objectives: after completing this module, you should be able to: write newton’s second law using...

Objectives: After completing Objectives: After completing this module, you should be this module, you should be able to:able to:• Write Write Newton’s second lawNewton’s second law using appropriate using appropriate

units for units for massmass, , forceforce, and , and accelerationacceleration..

• Demonstrate your understanding of the Demonstrate your understanding of the distinction between distinction between massmass and and weightweight..

• Draw Draw free-body diagramsfree-body diagrams for objects at rest and for objects at rest and in motion.in motion.

• Apply Apply Newton’s second lawNewton’s second law to problems to problems involving one or more bodies in constant involving one or more bodies in constant acceleration.acceleration.

Newton’s First Law Newton’s First Law ReviewedReviewed

Newton’s First Law:Newton’s First Law: An object at rest or an An object at rest or an object in motion at constant speed will object in motion at constant speed will remain at rest or at constant speed in the remain at rest or at constant speed in the absence of a resultant force.absence of a resultant force.


A glass is placed on a board and the board is jerked quickly to the right. The glass tends to remain at rest while the board is removed.

A glass is placed on a board and the board is jerked quickly to the right. The glass tends to remain at rest while the board is removed.

Newton’s First Law (Cont.)Newton’s First Law (Cont.)



Assume glass and board move together at constant speed. If the board stops suddenly, the glass tends to maintain its constant speed.

Assume glass and board move together at constant speed. If the board stops suddenly, the glass tends to maintain its constant speed.

Understanding the First Understanding the First Law:Law:

(a) The driver is forced to move forward. An object at rest tends to remain at rest.

Discuss what the driver experiences when a car accelerates from rest and then applies the brakes.

(b) Driver must resist the forward motion as brakes are applied. A moving object tends to remain in motion.

Newton’s Second Law:Newton’s Second Law:

• Second Law:Second Law: Whenever a resultant Whenever a resultant force acts on an object, it produces an force acts on an object, it produces an acceleration: an acceleration that is acceleration: an acceleration that is directly proportional to the force and directly proportional to the force and inversely proportional to the mass.inversely proportional to the mass.

• Second Law:Second Law: Whenever a resultant Whenever a resultant force acts on an object, it produces an force acts on an object, it produces an acceleration: an acceleration that is acceleration: an acceleration that is directly proportional to the force and directly proportional to the force and inversely proportional to the mass.inversely proportional to the mass.

Fa

m

Fa

m

Acceleration and Force Acceleration and Force With Zero Friction With Zero Friction

ForcesForces

Pushing the cart with twice the force produces twice the acceleration. Three times the force triples the acceleration.

Acceleration and Mass Acceleration and Mass Again With Zero Again With Zero

FrictionFriction

F F

aa/2

Pushing two carts with same force F produces one-half the acceleration. The acceleration varies inversely with the amount of material (the mass).

Measuring Mass and ForceMeasuring Mass and Force

The The SI unit of forceSI unit of force is the is the newton (N)newton (N) and the unit for and the unit for massmass is the is the kilogram kilogram (kg)(kg)..Before presenting formal definitions of Before presenting formal definitions of these units, however, we will conduct an these units, however, we will conduct an experiment by slowly increasing the force experiment by slowly increasing the force on a given object.on a given object.Although the force in Although the force in newtonsnewtons will will become our standard, we begin by using become our standard, we begin by using the more familiar unit of force--the the more familiar unit of force--the pound (lb).pound (lb).

Newton: The Unit of ForceNewton: The Unit of ForceOne One newtonnewton is that resultant force which is that resultant force which imparts an acceleration of imparts an acceleration of 11 m/sm/s22 to a mass to a mass ofof 1 kg1 kg..

F (N) = m (kg) a (m/s2)F (N) = m (kg) a (m/s2)

What resultant force will give a 3 kg mass What resultant force will give a 3 kg mass an acceleration of 4 m/san acceleration of 4 m/s22? ?

2(3 kg)(4 m/s )F

F = 12 NF = 12 N

Remember F = m aF = ?

a = 4 m/s2

3 kg

Consistent Units (Cont.)Consistent Units (Cont.)

When we say that the acceptable units for force and mass are the newton and the kilogram, we are referring to their use in physical formulas. ( Such as F = m a)

The centimeter, the millimeter, the milligram, the mile, and the inch may be useful occasionally in describing quantities. But they should not be used in formulas.

Problem Solving StrategyProblem Solving Strategy(For the Simpler Problems.)(For the Simpler Problems.)

• Read problem; draw and label sketch.Read problem; draw and label sketch.

• List all given quantities and state what is List all given quantities and state what is to be found.to be found.

• Make sure all given units are consistent Make sure all given units are consistent with Newton’s second law of motion (with Newton’s second law of motion (F = F = m m aa).).

• Determine two of the three parameters in Determine two of the three parameters in Newton’s law, then solve for the Newton’s law, then solve for the unknown.unknown.


• List all given quantities and state what is List all given quantities and state what is to be found.to be found.

• Make sure all given units are consistent Make sure all given units are consistent with Newton’s second law of motion (with Newton’s second law of motion (F = F = m m aa).).

• Determine two of the three parameters in Determine two of the three parameters in Newton’s law, then solve for the Newton’s law, then solve for the unknown.unknown.

Example 4.Example 4. A A 54-gm54-gm tennis ball is in tennis ball is in contact with the racket for a distance of contact with the racket for a distance of 40 cm40 cm as it leaves with a velocity of as it leaves with a velocity of 48 48 m/sm/s. What is the average force on the . What is the average force on the ball?ball?

Given: vGiven: voo = 0; v = 0; vff = 48 m/s = 48 m/s xx = 0.40 = 0.40 m; m = 0.0540 km; m; m = 0.0540 km; a a = ?= ?

First, draw sketch and list First, draw sketch and list given quantities:given quantities:

Given: vGiven: voo = 0; v = 0; vff = 48 m/s = 48 m/s xx = 40 cm; = 40 cm; mm = 54 = 54 gm gm a a = ?= ?

Consistent units require converting Consistent units require converting gramsgrams to to kilogramskilograms and and centimeterscentimeters to to metersmeters::

Cont. . .

Example 4 (Cont).Example 4 (Cont). A A 54-gm54-gm tennis ball tennis ball is in contact with the racket for a is in contact with the racket for a distance of distance of 40 cm40 cm as it leaves with a as it leaves with a velocity of velocity of 48 m/s48 m/s. What is the average . What is the average force on the ball?force on the ball?

2 202 ;fax v v

00

22(48 m/s)

; 2880 m/s2(0.40 m)

a a

F= F= (0.054 kg)(2880 m/s(0.054 kg)(2880 m/s22);); F = 156 N

F = maF = ma

2

2fv

ax

Knowing that Knowing that F = m F = m aa, we , we need first to find need first to find acceleration acceleration aa::

Weight and MassWeight and Mass

• WeightWeight is the force due to gravity. is the force due to gravity. It is directed downward and it varies It is directed downward and it varies from location to location.from location to location.

• MassMass is a universal constant which is a universal constant which is a measure of the inertia of a is a measure of the inertia of a body.body.

F = m F = m aa so that:so that: W = mgW = mg and and m m ==

WW

gg

Weight and Mass: Weight and Mass: ExamplesExamples

• What is the mass of a What is the mass of a 6464--lblb block?block?

W = mgW = mg

64 lb64 lb3232 ft/sft/s22

• What is the weight of a What is the weight of a 10-kg10-kg block?block?

9.89.8 m m/s/s22 WW

mm10 10 kgkg

W = mg = W = mg = (10 kg)(9.8 (10 kg)(9.8 m/sm/s22))

W = 98 NW = 98 N

2

64 lb

32 ft/2 s s

slugm

Mass is Constant; W Mass is Constant; W Varies.Varies.

EarthEarth

9898 NN

9.8 m/s9.8 m/s22

49 49 NN

4.9 m/s4.9 m/s22

m = = m = = 10 kg10 kgWW

gg

Description of ObjectsDescription of Objects• Objects described by mass or Objects described by mass or weight:weight:

• Conversions made by Newton’s 2nd Conversions made by Newton’s 2nd Law:Law:

W W (N) = (N) = mm (kg) x 9.8 m/s (kg) x 9.8 m/s22

W = mgW = mg m =m =WW

gg

Inconsistent Usage (Cont.)Inconsistent Usage (Cont.)

Even Even metric units metric units are used inconsistently. are used inconsistently. Mass in Mass in kgkg is often treated as if it were is often treated as if it were weight (weight (N)N). This is sometimes called the . This is sometimes called the kilogram-force. kilogram-force.

The kilogram is a mass - never a force - and it doesn’t have direction or vary with gravity.

A chemist might be A chemist might be asked to asked to weighweigh out 200 out 200 g of a certain element. g of a certain element. Also, you hear about a Also, you hear about a 10-kg 10-kg loadload as if it were as if it were weightweight. .

F

10 kg

Always Remember!!Always Remember!!In Physics, the use of Newton’s second law and many other applications makes it absolutely necessary to distinguish between mass and weight. Use the correct units!Metric SI units: Mass is in kg; weight is in N.

Example 5.Example 5. A A resultantresultant forceforce of of 40 40 NN gives a block an gives a block an accelerationacceleration of of 8 m/s8 m/s22. What is the . What is the weightweight of the of the block near the surface of the block near the surface of the Earth?Earth?

W=?W=?

F = F = 40 N40 Naa 8 8

m/sm/s22

To find weight, we must To find weight, we must first find the mass of the first find the mass of the block:block:

; F

F ma ma

2

40 N5 kg

8 m/sm

Now find weight of a 5-kg mass on earth.

Now find weight of a 5-kg mass on earth.

W = mgW = mg

= = (5 kg)(9.8 (5 kg)(9.8 m/sm/s22))

W = 49.0 NW = 49.0 N

Newton’s Third Law Newton’s Third Law (Reviewed):(Reviewed):

• Third Law:Third Law: For every action force, For every action force, there must be an equal and opposite there must be an equal and opposite reaction force. Forces occur in pairs.reaction force. Forces occur in pairs.

• Third Law:Third Law: For every action force, For every action force, there must be an equal and opposite there must be an equal and opposite reaction force. Forces occur in pairs.reaction force. Forces occur in pairs.

Action

Reaction

ActionReaction

Acting and Reacting Acting and Reacting ForcesForces

• Use the words Use the words byby and and onon to study to study action/reaction forces below as action/reaction forces below as they relate to the they relate to the handhand and the and the barbar::

• Use the words Use the words byby and and onon to study to study action/reaction forces below as action/reaction forces below as they relate to the they relate to the handhand and the and the barbar::

The The actionaction force is exerted force is exerted by by the _____ the _____ onon the _____. the _____.

The reaction force is The reaction force is exerted exerted by by the _____ the _____ onon the _____.the _____.

barbar

handhandss

barbar

handhandss

Action

Reaction

Review of Free-body Review of Free-body Diagrams:Diagrams:


• Construct force diagram for each object, Construct force diagram for each object, vectors at origin of x,y axes.vectors at origin of x,y axes.

• Dot in rectangles and label x and y Dot in rectangles and label x and y compo-nents opposite and adjacent to compo-nents opposite and adjacent to angles.angles.

• Label all components; choose positive Label all components; choose positive direction.direction.


• Construct force diagram for each object, Construct force diagram for each object, vectors at origin of x,y axes.vectors at origin of x,y axes.

• Dot in rectangles and label x and y Dot in rectangles and label x and y compo-nents opposite and adjacent to compo-nents opposite and adjacent to angles.angles.

• Label all components; choose positive Label all components; choose positive direction.direction.

Example of Free-body Example of Free-body DiagramDiagram

300 600

4 kg

A ABB

W = mg

300 600

Bx

By

Ax

Ay

1. Draw and label sketch.1. Draw and label sketch.

2. Draw and label vector force diagram.2. Draw and label vector force diagram.

3. Dot in rectangles and label x and y 3. Dot in rectangles and label x and y compo-nents opposite and adjacent to compo-nents opposite and adjacent to angles.angles.

Applying Newton’s Second Applying Newton’s Second LawLaw• Read, draw, and label problem.Read, draw, and label problem.

• Draw free-body diagram for each body.Draw free-body diagram for each body.

• Choose x or y-axis along motion and Choose x or y-axis along motion and choose direction of motion as positive.choose direction of motion as positive.

• Write Newton’s law for both axes:Write Newton’s law for both axes:

FFxx = m = m axx FFyy = m = m ayy

• Solve for unknown quantities. Solve for unknown quantities.

• Read, draw, and label problem.Read, draw, and label problem.






Example 7:Example 7: A cart and driver have a A cart and driver have a mass of mass of 120 kg120 kg. What force . What force FF is is required to give an acceleration of required to give an acceleration of 6 6 m/sm/s22 with no friction? with no friction?1. Read problem and draw a sketch.1. Read problem and draw a sketch.

2. Draw a vector force diagram and label 2. Draw a vector force diagram and label forces.forces.

Diagram for Cart:n

W

F

3. Choose x-axis along motion and 3. Choose x-axis along motion and indicate the right direction as positive indicate the right direction as positive (+).(+).

x+

Example 7 (Cont.)Example 7 (Cont.) What force What force FF is is required to give an acceleration of required to give an acceleration of 6 6 m/sm/s22? ?

Fy = 0; n - W = 0The normal force

n is equal to weight W

Fx = max; F = ma

F F = (120 kg)(6 m/s= (120 kg)(6 m/s22))

F = 720 N

Diagram for cart:

n

W

Fx

+

m =m = 120 kg120 kg

4. Write Newton's Law equation for both 4. Write Newton's Law equation for both axes.axes.

ay = 0

Example 8:Example 8: What is the tension What is the tension TT in in the rope below if the block the rope below if the block accelerates upward at accelerates upward at 4 m/s4 m/s22? (Draw ? (Draw sketch and free-body.)sketch and free-body.)

1010 kgkg

aa = = +4 m/s+4 m/s22

TT aa

TT

mgmg

++

FFxx = m = m aaxx = 0 (No info) = 0 (No info)FFyy = m = m aayy = m = m

aaT - mg = m T - mg = m aa

mg = mg = (10 kg)(9.8 m/s) = 98 (10 kg)(9.8 m/s) = 98 NN

m m aa= = (10 kg)(4 m/s) = 40 N(10 kg)(4 m/s) = 40 N

T - - 98 N98 N = = 40 40 NN

T = 138 N

T = 138 N

Example 9:Example 9: In the absence of friction, In the absence of friction, what is the acceleration down the what is the acceleration down the 303000 incline?incline?

303000

mgmg

nn

606000

nn

WW

mg mg cos 60cos 6000mg mg sin 60sin 6000

++

FFxx = m = m aaxx

mg mg cos 60cos 6000 = m = m aa

aa = g = g cos 60cos 6000

aa = = (9.8 m/s(9.8 m/s22) cos ) cos 606000

a = 4.9 m/s2

a = 4.9 m/s2

Example 10.Example 10. Two-Body Problem: Find tension Two-Body Problem: Find tension in the connecting rope if there is no friction on in the connecting rope if there is no friction on the surfaces.the surfaces.

2 2 kgkg 4 kg4 kg

12 N12 N Find acceleration Find acceleration of system and of system and tension in tension in connecting cord.connecting cord.

First apply First apply F = mF = maa to entire system ( to entire system (both both massesmasses).).

12 N12 Nnn

((mm22 + + mm44))gg

FFxx = = ((mm22 + m + m44) ) aa

12 N = (6 kg) 12 N = (6 kg) aa

aa = =1212 NN

6 kg6 kga = 2 m/s2

a = 2 m/s2

Example 10 (Cont.)Example 10 (Cont.) The two-body The two-body problem.problem.

2 kg

4 kg12 N Now find tension Now find tension

T in connecting T in connecting cord.cord.

Apply F = m Apply F = m aa to the to the 22 kgkg mass where mass where aa = = 2 2 m/sm/s22..

TTnn

mm22 g g

FFxx = = mm2 2 aa

TT = (2 kg)(2 m/s = (2 kg)(2 m/s22))

T = 4 NT = 4 N

Example 10 (Cont.)Example 10 (Cont.) The two-body problem.The two-body problem.

2 2 kgkg 4 kg4 kg

12 N12 N Same answer for Same answer for TT results from results from focusing on focusing on 4-kg4-kg by by itself.itself.

Apply Apply F = m F = m aa to the to the 4 kg4 kg mass where mass where aa = 2 = 2 m/sm/s22..

FFxx = = mm4 4 aa

12 N - 12 N - TT = (4 kg)(2 m/s = (4 kg)(2 m/s22))

T = 4 NT = 4 N

1212 NNnn

mm22 g g

TT

Example 11Example 11 Find acceleration of system Find acceleration of system and tension in cord for the arrangement and tension in cord for the arrangement shown.shown. First apply First apply F = m F = m aa to entire to entire

system system alongalong the line of the line of motion.motion.

FFxx = = ((mm22 + m + m44) ) aa

a = 6.53 m/s2

a = 6.53 m/s2

nn

mm22 gg

TT

mm44 gg

TT

+ + aa

Note Note mm22gg is balanced by is balanced by

nn..mm44g = g = ((mm22 + m + m44) )

aa(4 kg)(9.8 (4 kg)(9.8 m/sm/s22))

2 2 kgkg + + 4 kg4 kgaa = = = =

mm44gg

mm22 + m + m44

22 kgkg

44 kgkg

Example 11 (Cont.)Example 11 (Cont.) Now find the tension Now find the tension TT given that the acceleration is given that the acceleration is aa = 6.53 m/s= 6.53 m/s22..

To find To find TT, apply , apply F = m F = m aa to to just the just the 2 kg2 kg mass, ignoring 4 mass, ignoring 4

kg.kg.

T = T = (2 kg)(6.53 m/s(2 kg)(6.53 m/s22))

T = 13.1 NT = 13.1 N

Same answer if using 4 Same answer if using 4 kg.kg.

mm44g - T = mg - T = m4 4 aaT = mT = m44(g - (g - aa)) = = 13.1 N13.1 N

nn

mm22 gg

TT

mm44 gg

TT

+ + aa

22 kgkg

4 kg4 kg2 2 or xF m a T m a

Example 11.Example 11. Find the acceleration of the Find the acceleration of the system shown below. (The Atwood system shown below. (The Atwood machine.)machine.)First apply F = ma to entire system along the line of motion.

FFxx = = ((mm22 + m + m55) ) aa

a = 4.20 m/s2

a = 4.20 m/s2

TT

mm22 gg

mm55 gg

TT

+a+a

2 2 kgkg 5 kg5 kg

25 2

2 5

(5 kg 2 kg)(9.8 m/s )

2 kg + 5 kg

m g m ga

m m

5 2 2 5( )m g m g m m a

SummarySummary

Newton’s Second Law: A resultant force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass.

Newton’s Second Law: A resultant force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass.



Newton’s Third Law:Newton’s Third Law: For every action For every action force, there must be an equal and opposite force, there must be an equal and opposite reaction force. Forces occur in pairs.reaction force. Forces occur in pairs.

Newton’s Third Law:Newton’s Third Law: For every action For every action force, there must be an equal and opposite force, there must be an equal and opposite reaction force. Forces occur in pairs.reaction force. Forces occur in pairs.

Summary: ProcedureSummary: Procedure

• Read, draw and label problem.Read, draw and label problem.






• Read, draw and label problem.Read, draw and label problem.






; RR

FF ma a

m ; R

R

FF ma a

m N = (kg)(m/s2)N = (kg)(m/s2)

CONCLUSION: Chapter 7CONCLUSION: Chapter 7Newton’s Second Law of Newton’s Second Law of

MotionMotion

objectives: after completing this module, you should be able to: write newton’s second law using...

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