1.What is the weight of a 15 kg rock? 147 N 2.A skateboard (mass 12 kg ) accelerates from rest to 3.8 m/s over a distance of 9.4m. What is the skateboard ’s rate of acceleration? m/s 2 3.What force was applied to the skateboard in to achieve this rate of acceleration? 9.22 N 4.An unbalanced force of 25 N is applied to a kg mass. What should be the acceleration experienced by the mass? 2 m/s 2

11/13 Forces Assignment Parts 2 & 3 due Friday NOW: Pick up Force Notes II. ON Force ntoes one, answer example J. Last Friday: A quiz over Fun with Forces & Fun with Weight. Monday: finished lab. Tuesday: reviewed lab, turned it in and took a quiz. If you were absent for any of these you need to make them up today after school or tomorrow morning. PM Test corrections & retakes Mon-Thur this week. You must show completed PM notes to be eligible for corrections & retakes. Retakes pm only. I have pm duty until 2:55 There is an issue with the website. I will fix it today if possible

11/14 Have your HW out (Force, Friction, & Statics Part 2 and 3A) You will turn this in on Thursday next week, but may use it on your Quiz today. Quiz today. It is over anything in Forces that we covered through Wednesday (ie no friction). Hints: remember: –mass must be in kg, distance in m –Weight is a Force (ma g ) The quiz is keyed to 9.8 m/s 2 –When looking at horizontal motion, use horizontal acceleration –If velocity is constant, forces are balanced NOW: On Force Notes I, answer example K. You will have a quiz Monday over Newton’s Laws. No calculations Assignment 3B due Tuesday, Assignment 3c due Wednesday Forces Test Thursday 11/20 Thursday: We began friction and got up to example N. We will continue here today. There is an issue with the website. I am still trying to resolve it.

11/17 HW by tomorrow have Force, Friction, & Statics Part 2, 3A&B completed. Part C by Wednesday. We had a Quiz Friday. It was over anything in Forces that we covered through Wednesday (ie no friction). If you were absent you may make it up Today after school or tomorrow am or pm. Friday we worked through example K and N. Have Notes II out You will have a quiz Wednesday over Newton’s Laws at the end of the period. No calculations Forces Test Thursday 11/20 Thursday: We began friction and got up to example N. We will continue here today. Problem with the website has been resolved.

11/18 Pick up notes III if you did not get them Monday HW by tomorrow have Force, Friction, & Statics Part 2, 3A&B completed. Part C by Wednesday. We had a Quiz Friday. It was over anything in Forces that we covered through Wednesday (ie no friction). If you were absent you may make it up Today after school Friday we worked through example K and N. Yesterday we worked through Ex P (some classes we still need to finish it) Have Notes II out and Notes III You will have a quiz Wednesday over Newton’s Laws at the end of the period. No calculations I will not be available after school Wednesday due to a faculty meeting Forces Test Thursday 11/20

Ex H The brakes of a 1000-kg car exert 3000 N. a. How long will it take the car to come to a stop from a velocity of 30 m/s? m = 1000 kg F = N v i = 30 m/s v f = 0 m/s = - 3 m/s 2 = 10 s b. How far will the car travel during this time? d = v i t+.5at 2 = 30(10)+.5 (-3)(10) 2 = 150 m

Ex I A net horizontal force of 4000 N is applied to a car at rest whose weight is 10,000 N. What will the car's speed be after 8 s? F A = 4000 N F w = 10,000 N t = 8s = kg = 3.92 m/s 2 v f = v i + at = (8) = m/s

Forces Part II Friction

When surfaces are pressed together, we can identify four forces

Friction Force: F F Force opposing motion Measured in Newtons (N)

Applied Force: F A The push or pull applied to the object Measured in Newtons (N)

F w Force of Weight or Gravity (Mass in kg) (Acceleration due to Gravity) Kg x 9.8 m/s 2 Measured in Newtons (N)

Normal Force: F N Usually a 3 rd Law reaction to gravity, that is equal and opposite of Force of Weight (F w ) Perpendicular to the surface. Measured in Newtons (N) F N is NOT F NET Don’t confuse them just because they begin with N!

A 50kg object is moving horizontally at a constant velocity. Is there acceleration? Is there a net force? FWFW FNFN FAFA F

Ex J No Friction F net = ma A 50 kg object experiences an applied force of 400 N, what is the F net ? (Disregard friction) What is the acceleration?: F net = F A F net = Net Force, results in acceleration a = F net /m = (400N )/ (50 kg) = 8 m/s 2

Example J FAFA FWFW FNFN

Friction Is a force that opposes motion.

Ex K Friction and Constant velocity. Is there a net force? F net = 0 F A + - F F = ma or F A + - F F = F net A 50 kg object moves at a constant velocity when acted upon by an applied force of 400 N. What is the F F ? What is the F net ? What is the acceleration?: F F = F A F net = F A + - F F = 0 N If F net = 0 then acceleration = ? 0

Example K Friction and constant velocity FAFA FWFW F FNFN

How does friction affect net force? F net = ma ma = F A + F F Where: F net = Net Force F F = Friction Force F A = Applied Force You are actually subtracting F F from F A, since F f is in the opposite direction Friction will reduce the net force

Do all surfaces provide the same amount of friction? How is this described? FAFA FWFW F FNFN

Friction What does it depend on? Depends on the surface of the materials. Depends on how tightly the surfaces are pressed together. F F = Force of Friction

Coefficient of Friction The coefficient of friction is a measure of how difficult it is to slide a material of one kind over another; the coefficient of friction applies to a pair of materials, and not simply to one object by itself Coefficient of Friction Reference Table - Engineer's HandbookCoefficient of Friction Reference Table - Engineer's Handbook

When Surfaces are Pressed Together Coefficent of Friction µ can be calculated It is a ratio of F F and F N

Coefficient of friction What is it equal to? What is the unit for Coefficient of friction? If Coefficient of friction is small, what does that mean about the F F ? If Coefficient of friction is large, what does that mean about the F F ? The higher the coefficient of friction, the more difficult to slide

What forces cause the swimmer to move forward? Her push on the wall? Or The wall’s push on her?

I call these FAWN problems

Example L A 400 N force is applied horizontally to a 50kg object. Calculate the acceleration of the object if  = 0.3. FAFA FgFg F FNFN

Example Cont’d. F A = 400 N F g = m·g (50kg)(9.8m/s 2 ) F g = 490 N F N = 490 N also F F = (0.3)(490N) = 147 N

Example cont’d. F A = 400N F g = 490N F F = 147N F N = 490N

To solve for Acceleration must calculate Net Force F NET = ma= 400N – 147N ma = 253N F NET = ma = F A + - F F

Now use Net Force and mass of object in F=ma formula a = F/m a = 253N/50kg a =5.06 m/s 2

Ex M Putting it all together A 50 kg object accelerates horizontally at 0.2 m/s 2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the F net ? What is the F F ? What is the F A ?How far does it travel in 5 seconds?: F net = (50 kg) (0.2 m/s 2 ) = 10 N F F = μ F N F N = (50kg)(9.8 m/s 2 ) = 490 N F F = (0.01) (490 N) =4.9N F A = F net – - F F F A = 10 N N = 14.9 N

How does friction affect net force? Putting it all together A 50 kg object accelerates horizontally at 0.2 m/s 2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the F net ? What is the F F ? What is the F A ?How far does it travel in 5 seconds?: F net = (50 kg) (0.2 m/s 2 ) = 10 N Set up DVVAT d = ? v i = 0 m/s v f = ? a =0.2 m/s 2 d = (.5)(.2)(5 )2 d = 2.5m

As the coefficient of friction decreases and the object and the applied force remain the same, What happens to: F A F W F N F NET

TENSION aka F T is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object. It is the opposite of compression. It is a “response force”compression That is to say, if one pulls on the rope, the rope fights back by resisting being stretched Ropes, strings, and cables can only pull. They cannot push because they bend. is measured in newtonsnewtons is always measured parallel to the string on which it applies.

What does the rope provide? A lift (vertical force) and a pull (horizontal force) If there was no angle, would there be any vertical force? No If the angle was at 90°, how would that affect the force components? Force would only be in the vertical plane How would you calculate the horizontal and vertical force components if the angle of the rope with the floor was 57° and the Force of tension (F A ) in the rope was 400 N?

Ex N. This crate is be pulled with a rope across a friction based horizontal surface at a constant velocity. The rope exerts a tension of 400 N at an angle of 57°. What is the coefficient of friction? 50 kg 57°

A box is pulled into motion with a rope across a horizontal surface. The rope makes an angle of 57° to the floor. The Force of tension (F A ) in the rope is 400N 50 kg 57º F AX F AY 400 N A. Determine F AX or F Horiz = cos (57) (400N) = N B. Determine F AY or F Vert = sin (57) (400N) = N

I work in a circle Determine F W Determine F AX Determine F AY Determine F N The F AY supplies part of the upward force. The total upward force is F AY + F N and together these are equal but opposite the F W. F N = F W – F AY Determine F F Determine coefficient of friction

50 kg 57 ° F AX N F AY N 400 N Determine F w = (50 kg)(9.8 m/s 2 ) = 490 N Determine F N = F W - F AY = 490 N – N = N

50 kg 57 ° F AX N F AY N 400 N Determine F F F F = F ax (Constant Velocity) F F = N Determine µ= F F / F FN = N/ N = 1.41

1.Cameron has to clean up his room. He places all of his dirty clothes on a large sheet of cardboard and drags them to the laundry room His arms make an angle of 37° with the floor and he pulls with a force of 52.5N. If he pulls at a constant velocity, what is the force of friction? 2.Mehtab has a messy room too! She piles all her stuffed animals into a box. The box has a mass of 128 kg. She pulls into motion by a rope across the carpet. (μ = 0.264) with a force of 512 N at 35.9° above the horizontal. How far does the crate travel in 8 seconds?

Practice A sled with a weight of 80 N is dragged with a rope across the ground at a constant velocity. The Force Tension on the rope is 100N at 40º with the horizontal. What are the components? What is the normal force? What is the force of friction? What is the coefficient of friction between the crate and the surface?

Practice A sled with a weight of 80 N is dragged with a rope across the ground at a constant velocity. The Force Tension on the rope is 100N at 40º with the horizontal. What are the components? What is the normal force? What is the force of friction? What is the coefficient of friction between the crate and the surface?

The End