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. • To identify action/reaction pairs of forces. • To understand and use Newton’s third law. • To understand how to use propulsion forces and tension forces.

Newton’s Third Law A force results due to an interaction between two objects

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 a system of interacting objects

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

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

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 FAonB , FBonA for example.

Acceleration constraint An acceleration constraint is a well-defined relationship between the acceleration of 2 (or more) objects. In the case shown, we can assume ac =aT = ax Takes into account both magnitude and direction in an established coordinate system.

What is the acceleration constraint in this situation? aA = aB aA = - aB - aA = aB Both B and C None

What is the acceleration constraint in this situation? a2 = a1 a2 = - a1 a2 = 2a1 a2 = -.5a1

Problem-Solving Strategy: Interacting-Objects Problems

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-range attractive force between the magnets keeps the lower magnet in place. The upper magnet exerts a force of 6 Newtons on the lower magnet to keep it in place. a. Draw a fbd for each magnet and table. Determine all action/reaction pairs and connect them with dashed lines. b. Find the magnitude of all forces in your fbd and list them in a table.

EOC #8 Draw a fbd for each magnet and table. Determine all action/reaction pairs and connect them with dashed lines. How many action/ reaction pairs are there (don’t include the ground? 1 2 3 4

EOC #8 Upper Table Lower

FBDs for EOC 8

EOC #8 - Answer

Ranking Task – Pushing blocks Block 1 has a mass of m, block 2 has a mass of 2m, block 3 has a mass of 3m. 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 1,2,3 3,2,1 1,3,2 1=2=3

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

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

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 axes for the FBD for each? Acceleration constraints? Draw FBDs, with 3rd 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

Use subscripts to avoid mixups. I suggest starting with the equations for the box; its easier to deal with. Since friction is involved solve N’s 2nd Law in both y and x Now, identify quantities in sled equations that you already solved for in box equations. Use subscripts to avoid mixups. Plug and chug. 0.06

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.

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. TAonB and TBonA 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 3rd law action/reaction pair point in the same direction! T 100kg on m Tm on 100kg

All three 50 kg blocks are at rest All three 50 kg blocks are at rest. The tension in rope 2 is ______________ the tension in rope 1. Equal to Half of Twice that of Answer: A

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? Equal to Greater than Less than Answer: C

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: Which way will it go George? 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 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 3rd law pairs connected with dashed lines. 4.0 kg 350

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) a = - 0.48 m/s2, 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 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 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/s2 tmin = 1.75 s (time is important in this one)

EOC # 46 Find an expression for F, the magnitude of the horizontal force for which m1 does not slide up or down the wedge. This expression should be in terms of m1, m2 , θ, and any known constants, such as g. All surfaces are frictionless.

EOC # 46 .

EOC # 46 . F = (m1 + m2) g tan θ

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

car. Which of the following statements is true? 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? The truck exerts a larger force on the car than the car exerts on the truck. The truck exerts a force on the car but the car doesn’t exert a force on the truck. The car exerts a force on the truck but the truck doesn’t exert a force on the car. The car exerts a larger force on the truck than the truck exerts on the car. The car exerts the same amount of force on the truck as the truck exerts on the car. Answer: E 49

Boxes A and B are sliding to the right across a frictionless table 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. There should be 4 forces. Answer: C

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

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.