Chapter 7 Newtons Third Law Chapter Goal To

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Chapter 7. Newton’s Third Law Chapter Goal: To use Newton’s third law to understand

Chapter 7. Newton’s Third Law Chapter Goal: To use Newton’s third law to understand interacting objects.

Ch. 7 – Student Learning Objectives • To learn how two objects interact. •

Ch. 7 – Student Learning Objectives • To learn how two objects interact. • To identify action/reaction pairs of forces. • To understand use Newton’s third law. • To understand how to use propulsion forces and tension forces.

OMIT section 7. 2 (#1 -7) in workbook, but don’t xxxxx it out •

OMIT section 7. 2 (#1 -7) in workbook, but don’t xxxxx it out • We’ll do a modification during class.

Newton’s Third Law

Newton’s Third Law

Action-reaction Pair If object A exerts a force on object B, then object B

Action-reaction Pair If object A exerts a force on object B, then object B exerts a force on object A. The pair of forces (due to one interaction), is called an action/reaction pair. The action/reaction pair will never appear in the same free body diagram.

Tactics: Analyzing interacting objects

Tactics: Analyzing interacting objects

Example - analyzing interacting objects A person pushes a large crate across a rough

Example - analyzing interacting objects A person pushes a large crate across a rough surface. • Identify the objects that are systems of interest • Draw free-body diagrams for each system of interest. • Identify all action/reaction pairs with a dashed line.

Forces involved in pushing a crate – FBD of person and crate

Forces involved in pushing a crate – FBD of person and crate

Propulsion Force • The force label fp shows that the static friction force on

Propulsion Force • The force label fp shows that the static friction force on the person is acting as a propulsion force. • This is a force that a system with an internal source of energy uses to drive itself forward.

Propulsion forces

Propulsion forces

Freebody Diagrams – Workbook exercises 1 -7 Draw a freebody diagram of each object

Freebody Diagrams – Workbook exercises 1 -7 Draw a freebody diagram of each object in the interacting system. Show action/reaction pair with red/orange dotted lines. Draw force vectors in another color. Label vectors with standard symbols. Label action/reaction pairs FAon. B , FBon. A for example.

A fishing line of negligible mass lifts a fish upward at constant speed. The

A fishing line of negligible mass lifts a fish upward at constant speed. The line and the fish are the system, the fishing pole is part of the environment. What, if anything, is wrong with the free-body diagrams?

A fishing line of negligible mass lifts a fish upward at constant speed. The

A fishing line of negligible mass lifts a fish upward at constant speed. The line and the fish are the system, the fishing pole is part of the environment. What, if anything, is wrong with the free-body diagrams? The gravitational force and the tension force are incorrectly identified as an action/reaction pair. The correct action reaction pair is…? Action/reaction pairs are never on the same free body diagram. Mass of line considered negligible so no weight force necessary.

Acceleration constraint • An acceleration constraint is a welldefined relationship between the acceleration of

Acceleration constraint • An acceleration constraint is a welldefined relationship between the acceleration of 2 (or more) objects. • In the case shown, we can assume ac =a. T = ax

Is there an acceleration constraint in this situation? If so, what is it? The

Is there an acceleration constraint in this situation? If so, what is it? The pulley is considered to be massless and frictionless.

Answer: Acceleration constraint is: a. A = -a. B The actual signs may not

Answer: Acceleration constraint is: a. A = -a. B The actual signs may not be known until the problem is solved, but the relationship is known from the start.

Workbook exercises 12 -16 + “ 17” 17

Workbook exercises 12 -16 + “ 17” 17

answers a 2 kg = -. 5 a 1 kg

answers a 2 kg = -. 5 a 1 kg

Boxes A and B are sliding to the right across a frictionless table. The

Boxes A and B are sliding to the right across a frictionless table. The hand H is slowing them down. The mass of A is larger than the mass of B. Rank in order, from largest to smallest, the horizontal forces on A, B, and H. Ignore forces on H from objects not shown in the picture.

Boxes A and B are sliding to the right across a frictionless table. The

Boxes A and B are sliding to the right across a frictionless table. The hand H is slowing them down. The mass of A is larger than the mass of B. Rank in order, from largest to smallest, the horizontal forces on A, B, and H. FB on H = FH on B > FA on B = FB on A from Newton’s 2 nd and 3 rd Laws

Problem-Solving Strategy: Interacting. Objects Problems

Problem-Solving Strategy: Interacting. Objects Problems

EOC #8 Two strong magnets each weigh 2 N and are on opposite sides

EOC #8 Two strong magnets each weigh 2 N and are on opposite sides of the table. The table, by itself, has a weight of 20 N. The long range attractive force between the magnets keeps the lower magnet in place. The magnetic force on the lower magnet is 3 times its weight. a. Draw a fbd for each magnet and table. Use dashed lines to connect all action/reaction pairs. b. Find the magnitude of all forces in your fbd and list them in a table.

EOC #8 Upper Table Lower

EOC #8 Upper Table Lower

FBDs for EOC 8

FBDs for EOC 8

EOC #8 - Answer

EOC #8 - Answer

Ranking Task – Pushing blocks Block 1 has a mass of m, block 2

Ranking Task – Pushing blocks Block 1 has a mass of m, block 2 has a mass of 2 m, block 3 has a mass of 3 m. The surface is frictionless. Rank these blocks on the basis of the net force on each of them, from greatest to least. If the net force on each block is the same, state that explicitly

Ranking Task – Pushing blocks Answer: 3 2 1 Reason: ΣF = ma. Acceleration

Ranking Task – Pushing blocks Answer: 3 2 1 Reason: ΣF = ma. Acceleration is equal for all blocks.

EOC #10 Block 1 has a mass of 1 kg, block 2 has a

EOC #10 Block 1 has a mass of 1 kg, block 2 has a mass of 2 kg, block 3 has a mass of 3 kg. The surface is frictionless. a. Draw a fbd for each block. Use dashed lines to connect all action/ reaction pairs. b. How much force does the 2 -kg block exert on the 3 -kg block? c. How much force does the 2 -kg block exert on the 1 -kg block?

EOC #10 - Answer b. How much force does the 2 -kg block exert

EOC #10 - Answer b. How much force does the 2 -kg block exert on the 3 -kg block? – 6 N c. How much force does the 2 -kg block exert on the 1 -kg block? – 10 N

Page 163, 2 nd Ed (Found in Chapter 5, 1 st ed)

Page 163, 2 nd Ed (Found in Chapter 5, 1 st ed)

Interacting systems problem (EOC #35) A rope attached to a 20 kg wooden sled

Interacting systems problem (EOC #35) A rope attached to a 20 kg wooden sled pulls the sled up a 200 snow-covered hill. A 10 kg wooden box rides on top of the sled. If the tension in the rope steadily increases, at what value of tension will the box slip?

Interacting systems problem (EOC #35) What are the objects of interest? What kind of

Interacting systems problem (EOC #35) What are the objects of interest? What kind of axes for the FBD for each? Acceleration constraints? Draw FBDs, with 3 rd law pairs connected with dashed lines. Find the max tension in the rope, so the box does not slip.

Box: 3 forces Sled: 6 forces 0. 06

Box: 3 forces Sled: 6 forces 0. 06

0. 06 • I suggest starting with the equations for the sled, since the

0. 06 • I suggest starting with the equations for the sled, since the unknown of interest is found there. • Identify quantities in sled equations that you can find by solving box equations. • Solve box equations. • Return to sled equations with newfound booty. • Plug and chug.

Interacting systems problem (EOC #35) A rope attached to a 20 kg wooden sled

Interacting systems problem (EOC #35) A rope attached to a 20 kg wooden sled pulls the sled up a 200 snow-covered hill. A 10 kg wooden box rides on top of the sled. If the tension in the rope steadily increases, at what value of tension will the box slip? Answer: 155 N.

The Massless String Approximation A horizontal forces only fbd for the string: TAon. S

The Massless String Approximation A horizontal forces only fbd for the string: TAon. S ● TBon. S ΣF = TBon. S – TAon. S = ma. If string is accelerating to the right TBon. S = TAon. S + ma

The Massless String Approximation Often in physics and engineering problems the mass of the

The Massless String Approximation Often in physics and engineering problems the mass of the string or rope is much less than the masses of the objects that it connects. In such cases, we can adopt the following massless string approximation: This allows the objects A and B to be analyzed as if they exert forces directly on each other.

Pulleys If we assume that the string is massless and the pulley is both

Pulleys If we assume that the string is massless and the pulley is both massless and frictionless, no net force is needed to turn the pulley. TAon. B and TBon. A act “as if” they are an action/reaction pair, even though they are not acting in opposite directions.

Pulleys • In this case the Newton’s 3 rd law action/reaction pair point in

Pulleys • In this case the Newton’s 3 rd law action/reaction pair point in the same direction! T 100 kg on m Tm on 100 kg

All three 50 kg blocks are at rest. Is the tension in rope 2

All three 50 kg blocks are at rest. Is the tension in rope 2 greater than, less than, or equal to the tension in rope 1? A. Equal to B. Greater than C. Less than

All three 50 kg blocks are at rest. Is the tension in rope 2

All three 50 kg blocks are at rest. Is the tension in rope 2 greater than, less than, or equal to the tension in rope 1? A. Equal to B. Greater than C. Less than

In the (moving) figure to the right, is the tension in the string greater

In the (moving) figure to the right, is the tension in the string greater than, less than, or equal to the weight of block B? A. Equal to B. Greater than C. Less than

In the figure to the right, is the tension in the string greater than,

In the figure to the right, is the tension in the string greater than, less than, or equal to the weight of block B? A. Equal to B. Greater than C. Less than

Interacting systems problem (EOC #40) A 4. 0 kg box (m) is on a

Interacting systems problem (EOC #40) A 4. 0 kg box (m) is on a frictionless 350 incline. It is connected via a massless string over a massless, frictionless pulley to a hanging 2. 0 kg mass (M). When the box is released: a. Which way will it go George? b. What is the tension in the string? 4. 0 kg 350

Interacting systems problem (EOC #40) a. Which way will it go? Even if you

Interacting systems problem (EOC #40) a. Which way will it go? Even if you have no clue, follow the plan! What are the objects of interest? What kind of axes for the FBD for each? Acceleration constraints? ? Draw FBDs, with 3 rd law pairs connected with dashed lines. 4. 0 kg 350

Interacting systems problem (EOC #40) How do you figure out which way the system

Interacting systems problem (EOC #40) How do you figure out which way the system will move, once m is released from rest? massless string approx. allows us to join the tensions as an “as if” interaction pair

Interacting systems problem (EOC #40)

Interacting systems problem (EOC #40)

Interacting systems problem (EOC #40) a = - 0. 48 m/s 2, T =

Interacting systems problem (EOC #40) a = - 0. 48 m/s 2, T = 21 N. Which way is the system moving? How does the tension compare to the tension in the string while the box was being held? Greater than, less than, equal to?

EOC # 33 The coefficient of static friction is 0. 60 between the two

EOC # 33 The coefficient of static friction is 0. 60 between the two blocks in the figure. The coefficient of kinetic friction between the lower block and the floor is 0. 20. Force F causes both blocks to slide 5 meters, starting from rest. Determine the minimum amount of time in which the motion can be completed without the upper block slipping.

EOC # 33 The coefficient of static friction is 0. 60 between the two

EOC # 33 The coefficient of static friction is 0. 60 between the two blocks in the figure. The coefficient of kinetic friction between the lower block and the floor is 0. 20. Force F causes both blocks to slide 5 meters, starting from rest. Determine the minimum amount of time in which the motion can be completed without the upper block slipping.

EOC # 33 amax = 3. 27 m/s 2 tmin = 1. 75 s

EOC # 33 amax = 3. 27 m/s 2 tmin = 1. 75 s (time is important in this one)

EOC # 46 Find an expression for F, the magnitude of the horizontal force

EOC # 46 Find an expression for F, the magnitude of the horizontal force for which m 1 does not slide up or down the wedge. This expression should be in terms of m 1, m 2 , θ, and any known constants, such as g. All surfaces are frictionless.

EOC # 46.

EOC # 46.

EOC # 46. F = (m 1 + m 2) g tan θ

EOC # 46. F = (m 1 + m 2) g tan θ

EXAMPLE 7. 6 Comparing two tensions QUESTION:

EXAMPLE 7. 6 Comparing two tensions QUESTION:

EXAMPLE 7. 6 Comparing two tensions

EXAMPLE 7. 6 Comparing two tensions

EXAMPLE 7. 6 Comparing two tensions

EXAMPLE 7. 6 Comparing two tensions

EXAMPLE 7. 6 Comparing two tensions

EXAMPLE 7. 6 Comparing two tensions

EXAMPLE 7. 6 Comparing two tensions

EXAMPLE 7. 6 Comparing two tensions

A small car is pushing a larger truck that has a dead battery. The

A small car is pushing a larger truck that has a dead battery. The mass of the truck is larger than the mass of the car. Which of the following statements is true? A. The truck exerts a larger force on the car than the car exerts on the truck. B. The truck exerts a force on the car but the car doesn’t exert a force on the truck. C. The car exerts a force on the truck but the truck doesn’t exert a force on the car. D. The car exerts a larger force on the truck than the truck exerts on the car. E. The car exerts the same amount of force on the truck as the truck exerts on the car.

A small car is pushing a larger truck that has a dead battery. The

A small car is pushing a larger truck that has a dead battery. The mass of the truck is larger than the mass of the car. Which of the following statements is true? A. The truck exerts a larger force on the car than the car exerts on the truck. B. The truck exerts a force on the car but the car doesn’t exert a force on the truck. C. The car exerts a force on the truck but the truck doesn’t exert a force on the car. D. The car exerts a larger force on the truck than the truck exerts on the car. E. The car exerts the same amount of force on the truck as the truck exerts on the car.