page 136. Problems: 17, 23

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