Exam 1: Chapters 1-4

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Chapter 5 The Laws of Motion

5.1 Force – a push or pull.

CT1: The force of the elevator on Norbert and Zot is

A. up.

B. down.

CT2: The force of the non-sticky elevator surface on Norbert and Zot is

A. up.

B. down.

Assume the elevator is near the Earth’s surface and that it is right-side up in the cartoon.

CT3: The acceleration of Norbert and Zot is

A. up.

B. down.

C. zero

Assume the elevator is near the Earth’s surface and that it is right-side up in the cartoon.

Fig. 5.1, p.113

A.B.C.D.E.

CT4

5.2 Newton’s First Law: A body remains in uniform motion (or at rest) unless acted upon by a net external force.

acceleration due to Earth’s rotation 0.03 m/s2

acceleration due to Earth’s orbit 0.006 m/s2

acceleration due to Sun’s orbit 2 x 10-10 m/s2

We will assume that the Earth’s surface is an inertial frame and not make errors greater than 0.03/10 = 0.3%.

An inertia frame of reference is a coordinate system (or frame) in which Newton’s Law’s hold.

Newton’s First Law involves force which is a vector so we can look at it separately in both the x and y directions. Remember Galileo!

A mass which has inertia won’t move unless a force is applied.

Demonstration

What will happen to the egg, which is currently in a state of rest?

5.3 Mass: Mach’s definition uses the fact that for two isolated masses acting on each other

muau = msas

(where a1 and a2 are magnitudes)

This fact is consistent with Newton’s formulation – in particular the 2nd Law.

Then mu = msas/au

5.4 Newton’s Second Law: The net external force is equal to the mass times the acceleration.

F = ma Fx = max

Fy = may remember Galileo!

Fz = maz

Normally we will do problems in a plane with only x and y components.

A.

B.

C.

D.

E.

CT5

A.

B.

C.

D.

E.

CT6

5.5 Gravitational Force

Fg = mg (weight)

g = -9.8 j (m/s2 )

x

y

5.6 Newton’s Third Law: If body A acts on body B, then body B acts back on body A with a force equal in magnitude but opposite in direction.

A BFBAFAB

A.

C.

B.

D.

E.

CT7

5.7 Applications of Newton’s Laws

Equilibrium is defined as F = 0.

Remember Galileo!

Ropes, strings, cords, etc. are assumed massless unless otherwise stated.

Thus tensions are the same throughout the rope, string, cord, etc.

Pulleys are assumed massless and mounted on frictionless bearings unless otherwise stated.

Thus pulleys only change the direction of the force.

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Applications of Newton’s Laws - Method Draw picture of the problem.

Choose body (bodies) to isolate.

Draw Free Body Diagrams (FBDs) for isolated bodies.

Choose and label coordinate axes.

Apply Newton’s 2nd Law: Fx = max and Fy = may

Solve for F, m or a.

Work out kinematics.

Check solution is reasonable.P5.2 (p.128)

F1 = 20 N; F2 = 15 N; m = 5.00 kgP5.9 (p.128)

CT8: What is ax in 6.9a in m/s2?

A. 1

B. 2

C. 3

D. 4

E. 5

CT9: What is ay in 6.9a in m/s2?

A. 1

B. 2

C. 3

D. 4

E. 5

Applications of Newton’s Laws - Method Draw picture of the problem.

Choose body (bodies) to isolate.

Draw Free Body Diagrams (FBDs) for isolated bodies.

Choose and label coordinate axes.

Apply Newton’s 2nd Law: Fx = max and Fy = may

Solve for F, m or a.

Work out kinematics.

Check solution is reasonable.P5.17 (p.129)

F = 18.0 N

m1 = 2.00 kg

m2 = 3.00 kg

m3 = 4.00 kg

P5.54 (p.133)

P5.31 (p.131)

a vs. Fx

-25

-20

-15

-10

-5

0

5

10

-200 -150 -100 -50 0 50 100 150

Fx (N)

a (

m/s

2)

Fx (N) a (m/s2)

100 8.04

80 6.04

60 4.04

40 2.04

20 0.04

0 -1.96

-20 -3.96

-40 -5.96

-60 -7.96

-78.4 -9.8

-78.4 -9.8

-80 -10

-100 -12.5

-120 -15

-140 -17.5

-160 -20

a = Fx/8

a = (Fx -2g)/10

a = -g

D.

C.

B.

A.

CT10

X The engine or battery exerts a force on the object.

X If an object is moving there is a “force of motion.”.

X An object can’t exert a force on itself.

X

X

X

D.

C.

B.

A.

CT11

X If an object moves, the third law pair forces must be unbalanced.

X The moving object or a faster moving object exerts a greater force.

X The student believes that inanimate/passive objects cannot exert a force.

X

X

X

X

Newton’s Third Law!

X

D.

C.

B.

A.

CT12

X The more active or energetic object exerts more force.

X The bigger or heavier object exerts more force.

X The student uses the effects of a force as an indication of the relative magnitudes of the forces in an interaction.

X

X

X

Newton’s Third Law!

D.

C.

B.

A.

CT13

X If an object moves, the third law pair forces must be unbalanced.

X The student identifies equal force pairs, but indicates that both forces act on the same object. (For the example of a book at rest on a table, the gravitational force down on the book and the normal force up by the table on the book are identified as an action-reaction pair.)

X The bigger or heavier object exerts more force.

Newton’s Third Law!

X

X

X

A.

B.

CT14

P5.27 (p.130)

5.8 Force of Friction

Force of Friction - Model

Static Friction

fs sn

(as needed to

maintain

equilibrium)

Kinetic Friction

fk = kn

(opposing

motion)

A.

B.

C.

D.

E.

CT15

s = 0.25

m = 3.00 kg

P5.39 (p.131)

P5.44 (p.132)

Fig. P5.45, p.145

F = 68.0 N

m1 = 12.00 kg

m2 = 18.00 kg

k = 0.100

P5.43 (p.132)

Applications of Newton’s Laws - Method

Draw picture of the problem.

Choose body (bodies) to isolate.

Draw Free Body Diagrams (FBDs) for isolated bodies.

Choose and label coordinate axes.

Apply Newton’s 2nd Law: Fx = max and

Fy = may

Solve for F, m or a.

Work out kinematics.

Check solution is reasonable.