Presentation on theme: "Chapter 7. Newton’s Third Law"— Presentation transcript:
1Chapter 7. Newton’s Third Law Chapter Goal: To use Newton’s third law to understand interacting objects.
2Ch. 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.
3Newton’s Third Law A force results due to an interaction between two objects
4Action-reaction PairIf 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.
5Tactics: Analyzing a system of interacting objects
6Example - analyzing interacting objects A person pushes a large crate across a rough surface.Identify the objects that are systems of interestDraw free-body diagrams for each system of interest.Identify all action/reaction pairs with a dashed line.
7Forces involved in pushing a crate – FBD of person and crate
8Propulsion ForceThe 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.
10Freebody 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.
11Acceleration 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 = axTakes into account both magnitude and direction in an established coordinate system.
12What is the acceleration constraint in this situation? aA = aBaA = - aB- aA = aBBoth B and CNone
13What is the acceleration constraint in this situation? a2 = a1a2 = - a1a2 = 2a1a2 = -.5a1
15EOC #8Two 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.
16EOC #8Draw 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?1234
20Ranking 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 explicitly1,2,33,2,11,3,21=2=3
21EOC #10Block 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?
22EOC #10- Answerb. How much force does the 2-kg block exert on the 3-kg block? – 6Nc. How much force does the 2-kg block exert on the 1-kg block? – 10N
24Interacting 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?
25Interacting 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.
27Use 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 xNow, identify quantities in sled equations that you already solved for in box equations.Use subscripts to avoid mixups.Plug and chug.0.06
28Interacting 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.
29PulleysIf 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.
30PulleysIn this case the Newton’s 3rd law action/reaction pair point in the same direction!T 100kg on mTm on 100kg
31All 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 toHalf ofTwice that ofAnswer: A
32In 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 toGreater thanLess thanAnswer: C
33Interacting 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 kg350
34Interacting 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 kg350
35Interacting 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
37Interacting systems problem (EOC #40) a = 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?
38EOC # 33The 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 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.
39EOC # 33The 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 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.
40EOC # 33amax = 3.27 m/s2tmin = 1.75 s (time is important in this one)
41EOC # 46Find 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.
49car. 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 thecar. 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: E49
50Boxes 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
51The Massless String Approximation A horizontal forces only fbd for the string:TAonSTBonS●ΣF = TBonS – TAonS = ma. If string is accelerating to the rightTBonS = TAonS + ma
52The 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.