chapter 4 force and motion

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Day 7 – May 21 – WBL 4.1- 4.4 Chapter 4 Force and Motion PC141 Intersession 2013 Slide 1 We have completed our study of kinematics. Now it’s time to consider why objects should accelerate. In this chapter, we introduce Newton’s laws, on which all of classical mechanics is based. These laws describe the relation between force and acceleration. Newto n Force

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Chapter 4 Force and Motion. We have completed our study of kinematics. Now it’s time to consider why objects should accelerate. In this chapter, we introduce Newton’s laws, on which all of classical mechanics is based. These laws describe the relation between force and acceleration. Newton. - PowerPoint PPT Presentation

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Page 1: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Chapter 4Force and Motion

PC141 Intersession 2013 Slide 1

We have completed our study of kinematics. Now it’s time to consider why objects should accelerate. In this chapter, we introduce Newton’s laws, on which all of classical mechanics is based. These laws describe the relation between force and acceleration.

Newton

Force

Page 2: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

The relationship between a force and the acceleration that it causes was first described by Isaac Newton in a 1687 publication, Philosophiae Naturalis Principia Mathematica, or “Mathematical Principles of Natural Philosophy.”

4.0 Newtonian Mechanics

PC141 Intersession 2013 Slide 2

The field of study now known as classical mechanics or Newtonian mechanics is based on these laws of motion.

Page 3: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Newtonian mechanics does not apply to all situations:• If the objects in question are moving at speeds that are an

appreciable fraction of the speed of light, we must invoke principles of relativity (PC242).

• If the objects are very small ( < 10 nm or so), their motion is described by quantum mechanics (PC321/PC331).

However, it is valid over a vast range of sizes and speeds

4.0 Newtonian Mechanics

PC141 Intersession 2013 Slide 3

Micro-ratchet. The gears are about 10-5 m in diameter. (Sandia Nat’l Labs)

Hercules galaxy cluster A2151. Distance between galaxies is about 1022 m. (Jim Misti)

Page 4: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

It’s not easy to precisely define “force”. The text provides a fairly good definition:

The qualifier “…is capable of…” is important. A force doesn’t necessarily produce an acceleration – it’s possible for multiple forces to cancel each other out such that no acceleration occurs (picture a tug-of-war, for example). However, if there is precisely one force acting on an object, acceleration must occur. This statement might seem odd to anyone who has tried to push a heavy box across a floor with no success…can this be explained?

4.1 The Concepts of Force and Net Force

PC141 Intersession 2013 Slide 4

A force is something that is capable of changing an object’s state of motion, that is, changing its velocity or producing an acceleration

Page 5: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

If a force is to produce an acceleration (which is a vector quantity), the force must also be a vector quantity. That is to say, every force has both a magnitude and a direction. Forces are generally indicated by the symbol , although our text uses other symbols for particular forces.When several forces act on an object, we are often interested in their sum, called the net force, . Note carefully that this is a vector sum, which must be calculated using the methods of chapter 3.Forces can be classified as contact forces, in which objects apply forces on each other due to physical contact, or action-at-a-distance or non-contact forces, in which no physical contact is required. Gravity is an example of the latter.

4.1 The Concepts of Force and Net Force

PC141 Intersession 2013 Slide 5

Page 6: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

The concept of net force allows us to consider balanced and unbalanced forces. Balanced forces have , and can not

4.1 The Concepts of Force and Net Force

PC141 Intersession 2013 Slide 6

lead to an acceleration. Unbalanced forces, on the other hand, will cause an acceleration.

Page 7: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Aristotle (~350 BC) believed that the “natural state” of an object is to be at rest, and that a force was needed in order to keep an object moving at constant velocity. Let’s return to the idea of pushing a heavy box along the floor…when you stop pushing, the box stops moving. We understand now that this is a result of friction. If you remove the friction (as in an ice rink or air hockey table) then the object will keep moving even when the force is no longer applied.Newton’s first law states that

4.2 Inertia and Newton’s First Law of Motion

PC141 Intersession 2013 Slide 7

In the absence of an unbalanced applied force (), a body at rest remains at rest and a body in

motion remains in motion with a constant velocity

Page 8: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

This is commonly phrased “a body at rest stays at rest, while a body in motion continues to move with the same velocity.”This law was actually described as early as the 3rd century BC by the Chinese philosopher Mo Tzu, and further developed by Descartes, Galileo, and others. In fact, Newton’s 1st law is often referred to as the law of inertia, after Galileo’s earlier assertion that…

4.2 Inertia and Newton’s First Law of Motion

PC141 Intersession 2013 Slide 8

Inertia is the natural tendency of an object to maintain a state of rest or to remain in uniform

motion in a straight line

Page 9: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Newton’s first law is not true in all reference frames (these were defined in the last lecture). An inertial frame is one in which Newton’s laws hold.Frames that are accelerating are called noninertial. For instance, Newton’s first law will fail on an airplane that is taking off. We’ll discuss this situation in class.In PC141, we assume that the ground is an inertial frame. In reality this is not exactly true; because of the earth’s rotation, every point on its surface is actually accelerating. The effects of this rotation are small but they can be noticeable in some circumstances (PC235!)Not to worry though – unless otherwise specified, in this course we will assume that all reference frames are inertial.

4.2 Inertia and Newton’s First Law of Motion

PC141 Intersession 2013 Slide 9

Page 10: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #1: Constant Velocity

PC141 Intersession 2013 Slide 10

If an object is moving at constant velocity…

A …there must be a force in the direction of the velocity

B …there must be no force in the direction of the velocity

C …there must be no net force

D …there must be a net force in the direction of the velocity

WBL LP 4.3

Page 11: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #2: A Block at Rest

PC141 Intersession 2013 Slide 11

WBL EX 4.7

A 5.0 kg block at rest on a frictionless surface is acted on by forces F1 = 5.5 N and F2 = 3.5 N as shown in the figure. What additional force will keep the block at rest? Consider only the horizontal motion for now.

Solution: In class

Page 12: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Newton’s 2nd law of motion is commonly written

From this definition, we can see that the SI unit for force is kg·m/s2. We call this derived unit the Newton, with symbol N.Rearranging this equation, we see that. In other words, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The

4.3 Newton’s 2nd Law of Motion

PC141 Intersession 2013 Slide 12

direction of the acceleration is in the direction of the net force.

Page 13: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

One Newton of net force will give a mass of 1 kg an acceleration of 1 m/s2, a mass of 2 kg an acceleration of 0.5 m/s2, and a mass of 0.5 kg an acceleration of 2 m/s.If the net force on an object is zero, the object’s acceleration must also be zero. This agrees with Newton’s first law.

4.3 Newton’s 2nd Law of Motion

PC141 Intersession 2013 Slide 13

Component formNewton’s second law is a vector equation. For 2D and 3D problems, it is valid in each dimension separatelyThat is,

Later in this chapter, we will solve higher-dimension problems using these equations.

Page 14: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Mass and WeightIn everyday speech, the terms “mass” and “weight” are used interchangeably. However, these are actually different quantities entirely. In section 1.2 of the text (during the initial discussion of mass), the weight of an object was defined as the gravitational attraction that the Earth exerts on the object. We didn’t have a good understanding of this “attraction” at the time, but now we can elaborate – this attraction is the gravitational force. And, since this force causes the object to accelerate downward with a magnitude g, we can use Newton’s second law to write

Our textbook uses the symbol for weight (I find this unfortunate…it’s a force, so it should be indicated with an “F”. Many other texts indicate it as ).

4.3 Newton’s 2nd Law of Motion

PC141 Intersession 2013 Slide 14

Page 15: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Mass and Weight cont’The result of this discussion is simply that an object’s weight is given by the magnitude of its gravitational force,

That is, an object with a mass of 1.0 kg has a weight of

(1.0 kg)(9.8 m/s2) = 9.8 kg·m/s2 = 9.8 N

This raises a couple of issues. The first is that whereas an object’s mass is an intrinsic property of the object, its weight depends on the magnitude of gravitational acceleration. Remember, g = 9.80 m/s2 only near the surface of the Earth. On the moon, g is only about one-sixth as large, so the same 1.0-kg object will weigh must less there (about 1.6 N). In deep space, g is essentially zero, so the same object will be “weightless”.

4.3 Newton’s 2nd Law of Motion

PC141 Intersession 2013 Slide 15

Page 16: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Mass and Weight cont’The second issue is that mass and weight don’t even have the same dimensions (in the notation of chapter 1, weight has dimensions of ML/T2). But how is this possible? When I step on my bathroom scale, I get a reading in kg (or pounds...different units, but same dimensions).So what’s happening here? Is my scale actually telling me my mass, rather than my weight? No! If I took that scale to the moon, it would correctly give me a reading 1/6th of that on the Earth.The explanation is actually rather dull. My scale simply gives me a reading that is calibrated in mass units. That is, it has the conversion factor (g) already built in.

4.3 Newton’s 2nd Law of Motion

PC141 Intersession 2013 Slide 16

Page 17: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #3: Acceleration

PC141 Intersession 2013 Slide 17

The acceleration of an object is…

A …inversely proportional to the acting net force

B …directly proportional to its mass

C …directly proportional to the net force and inversely proportional to its mass

D …none of these

WBL LP 4.7

Page 18: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #4: Emergency Stop

PC141 Intersession 2013 Slide 18

WBL EX 4.26

In an emergency stop to avoid an accident, a shoulder-strap seatbelt holds a 60 kg passenger in place. If the car was initially traveling at 90 km/h and came to a stop in 5.5 s along a straight, level road, what was the average force applied to the passenger by the seatbelt?

Solution: In class

Page 19: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #5: Two Boxes

PC141 Intersession 2013 Slide 19

WBL EX 4.28

A force of 10 N acts on two boxes on a frictionless surface. a) What is the acceleration of the system?b) What force does block A exert on block B?c) What force does block B exert on block A?

Solution: In class

Page 20: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #6: Your Weight in an Elevator

PC141 Intersession 2013 Slide 20

A passenger of mass m = 72.2 kg stands on a scale in an elevator.

a) Find a general solution for the scale reading (in Newtons), for any arbitrary vertical motion of the elevator.

b) What does the scale read if the elevator is stationary? Moving upward at a constant 0.50 m/s? Accelerating downward at 3.20 m/s2?

Solution: In class

Page 21: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #7: 2D Tug of War

PC141 Intersession 2013 Slide 21

In a 2D tug-of-war, Alex, Betty, and Charles pull horizontally on a tire at the angles shown in the overhead view shown below. The tire remains stationary in spite of the three pulls. Alex pulls with force of magnitude 220 N and Charles pulls with force of magnitude 170 N (note that the direction is not specified). What is the magnitude of Betty’s force ?

Solution: In class

Page 22: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Two objects interact when they push or pull on each other. In the figure to the right, a book rests on a crate. The book exerts a horizontal force on

4.4 Newton’s 3rd Law of Motion

PC141 Intersession 2013 Slide 22

the crate and crate exerts a horizontal force on the book. These forces are notated (“force on B due to C”) and (“force on C due to B”). Newton’s 3rd law states that…

This is more commonly stated “for every force (action), there is an equal and opposite force (reaction)”. In this example, we can write . The choice of which is the “action” force and which is the “reaction” force is arbitrary. We refer to these two forces as “third-law force pairs”.

when two objects interact, the forces on the objects from each other are always equal in

magnitude and opposite in direction

Page 23: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Normal ForceThis is a good time to introduce the normal force. In this context, “normal” means “perpendicular to”.Consider the following thought experiment: a block sitting motionless on the sidewalk is subjected to a downward gravitational force. According to Newton’s 2nd law, the block should therefore experience a downward acceleration,

4.4 Newton’s 3rd Law of Motion

PC141 Intersession 2013 Slide 23

equal to . But it doesn’t…remember, it’s just sitting there! The reason that its acceleration is zero is that the gravitational force is balanced by a normal force, .

Page 24: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Normal Force cont’It might seem as if the normal force is simply a mathematical trick, used in order to make Newton’s 2nd law work. This is not true – is a real, physical force.Although it seems rigid, the sidewalk (and any other surface) deforms very slightly when the block tries to accelerate downward into it. The normal force represents the effort of the sidewalk to restore its original shape.For this example, Newton’s 2nd law in the vertical direction tells us that

+ = If , the normal force simply has a magnitude of .

4.4 Newton’s 3rd Law of Motion

PC141 Intersession 2013 Slide 24

Page 25: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Does this mean that the normal force is always simply equal in magnitude to an object’s weight, but opposite in direction?

No!The preceding example only described an object placed on a level surface, with zero acceleration and no other applied forces.Fig. 4.11 of the text illustrates some more interesting aspects of the normal force. We will discuss them in class.

4.4 Newton’s 3rd Law of Motion

PC141 Intersession 2013 Slide 25

Page 26: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

4.4 Newton’s 3rd Law of Motion

PC141 Intersession 2013 Slide 26

Page 27: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #8: Head-On Collision

PC141 Intersession 2013 Slide 27

A fly collides with a car’s windshield, demolishing the fly with no discernible damage to the car. Based on this delightful experiment, we can conclude that…

A …the magnitude of the force of the car on the fly is greater than the magnitude of the force of the fly on the car

B …the magnitude of the force of the car on the fly is equal to the magnitude of the force of the fly on the car

C …the magnitude of the force of the car on the fly is smaller than the magnitude of the force of the fly on the car

D None of these can be concluded

WBL LP 4.11

Page 28: Chapter 4 Force and Motion

Day 7 – May 21 – WBL 4.1-4.4

Problem #9: Slapshot to the Face

PC141 Intersession 2013 Slide 28

WBL EX 4.87

A hockey puck impacts a goalie’s plastic mask horizontally at 122 mph, and rebounds horizontally off of the mask at 47 mph. If the puck has a mass of 170 g and it is in contact with the mask for 25 ms…

a) what is the average force (magnitude and direction) that the puck exerts on the mask?

b) if this average force accelerates the goalie, with what speed will the goalie move, assuming she was initially at rest and has a total mass (including padding) of 85 kg? Neglect friction with the ice.

Solution: In class