February 18, Week 6 Today: Chapter 4, Forces Exam #1 is in - - PowerPoint PPT Presentation

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February 18, Week 6 Today: Chapter 4, Forces Exam #1 is in - - PowerPoint PPT Presentation

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February 18, Week 6 Today: Chapter 4, Forces Exam #1 is in mailboxes Homework Assignment #5 - Due March 1. Mastering Physics: 10 problems from chapters 4 and 5. Written Questions: 5.74 Help sessions with Jonathan: M: 1000-1100, RH 111 T:

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SLIDE 1

Forces February 18, 2013 - p. 1/9

February 18, Week 6

Today: Chapter 4, Forces Exam #1 is in mailboxes Homework Assignment #5 - Due March 1.

Mastering Physics: 10 problems from chapters 4 and 5. Written Questions: 5.74

Help sessions with Jonathan: M: 1000-1100, RH 111 T: 1000-1100, RH 114 Th: 0900-1000, RH 114

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SLIDE 2

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem:

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SLIDE 3

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem: Weight - − → w

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SLIDE 4

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem: Weight - − → w, the downward force on an object due to gravity.

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SLIDE 5

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem: Weight - − → w, the downward force on an object due to gravity. Normal Force - − → n

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SLIDE 6

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem: Weight - − → w, the downward force on an object due to gravity. Normal Force - − → n , the perpendicular force exerted by one solid

  • bject onto another solid object.
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SLIDE 7

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem: Weight - − → w, the downward force on an object due to gravity. Normal Force - − → n , the perpendicular force exerted by one solid

  • bject onto another solid object.

Friction - − → f

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SLIDE 8

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem: Weight - − → w, the downward force on an object due to gravity. Normal Force - − → n , the perpendicular force exerted by one solid

  • bject onto another solid object.

Friction - − → f , force which slows a moving object, always

  • pposed to the motion ⇒ opposite to −

→ v .

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SLIDE 9

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem: Weight - − → w, the downward force on an object due to gravity. Normal Force - − → n , the perpendicular force exerted by one solid

  • bject onto another solid object.

Friction - − → f , force which slows a moving object, always

  • pposed to the motion ⇒ opposite to −

→ v . Tension - − → T

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SLIDE 10

Forces February 18, 2013 - p. 2/9

Force Examples

Forces to be identified in any problem: Weight - − → w, the downward force on an object due to gravity. Normal Force - − → n , the perpendicular force exerted by one solid

  • bject onto another solid object.

Friction - − → f , force which slows a moving object, always

  • pposed to the motion ⇒ opposite to −

→ v . Tension - − → T, pulling force exerted by rope, chain, or spring, always at same angle as rope.

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SLIDE 11

Forces February 18, 2013 - p. 3/9

Newton’s First Law

First Law - The Law of Inertia An object at rest stays at rest, an object in uniform motion stays if uniform motion if (and only if) the net force acting on the object is zero.

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SLIDE 12

Forces February 18, 2013 - p. 3/9

Newton’s First Law

First Law - The Law of Inertia An object at rest stays at rest, an object in uniform motion stays if uniform motion if (and only if) the net force acting on the object is zero. Uniform motion - Straight line and constant speed, i.e, constant velocity.

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SLIDE 13

Forces February 18, 2013 - p. 3/9

Newton’s First Law

First Law - The Law of Inertia An object at rest stays at rest, an object in uniform motion stays if uniform motion if (and only if) the net force acting on the object is zero. Uniform motion - Straight line and constant speed, i.e, constant velocity. Inertia - The property of all matter to stay in motion if already in motion; to stay at rest if already at rest.

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SLIDE 14

Forces February 18, 2013 - p. 4/9

First Law Example

Example: A 6860 N car is traveling with a constant 30 m/s speed on a straight road. If the ground is exerting a forward 350 N force∗, what is the magnitude and direction of all forces acting on the car? (∗ We’ll learn later that this is due to the car’s engine.)

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SLIDE 15

Forces February 18, 2013 - p. 4/9

First Law Example

Example: A 6860 N car is traveling with a constant 30 m/s speed on a straight road. If the ground is exerting a forward 350 N force∗, what is the magnitude and direction of all forces acting on the car? (∗ We’ll learn later that this is due to the car’s engine.) Free-Body Diagram - f. b. d. sketch of all the forces acting on an object using a convenient coordinate system.

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SLIDE 16

Forces February 18, 2013 - p. 5/9

First Law Exercise

A 5 kg mass is hung from the ceiling using a "massless" rope. What is the magnitude of the tension force exerted by the rope

  • n the mass? Hint: A 5 kg mass has a weight of 49 N on earth

where this problem is taking place.

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SLIDE 17

Forces February 18, 2013 - p. 5/9

First Law Exercise

A 5 kg mass is hung from the ceiling using a "massless" rope. What is the magnitude of the tension force exerted by the rope

  • n the mass? Hint: A 5 kg mass has a weight of 49 N on earth

where this problem is taking place. (a) 0 N

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SLIDE 18

Forces February 18, 2013 - p. 5/9

First Law Exercise

A 5 kg mass is hung from the ceiling using a "massless" rope. What is the magnitude of the tension force exerted by the rope

  • n the mass? Hint: A 5 kg mass has a weight of 49 N on earth

where this problem is taking place. (a) 0 N (b) 24.5 N

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SLIDE 19

Forces February 18, 2013 - p. 5/9

First Law Exercise

A 5 kg mass is hung from the ceiling using a "massless" rope. What is the magnitude of the tension force exerted by the rope

  • n the mass? Hint: A 5 kg mass has a weight of 49 N on earth

where this problem is taking place. (a) 0 N (b) 24.5 N (c) 49 N

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SLIDE 20

Forces February 18, 2013 - p. 5/9

First Law Exercise

A 5 kg mass is hung from the ceiling using a "massless" rope. What is the magnitude of the tension force exerted by the rope

  • n the mass? Hint: A 5 kg mass has a weight of 49 N on earth

where this problem is taking place. (a) 0 N (b) 24.5 N (c) 49 N (d) 98 N

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SLIDE 21

Forces February 18, 2013 - p. 5/9

First Law Exercise

A 5 kg mass is hung from the ceiling using a "massless" rope. What is the magnitude of the tension force exerted by the rope

  • n the mass? Hint: A 5 kg mass has a weight of 49 N on earth

where this problem is taking place. (a) 0 N (b) 24.5 N (c) 49 N (d) 98 N (e) Not enough information to determine

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SLIDE 22

Forces February 18, 2013 - p. 5/9

First Law Exercise

A 5 kg mass is hung from the ceiling using a "massless" rope. What is the magnitude of the tension force exerted by the rope

  • n the mass? Hint: A 5 kg mass has a weight of 49 N on earth

where this problem is taking place. (a) 0 N (b) 24.5 N (c) 49 N (d) 98 N (e) Not enough information to determine

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SLIDE 23

Forces February 18, 2013 - p. 6/9

First Law Exercise

Two 5 kg masses are connected to each over pulleys using a

  • rope. What is the tension force that the rope exerts on the

right-hand mass if they are both at rest?

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SLIDE 24

Forces February 18, 2013 - p. 6/9

First Law Exercise

Two 5 kg masses are connected to each over pulleys using a

  • rope. What is the tension force that the rope exerts on the

right-hand mass if they are both at rest? (a) 0 N

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SLIDE 25

Forces February 18, 2013 - p. 6/9

First Law Exercise

Two 5 kg masses are connected to each over pulleys using a

  • rope. What is the tension force that the rope exerts on the

right-hand mass if they are both at rest? (a) 0 N (b) 24.5 N

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SLIDE 26

Forces February 18, 2013 - p. 6/9

First Law Exercise

Two 5 kg masses are connected to each over pulleys using a

  • rope. What is the tension force that the rope exerts on the

right-hand mass if they are both at rest? (a) 0 N (b) 24.5 N (c) 49 N

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SLIDE 27

Forces February 18, 2013 - p. 6/9

First Law Exercise

Two 5 kg masses are connected to each over pulleys using a

  • rope. What is the tension force that the rope exerts on the

right-hand mass if they are both at rest? (a) 0 N (b) 24.5 N (c) 49 N (d) 98 N

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SLIDE 28

Forces February 18, 2013 - p. 6/9

First Law Exercise

Two 5 kg masses are connected to each over pulleys using a

  • rope. What is the tension force that the rope exerts on the

right-hand mass if they are both at rest? (a) 0 N (b) 24.5 N (c) 49 N (d) 98 N (e) Not enough information to determine

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SLIDE 29

Forces February 18, 2013 - p. 6/9

First Law Exercise

Two 5 kg masses are connected to each over pulleys using a

  • rope. What is the tension force that the rope exerts on the

right-hand mass if they are both at rest? (a) 0 N (b) 24.5 N (c) 49 N (d) 98 N (e) Not enough information to determine

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SLIDE 30

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v

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SLIDE 31

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0.

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SLIDE 32

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0. So if Σ− → F = 0 ⇒?

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SLIDE 33

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0. So if Σ− → F = 0 ⇒ − → a = 0.

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SLIDE 34

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0. So if Σ− → F = 0 ⇒ − → a = 0. Forces cause acceleration

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SLIDE 35

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0. So if Σ− → F = 0 ⇒ − → a = 0. Forces cause acceleration Newton found that the acceleration is:

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SLIDE 36

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0. So if Σ− → F = 0 ⇒ − → a = 0. Forces cause acceleration Newton found that the acceleration is: (a) In the same direction as the net force

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SLIDE 37

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0. So if Σ− → F = 0 ⇒ − → a = 0. Forces cause acceleration Newton found that the acceleration is: (a) In the same direction as the net force (b) Directly proportional to the net force

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SLIDE 38

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0. So if Σ− → F = 0 ⇒ − → a = 0. Forces cause acceleration Newton found that the acceleration is: (a) In the same direction as the net force (b) Directly proportional to the net force (c) Inversely proportional to the mass

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SLIDE 39

Forces February 18, 2013 - p. 7/9

Newton’s Second Law

The first law tells us that if Σ− → F = 0 then we have a constant − → v Constant − → v ⇒ − → a = 0. So if Σ− → F = 0 ⇒ − → a = 0. Forces cause acceleration Newton found that the acceleration is: (a) In the same direction as the net force (b) Directly proportional to the net force (c) Inversely proportional to the mass Measure

  • f

the amount of matter inside an object

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SLIDE 40

Forces February 18, 2013 - p. 8/9

Second Law II

− → a = Σ− → F M

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SLIDE 41

Forces February 18, 2013 - p. 8/9

Second Law II

− → a = Σ− → F M ⇒ Σ− → F = M− → a

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SLIDE 42

Forces February 18, 2013 - p. 8/9

Second Law II

− → a = Σ− → F M ⇒ Σ− → F = M− → a Units: Newton is a unit simplification.

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SLIDE 43

Forces February 18, 2013 - p. 8/9

Second Law II

− → a = Σ− → F M ⇒ Σ− → F = M− → a Units: Newton is a unit simplification. Ma

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SLIDE 44

Forces February 18, 2013 - p. 8/9

Second Law II

− → a = Σ− → F M ⇒ Σ− → F = M− → a Units: Newton is a unit simplification. Ma ⇒ kg · m/s2

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SLIDE 45

Forces February 18, 2013 - p. 8/9

Second Law II

− → a = Σ− → F M ⇒ Σ− → F = M− → a Units: Newton is a unit simplification. Ma ⇒ kg · m/s2 ΣF

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SLIDE 46

Forces February 18, 2013 - p. 8/9

Second Law II

− → a = Σ− → F M ⇒ Σ− → F = M− → a Units: Newton is a unit simplification. Ma ⇒ kg · m/s2 ΣF ⇒ N

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SLIDE 47

Forces February 18, 2013 - p. 8/9

Second Law II

− → a = Σ− → F M ⇒ Σ− → F = M− → a Units: Newton is a unit simplification. Ma ⇒ kg · m/s2 ΣF ⇒ N N = kg · m/s2

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SLIDE 48

Forces February 18, 2013 - p. 9/9

Second Law Examples

Example: A 6860 N car is in free-fall, what it its mass?

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SLIDE 49

Forces February 18, 2013 - p. 9/9

Second Law Examples

Example: A 6860 N car is in free-fall, what it its mass? Example: A 6860 N car is sitting stationary on the ground, what is its mass?

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SLIDE 50

Forces February 18, 2013 - p. 9/9

Second Law Examples

Example: A 6860 N car is in free-fall, what it its mass? Example: A 6860 N car is sitting stationary on the ground, what is its mass? w = Mg