A large model rocket engine can produce a thrust of 120 N upon ignition The engine part has a total mass of 0288 kg when launched Draw a freebody diagram of the rocket What is the net force that is acting on the model rocket just after it leaves the ground ID: 760246
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Slide1
QQ:
Five people are playing tug-of-war. Greg and Jacob pull to the right with 45 N and 35 N, respectively. Brandon and Sharon pull to the left with 53 N and 38 N, respectively. With what force and in what direction does Victor pull if the game is tied?
A large model rocket engine can produce a thrust of 12.0 N upon ignition. The engine part has a total mass of 0.288 kg when launched.
Draw a free-body diagram of the rocket .
What is the net force that is acting on the model rocket just after it leaves the ground?
What is the initial acceleration of the rocket?
Slide2Today’s Objective: I can use Newton’s Third Law to explain forces and motion.
Slide3Isaac Newton and the Bike
Slide4Newton’s third law describes something else that happens when one object exerts a force on another object.
According to
Newton’s third law of motion
, forces always act in equal but opposite pairs
.
Slide5Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.
Newton’s Third Law of Motion
Slide6Example of 3rd Law
Space shuttle’s rocket boosters propel the orbiter into space by exerting an equal and opposite force to exhaust gasses.
Slide7Forces always come in pairs.
Slide8Interaction Pairs:
Two forces acting in opposite directions AND with equal magnitudes
The greater the force of the oars in this direction
The greater force the boat moves in this direction
Slide9The forces exerted by two objects on each other are often called
an action-reaction force pair.
Either force can be considered the action force or the reaction force.
Slide10Beware of these common misconceptions:1. They do NOT cancel each other out. If they did, there would never be any movement.
Slide11Forces that act on different objects don't cancel - after all, they affect the motion of different objects!
Fins push water backwards
.
The push on water moves the fish.
Water pushes the fish forward.
Size of force on water = size of force on fish
The direction of force on water is in the opposite of the direction the fish moves
Action pairs make it possible for the fish to move.
Slide12Mathematically, if the forces cancel each other, the net force would be zero. Think…. f = 0If the net forces = 0, think how this would affect objects using Newton’s 2nd Law: f = ma or rewritten as a = f/m If f is 0, acceleration would always be 0…therefore nothing would move.
0
Slide13Action and reaction force pairs don’t cancel because they act on different objects.
When you jump, you push down on the ground.
The ground then pushes up on you. It is this upward force that pushes you into the air.
Slide14Object A accelerates (or not) because of the forces that push or pull on it. (Newton's 2nd Law) Forces that push or pull on some other object have no effect on object A's motion - even if object A exerts them.
Slide15Only the forces that act on an object can cancel
.
Forces that act on different objects don't cancel - after all, they affect the motion of different objects
!
Wheels grip the road and push the road backwards. At the same time, the road pushes the wheels forward. The sizes of the forces are equal but in opposite directions. This moves the car forward.
Slide16These forces cancel each other out.
Both of these forces are acting on the box.
Slide17The normal upwards force and the force of gravity cancel each other because they are both acting on the same object.
Slide18Another common misconception:In the beginning it is often assumed that the more powerful entity in an interaction – the faster, bigger, or stronger one – will exerts a larger force. According to Newton’s third law, this is NOT true. The objects exert forces of identical magnitude.
Slide19The 3rd Law Expert --- Bill Nye
Slide20Solving Problems:Determine the system.Draw a system schema.
Slide21What is a System Schema?
Slide22Solving Problems:Determine the system.Draw a system schema.Create free-body diagrams for each system.Connect interaction pairs by dashed lines.Use 2nd law to relate net force & accelerationUse 3rd law to find magnitudes & direction
Slide23When a softball with a mass of 0.18 kg is dropped, its acceleration toward Earth is equal to the variable g. What is the force on Earth due to the ball, and what is Earth’s resulting acceleration? The mass of Earth is 6.0 x 10 24 kg.
Slide24Determine the system: Ball and earthDraw a system schema. 3) Create free-body diagrams for each system.4) Connect interaction pairs by dashed lines.Solve problem.mball = 0.18 kg m earth = 6.0 x 10 24 kgg = 9.8 m/s2 F earth on ball = ? aearth = ?
B
E
F
ball
on earth
F
earth
on ball
Slide25SOLVE:mball = 0.18 kg m earth = 6.0 x 10 24 kg g = 9.8 m/s2 F earth on ball = ? aearth = ? Use Newton’s 2nd law to find F earth on ball which is known as weight. F earth on ball = mball a Acceleration due to gravity is -9.8 m/s2F earth on ball = 0.18 kg (- 9.8 m/s2)F earth on ball = -1.8 kg *m/s2F earth on ball = -1.8 N
Slide26Use Newton’s third Law to find F ball on earth mball = 0.18 kg m earth = 6.0 x 10 24 kgF earth on ball = -1.8 N (from 1st problem solved)F ball on earth = -F earth on ballF ball on earth = -(-1.8 N)F ball on earth = 1.8 NF earth on ball =
Slide27Use Newton’s Second law to find the aearth because of the net force caused by the ball hitting it.Fnet = 1.8 N m earth = 6.0 x 10 24 kg Substitute: Fnet = m earth a earth 1.8 N = 6.0 x 10 24 kg * a earth 6.0 x 10 24 kg 6.0 x 10 24 kg 3 X 10 -25 N/kg = a earth 3 X 10 -25 m/s 2 = a earth
k
g * m/s
2
kg
Slide28You are walking along when you slip on some ice and fall. For a moment you are in free fall. During this time, what force do you exert on Earth if your mass is 55.0 kg? Hint: find the force Earth exerts on you first.
Slide29Determine the system.Draw a system schema.Create free-body diagrams for each system.Connect interaction pairs by dashed lines.Solve problems: Your mass = 55 kg Acceleration due to gravity = 9.8 m/s2 Choose formula: f = ma Solve: Fearth on you = 539 N
Slide30Solve: Fearth on you = 539 NThe force you exert on Earth is the same magnitude. Fyou on earth = 539 N
Slide31Newton’ s Three Laws of Motion
Slide32online problems
https://www.sophia.org/tutorials/newtons-3rd-law-of-motion-actionreaction-law
Slide33Two 100-N weights are attached to a spring scale as shown. Does the scale read zero, 100 N, or 200 N -- or some other reading?
Slide34The scale reads the tension in the string. The tension in the string is 100 N. This is the force the string must exert up on either of the 100-N weights at either end of the string.
Slide35Nothing is moving, nothing is accelerating, so the net force on the spring is zero. Likewise, the net force on either of the 100-N weights is also zero. But that is another question. The spring scale does not measure the net force. The spring scale simply measures the tension, the magnitude of the force exerted by the string.
Slide36Concept Question 1
Why are we able to walk?
Slide37Concept Question Answer
We walk forward because when one foot pushes backward against the ground, the ground pushes forward on that foot.
Force exerted on the ground
by person’s foot. Fgp
Force exerted
on the person’s foot by the ground. Fpg
Fgp=-Fpg
Slide38Concept Question 2
What makes a car go forward?
Slide39Concept Question answer
By Newton’s third law, the ground pushes on the tires in the opposite direction, accelerating the car forward.
Slide40Concept Question
Which is stronger, the Earth’s pull on an orbiting space shuttle or the space shuttle’s pull on the earth?
Slide41Concept Question Answer
According to Newton’s Third Law, the two forces are equal and opposite. Because of the huge difference in masses, however the space shuttle accelerates much more towards the Earth than the Earth accelerates toward the space shuttle. a = F/m
Slide42Problem 1
What force is needed to accelerate the 60 kg cart at 2m/s^2?
Slide43How to solve Problem 1
What force is needed to accelerate the 60kg cart at 2 m/s^2? Force = mass times acceleration F = m * aF = 60kg * 2m/s^2F = 120kgm/s^2Kgm/s^2 = NewtonNewton = NF =120 N
Slide44Problem 2
A force of 200 N accelerates a bike and rider at 2 m/s^2. What is the mass of the bike and rider?
Slide45How to solve Problem 2
A force of 200 N accelerate a bike and rider at 2m/s^2. What is the mass of the bike and rider?F = ma therefore:m =F/a m = 200N/2m/s^2N= kgm/s^2 so when divide your answer will be kg left. m = 100kg
Slide46Why can you exert greater force on the pedals of a bicycle if you pull up on the handlebars?The handlebars then pull down on you, somewhat as if someone were pushing down on your shoulders. This lets you exert a greater downward force on the pedals.
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