By the end of the day today IWBAT

By the end of the day today, IWBAT… • explain Newton’s 3 rd Law of Motion Do Now: Wed/Thurs, Jan 6/7, 2021

By the end of the day today, IWBAT… • explain Newton’s 3 rd Law of Motion Do Now: 1. Paiten is grocery shopping and pulls the 29 kg shopping cart with a force of 22 N at 38 degrees above the horizontal. Assume no friction. A) Draw the FBD. B) Draw an determine the X and Y components. C) Using only variables determine the sum of the forces. D) Find the acceleration for the X-component Thursday, Jan 7, 2021

By the end of the day today, IWBAT… • explain Newton’s 3 rd Law of Motion Do Now: You push a 25. 0 kg box across a floor with a coefficient of friction of. 20 at a speed of 1. 5 m/s. A) Determine the required force. B) If the force is tripled, what would the resulting acceleration be? Friday, Jan 8, 2021

By the end of the day today, IWBAT… • explain Newton’s 3 rd Law of Motion Do Now: Draw the FBD for a 17 kg box being pushed with 105 N at 38 degrees below the horizontal, ignoring friction. A) Draw the FBD. B) Draw an determine the X and Y components. C) Using only variables determine the sum of the forces. D) Find the acceleration for the X-component. E) If the angle were increased, how would it influence the acceleration. Monday, Jan 11, 2021

By the end of the day today, IWBAT… • explain Newton’s 3 rd Law of Motion Do now: Three blocks, M 1 to M 2 to M 3, are connected on massless strings and pulled along a horizontal surface. M 1= 2 kg M 2=4 kg M 3=6 kg and net force is 36. 0 N on M 3. a. Find the acceleration of each block. b. Find the tension forces. Tuesday, Jan 12, 2021

By the end of the day today, IWBAT… • explain Newton’s 3 rd Law of motion Why it matters in LIFE: This is why normal force exists! Why it matters in THIS CLASS: This is a HUGE topic that will not only help us reach our goal of 85%, but will also be test on the midterm

What it might look like on a test: A student pulls a 60 -newton sled with a force having a magnitude of 20 newtons. What is the magnitude of the force that the sled exerts on the student ? (1) 20 N (3) 60 N (2) 40 N (4) 80 N

Agenda + Announcements Do Now 5 min Newton’s 2 nd Law Independent Problems 10 min Newton’s 3 rd Law 30 min

Independent Problems Complete Silence 100% Participation - Stay on-task! Show all of your work, box your answer Don’t forget units! Raise your hand if you have a question

What is Newton’s 3 rd Law of Motion? For every action, there is an EQUAL and OPPOSITE reaction Forces always exist in pairs

Basically… For every action, there is a reaction that is EQUAL in magnitude and OPPOSITE in direction

Let’s see an example… ACTION: Rocket Gases push DOWN on air. REACTION: Air pushes UP on rocket.

Another example of Newton’s 3 rd Law ACTION: Wings push DOWN on the air. REACTION: Air pushes UP on wings.

Another example of Newton’s 3 rd Law ACTION: Fins push BACK on water REACTION: Water pushes Forward on fins.

What is an action-reaction pair? Action – reaction pair = A pair of simultaneous EQUAL but OPPOSITE forces resulting from the interaction of two objects If the same force is applied to both the hammer and the nail, why does only the nail accelerate?

Now let’s connect this to something we already know… Use Newton’s 3 rd law to explain normal force Every object possessing mass has weight. Weight is a force directed downward. Therefore, it needs to have an equal and opposite force. In other words, if an object pushes down on the ground (weight) the ground must push back up on the object (normal force). How would you calculate normal force?

Identifying action-reaction pairs a. A person taking a step Action: person pushing down on ground Reaction: ground pushing up on person Do b. and c. with your shoulder partner Do d. and e. alone

Question #2 A 600 -N man sits on the ground. Use Newton’s 3 rd law to explain the forces at play. The man exerts a force of 600 N on the ground (weight). The ground exerts an equal and opposite force of 600 N on the man (normal force).

What it might look like on a test: A student pulls a 60 -newton sled with a force having a magnitude of 20 newtons. What is the magnitude of the force that the sled exerts on the student ? (1) 20 N (3) 60 N (2) 40 N (4) 80 N

Working 2 nd Law Problems 1. 2. Draw a force or free body diagram. Set up 2 nd Law equations in each dimension. Fx = max and/or 3. Fy = may Identify numerical data. x-problem and/or y-problem 4. Substitute numbers into equations. “plug-n-chug” 5. Solve the equations.

Forces in 2 -D

TENSION

4. 11 Equilibrium Application of Newton’s Laws of Motion Reasoning Strategy • Select an object(s) to which the equations of equilibrium are to be applied. • Draw a free-body diagram for each object chosen above. Include only forces acting on the object, not forces the object exerts on its environment. • Choose a set of x, y axes for each object and resolve all forces in the free-body diagram into components that point along these axes. • Apply the equations and solve for the unknown quantities.

T 1 sin T 2 cos T 1 cos M 3 g

Vertical Components Equation T 1 sin T 2 sin M 3 g T 2 cos sin T 2 sin cos M 3 g

Static Equilibrium

Horizontal forces balance T 1 cos T 2 cos

Vertical forces balan ce. T 1 sin T 2 sin M g

A mass, ( M 23 kg ), is suspended by two ropes from a ceiling. Rope 1 makes an angle of 60 deg from the ceiling, while Rope 2 makes an angle of 40 deg. What are the tensions T 1 and T 2?

Tension Horizontal Force Components Equation T 1 cos T 1 T 2 cos cos

T 1 T 2 cos T 1 sin T 2 sin M g T 2 cos sin T 2 sin M g cos

TL is the tension of the horizontal cable on the left, attached to the wall. TR is the tension of the angled cable on the right attached to the ceiling.

TL is the tension of the horizontal cable on the left, attached to the wall. TR is the tension of the angled cable on the right attached to the ceiling. Fy 0 TR sin(37 deg) 625 N Fx 0 TR cos (37 deg) TL

Setting up a T 1 T 2 Problem

Find the tensions. T 1 for the rope attached to the left, an. Td 2 for the rope attached to the right. Fx 0 T 1 cos (43. 0 deg ) T 2 cos (55 deg ) Fy 0 T 1 sin(43. 0 deg ) T 2 sin(55 deg ) 43. 8 kg 9. 8 N kg

T 1 T 2 cos (55 deg ) cos (43. 0 deg ) This horizontal equation substitutes into the vertical equation, enab ling us solve for T 2 cos (55 deg ) sin(43. 0 deg) T 2 sin(55 deg ) cos (43. 0 deg ) 0. 574 T 0. 682 0. 819 T 2 2 0. 731 0. 535 T 2 0. 819 T 2 1. 354 T 2 43. 8 kg 9. 8 N kg 429. 24 N 1. 354 T 1 T 2 cos (55 deg ) cos (43. 0 deg ) T 2 317 N T 1 (317 N) cos (55 deg ) cos (43. 0 deg ) T 1 249 N

4. 11 Equilibrium Application of Newton’s Laws of Motion T 1 sin 35� T 2 sin 35� 0 T cos 35 F 0 � 1 � 2

4. 11 Equilibrium Application of Newton’s Laws of Motion

4. 11 Equilibrium Application of Newton’s Laws of Motion Force x component y component T 1 sin 10. 0� T 1 cos 10. 0� T 2 sin 80. 0� T 2 cos 80. 0� W 0 W W 3150 N

4. 11 Equilibrium Application of Newton’s Laws of Motion F T sin 10. 0� T sin 80. 0� 0 x 1 2 F T cos 10. 0� T cos 80. 0� W 0 y 1 The first equation gives 2 sin 80. 0� T T 1 � 2 sin 10. 0 Substitution into the second gives sin 80. 0� � � T cos 10. 0 T cos 80. 0 W 0 2 � 2 sin 10. 0

4. 11 Equilibrium Application of Newton’s Laws of Motion T 2 W sin 80. 0� � � cos 10. 0 cos 80. 0 � sin 10. 0 T 2 582 N T 1 3. 30 103 N

Page 128 Problem 11

4. 11 Equilibrium Application of Newton’s Laws of Motion T 1 sin 35� T 2 sin 35� 0 T cos 35 F 0 � 1 � 2

4. 12 Nonequilibrium Application of Newton’s Laws of Motion When an object is accelerating, it is not in equilibrium. F max F ma y x y

4. 12 Nonequilibrium Application of Newton’s Laws of Motion The acceleration is along the x axis so ay 0

4. 12 Nonequilibrium Application of Newton’s Laws of Motion Force x component y component T 1 cos 30. 0 T 2 cos 30. 0 D D 0 R � T sin 30. 0 � 1 � T sin 30. 0� 2

4. 12 Nonequilibrium Application of Newton’s Laws of Motion F T sin 30. 0 0 � y 1 2 T 1 T 2 F T cos 30. 0 D R � x max 1 2

4. 12 Nonequilibrium Application of Newton’s Laws of Motion T 1 T 2 T max R D 5 T 1. 53 10 N � 2 cos 30. 0

4. 12 Nonequilibrium Application of Newton’s Laws of Motion When an object is accelerating, it is not in equilibrium. F max F ma y x y

4. 12 Nonequilibrium Application of Newton’s Laws of Motion The acceleration is along the x axis so ay 0

4. 12 Nonequilibrium Application of Newton’s Laws of Motion Force x component y component T 1 cos 30. 0 T 2 cos 30. 0 D D 0 R � T sin 30. 0 � 1 � T sin 30. 0� 2

4. 12 Nonequilibrium Application of Newton’s Laws of Motion F T sin 30. 0� T sin 30. 0 0 y 1 2 T 1 T 2 F T cos 30. 0 D R � x max 1 2

4. 12 Nonequilibrium Application of Newton’s Laws of Motion T 1 T 2 T max R D 5 T 1. 53 10 N � 2 cos 30. 0

Page 128 Problem 11 • A net force causes acceleration of the block. • Since the block is moving along the horizontal, it is accelerating along the horizontal.

Page 128 Problem 11 • Therefore, only the horizontal components of the force vectors are taken into consideration.

F a a m a F 45. 0 N cos (65°) 25. 0 N m 5. 00 kg 1. 20 m s 2 A negative acceleration is a deceleration.

4. 4 Vector Nature of 2 nd Law

Sample Problem Jack & Jill lift upward on a 1. 3 kg pail of water, with Jack exerting a force of 7 N and Jill exerting a force of 11 N. Jill’s force is exerted at an angle of 28 o with the vertical. At what angle should Jack exert his force if the pail is to accelerate straight upward?

Sample Problem 1) The minimum vertical upward force has to be slightly greater than the weight of the pail of water. 2) Weight of the pail of water = mass * g = 1. 3 kg * 9. 8 m/s^2 = 12. 74 N 3) Jill's upward vertical force = 11. 0 N cos(28°) = 9. 71 N 4) Jack's upward vertical force = 7. 0 N cos (x) 5) Net upward vertical force = weight of the pail of water => 9. 71 N + 7. 0 N cos(x) = 12. 74 N => cos(x) = [12. 74 N - 9. 71 N ] / (7. 0 N) = 0. 4329 6) x = arc cos(0. 4329) = 64. 4°

SWBAT: Explain what factors influence the force of friction Calculate the coefficient of friction between two surfaces

Todays class is brought to you by 1

Word of the day fric tion

Friction What is friction What is the difference between kinetic friction and static friction Determining frictional forces Examples

What is friction? Friction is a force that exists between two surfaces that are in contact and works to resist motion between those surfaces.

TWO types of Friction Static – Friction that keeps an object at rest and prevents it from moving Kinetic – Friction that acts during motion

Force of Friction The Force of Friction is directly related to the Force Normal. Mostly due to the fact that BOTH are surface forces Very slick surfaces have a low coefficient of friction (glass, ice) Rough or sticky surfaces have a high coefficient of friction (sandpaper, carpet) The coefficient of friction is a unitless constant that is specific to the material type and usually less than one. Note: Friction ONLY depends on the MATERIALS sliding against each other, NOT on surface area.

0< <1 Greek symbol “mu” coefficient of friction Unitless quantity

Kinetic Friction Equation Kinetic or sliding friction acts when objects are rubbing against each other while moving. REMEMBER: Friction is FUN! Ff = FN Greek symbol “mu” coefficient of friction

Orientation Effects on

FN Fs Fk data 0 1 2 0 0 2. 5 FN data 1 3. 5 2 4. 3 3 7. 1 4 12 Fs data 1 N Fk data 2 N 5. 5 N Normal Force against block. Static Force against block. Kinetic Force against block. Friction as a Function of Normal Force Static & Kinetic Friction Fs slopes N slopek Fk newton b est_fits best_fitk 0 0 FN 12 newton Normal Force (Weight of Wood Block)

5. 5 Friction as a Function of Normal Force Static & Kinetic Friction Fs slopes N slopek Fk newton best_fits best_fitk 0 0 FN 12 newton Normal Force (Weight of Wood Block)

6 Static & Kinetic Friction Fs N Fk newton Friction as a Function of Normal Force 5 slope s 0. 465 slopek 0. 331 4 3 best_fits best_fitk 2 1 0 0 5 10 15 FN newton Normal Force (Weight of Wood Block)

What information do the slop Fs Fk of and tell us from FN FN the straight-line graph?

4. 9 Static and Kinetic Frictional Forces

Coefficient of Friction #41 Fw 45. 0 N s 0. 650 Fa 36. 0 N k 0. 420 If calculated. Fa Fs , then block will move.

Coefficient of Friction #41 s Fs FN Fa Fw Holds true if block is static or moves at constant speed

Coefficient of Friction #41 If calculated Fa Fs , then block will move. Fs s FN 36. 0 N 29. 3 N s Fw 29. 3 N Block moves.

Coefficient of Friction #41 k Fk Fw F m a a F Fa Fk m m

Coefficient of Friction #41 a Fa Fw k Fw N 9. 8 kg a 3. 72 m s 2

Static & Kinetic Friction s Fs FN k Fk FN

Solving Friction Problems Step 1: FBD Step 2: Fg given: Calculate mass Fg=ma Fg not given: Fg=ma Step 3: Fn or N: opposes gravity If a is horizontal: Fn=Fg If a is vertical: then Fn DOES NOT equal Fg, so use the net force equation. Step 4: Ff=u. Fn or Net Force Eq Find Ff and Find mu Good for finding a, Fappl, or Ff, use GUESS method

What causes frictional forces?

A box has a mass of 3. 0 kg. A force of 10 N is applied causing the crate to move with a constant velocity. Find the coefficient of kinetic friction. FN Ff F = 10 N Fg

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Situations using kinetic friction Velocity is (changing / constant) FN Ff=10 N FA=10 N Fg

Situations using kinetic friction Velocity is (changing / constant) FN Ff=10 N Fg

Situations using kinetic friction Velocity is (changing / constant) FN Ff=10 N FA=13 N Fg

Static Friction Equation Static friction acts when an object is stationary Ff < FN

Situations using static friction Object is stationary FN FA Ff Fg

Situations using static friction Object is stationary FN FA Ff Fg

Interesting situation Stationary object pressed against a wall. fs N F W

Venn Diagram Static verses Kinetic

Inclined Planes Label the direction of N and mg. N θ mg Use slide show mode (press F 5) for animations.

Inclined Planes Mark the direction of acceleration a. N a θ mg

Inclined Planes Choose the coordinate system with x in the same or opposite direction of acceleration and y y perpendicular to x. N x a θ mg

Inclined Planes Now some trigonometry y N x a θ 90 - θ θ mg

Inclined Planes Replace the force of gravity with its components. y N x a mg θ n i s θ θ mg mg θ s o c

Inclined Planes Use Newton’s second law for both the x and y directions y N x a mg θ n i s θ θ mg mg θ s o c The force and acceleration in the x-direction have a negative sign because they point in the negative x-direction.

Inclined Planes Why is the component of mg along the x-axis –mgsinθ Why is the component of mg along the y-axis –mgcosθ y N x a mg θ n i s θ θ mg mg θ s o c

Inclined Planes Why is the component of mg along the x-axis: –mgsinθ Why is the component of mg along the y-axis: –mgcosθ y N a mg sinθ θ mg x θ mg cosθ

Inclined Planes Why is the component of mg along the x-axis: –mgsinθ y along the y-axis: –mgcosθ Why is the component of mg N a mg sinθ θ mg x θ mg cosθ

Inclined Planes Why is the component of mg along the x-axis: –mgsinθ Why is the component of mg y along the y-axis: –mgcosθ a mg sinθ x θ mg θ N mg cosθ sinθ = opposite hypotenuse cosθ = adjacent hypotenuse

Friction & N. F. L If the coefficient of kinetic friction between a 35 -kg crate and the floor is 0. 30, what horizontal force is required to move the crate to the right at a constant speed across the floor? Fn Fa Ff mg 102. 9 N

Inclines Ff FN q mg Tips • Rotate Axis • Break weight into components • Write equations of motion or equilibrium • Solve

Friction & Inclines A person pushes a 30 -kg shopping cart up a 10 degree incline with a force of 85 N. Calculate the coefficient of friction if the cart is pushed at a constant speed. Fa Fn Ff mg 0. 117

Example A 5 -kg block sits on a 30 degree incline. It is attached to string that is thread over a pulley mounted at the top of the incline. A 7. 5 -kg block hangs from the string. a) Calculate the tension in the string if the acceleration of the system is 1. 2 m/s/s b) Calculate the coefficient of kinetic friction. T FN m 2 gcos 30 30 T m 2 g m 1 m 2 gsin 30 m 1 g Ff 30

Example 64. 5 N 0. 80 N

Problem 59 (Honors)