Quick Questions: Quick Question: Go to the fun and helpful links part of my webpage and click on Newton’s 3rd Law self-quiz. Be sure to check hints to help you understand when you miss instead of just guessing again. https://www.khanacademy.org/science/physics/forces-newtons-laws/newtons-laws-of-motion/e/newtons-third-law

Solving Problems: Determine the system. Draw a system schema.

What is a System Schema?

Solving 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 & acceleration Use 3rd law to find magnitudes & direction

When a softball with a mass of 0 When 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.

F earth on ball = ? aearth = ? Determine the system: Ball and earth Draw 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 kg g = 9.8 m/s2 F earth on ball = ? aearth = ? Fball on earth B E Fearth on ball

Use Newton’s 2nd law to find F earth on ball which is known as weight. SOLVE: 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/s2 F earth on ball = 0.18 kg (- 9.8 m/s2) F earth on ball = -1.8 kg *m/s2 F earth on ball = -1.8 N

F earth on ball = -1.8 N (from 1st problem solved) Use Newton’s third Law to find F ball on earth mball = 0.18 kg m earth = 6.0 x 10 24 kg F earth on ball = -1.8 N (from 1st problem solved) F ball on earth = -F earth on ball F ball on earth = -(-1.8 N) F ball on earth = 1.8 N F earth on ball =

Substitute: Fnet = m earth a earth 1.8 N = 6.0 x 10 24 kg * a earth Use 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 kg * m/s2 kg

You are walking along when you slip on some ice and fall You 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.

Determine 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

Solve: Fearth on you = 539 N The force you exert on Earth is the same magnitude. Fyou on earth = 539 N

QQ: What is the weight of a 65 kg. person on the Earth? What is the weight of a 65 kg. person on the moon where the force of gravity is 1.6 N kg-1? 3. A boxer punches a sheet of tissue paper with mass of .0001 kg in midair. In only 0.05 seconds, it gains a speed of 30 m/s. Find the force the boxer places on the paper.

QQ: N What is the weight of a 65 kg. person on the Earth? m = 65 kg g = 9.8 m/s2 w = fg = m*g fg = 65 kg*9.81 m/s2 fg = 637 kg * m/s2 N

What is the weight of a 65 kg What is the weight of a 65 kg. person on the moon where the force of gravity is 1.6 N kg-1? M = 65 kg g = 1.6 N kg-1 W = fg = m*g W = fg = 65 kg * 1.6 N/kg fg = 104 kg N/kg = 104 N = 1.6 N/kg

3. A boxer punches a sheet of tissue paper with mass of .001 kg in midair. In only 0.05 seconds, it gains a speed of 30 m/s. Find the force the boxer places on the paper. m = .001 kg t = 0.05 s s = 30 m/s f = ma f = .001kg * 600 m/s2 f = .6 kg*m/s2 a = s/t a = 30 m/s .05 s = 600 m/s2 = 30 m 0 .05 s2

Let’s all jump!

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Newton’ s Three Laws of Motion

online problems https://www.sophia.org/tutorials/newtons-3rd-law-of-motion-actionreaction-law

Two 100-N weights are attached to a spring scale as shown Two 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?

The scale reads the tension in the string The 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.

Nothing 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.

Concept Question 1 Why are we able to walk?

Concept 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 person’s foot by the ground. Fpg Force exerted on the ground by person’s foot. Fgp Fgp=-Fpg

Concept Question 2 What makes a car go forward?

Concept Question answer By Newton’s third law, the ground pushes on the tires in the opposite direction, accelerating the car forward.

Concept Question Which is stronger, the Earth’s pull on an orbiting space shuttle or the space shuttle’s pull on the earth?

Concept 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

Problem 1 What force is needed to accelerate the 60 kg cart at 2m/s^2?

How to solve Problem 1 Force = mass times acceleration F = m * a What force is needed to accelerate the 60kg cart at 2 m/s^2? Force = mass times acceleration F = m * a F = 60kg * 2m/s^2 F = 120kgm/s^2 Kgm/s^2 = Newton Newton = N F =120 N

Problem 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?

How to solve Problem 2 F = ma therefore:m =F/a m = 200N/2m/s^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^2 N= kgm/s^2 so when divide your answer will be kg left. m = 100kg

Why 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.