100 90 80 70 vel 60 (m/s) 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 time (sec)

Ch7 – More Forces Equilibrium - net force = 0 In this chapter there can be 3 or more forces present in a problem. Equilibrium results in an object that is stationary (static) or moving at constant speed (dynamic). Ex1) What is the tension in the rope? 1 kg

Ch7 – More Forces Equilibrium - net force = 0 In this chapter there can be 3 or more forces present in a problem. Equilibrium results in an object that is stationary (static) or moving at constant speed (dynamic). Ex1) What is the tension in the rope? 1 kg FTFT FgFg F net = F g – F T 0 = 10N – F T F T = 10N

1 kg Ex2) What is the tension in each rope?

1 kg FTFT FTFT F net = F g – 2F T 0 = 10N – 2F T F t = 5N Ex2) What is the tension in each rope? F g

Ex3) What is the tension in each rope if they are pulled to an angle of 45°? 45° 1 Kg

Ex3) What is the tension in each rope if they are pulled to an angle of 45°? 45° F g F net = F g – 2F Ty 0 = 10N – 2(F T ∙ sinθ) 0 = 10N – 2(F T ∙ sin45°) F T = 7N 1 Kg 45° F Ty F Tx F Ty FTFT F Ty = F T ∙ sinθ F Tx = F T ∙ cosθ F T

Ex 4) A 168 N sign is held by 2 ropes that make 22˚ angles as shown. What is the tension in each? OPEN 22˚

Ex 4) A 168 N sign is held by 2 ropes that make 22˚ angles as shown. What is the tension in each? OPEN FgFg F Ty FTFT 22˚ F net = F g − 2F Ty 0 = 168N – 2(F T ∙ sinθ) 0 = 168N – 2(F T ∙ sin22˚) F T = 224N FTFT

Ex 5) A 168N sign is held by 2 ropes as shown. What is the tension in rope 2 if F T1 = 86N? Open F T1 = 86N 60˚ 20˚ F T2

Ex 5) A 168N sign is held by 2 ropes as shown. What is the tension in rope 2 if F T1 = 86N? F Ty2 F Ty1 F g Ch7 HW#1 1 – 5 (Sep paper) F net = F g − F T1y − F T2y 0 = 168N − F T1 ∙sinθ 1 − F T2 ∙sinθ 2 0 = 168N − 86N∙sin20˚ − F T2 ∙sin60˚ F T2 = 160N Open F T1 = 86N 60˚ 20˚ F T2

Lab7.1 – Vector Addition - due in tomorrow - Ch7 HW#1 due at beginning of period

Ch7 HW#1 1 – 5 1.What is the tension in each rope? F g 20 kg FTFT FTFT

Ch7 HW#1 1 – 5 1. What is the tension in each rope? F net = F g – 2F T 0 = 200N – 2F T F t = 100N F g 20 kg FTFT FTFT

2. What is the tension in each rope? 45° F g 20kg F Ty F T

2. What is the tension in each rope? F net = F g − 2F Ty 0 = 200N – 2(F T ∙ sinθ) 0 = 200N – 2(F T ∙ sin45˚) 45° F T = 141N F g 20kg F Ty F T

3. What is the tension in each rope? F g 50kg F Ty F T

3. What is the tension in each rope? F net = F g − 2F Ty 0 = 500N – 2(F T ∙ sinθ) 0 = 500N – 2(F T ∙ sin15˚) F T = 965N F g 50kg F Ty F T

4. What is the tension in each rope? 45° 60° F g 50kg F Ty1 F Ty2 F T1 =259N F T2

4. What is the tension in each rope? F net = F g − F Ty1 – F Ty2 0 = 500N – (F T1 ∙ sinθ) – (F T2 ∙ sinθ) 0 = 500N – (259N ∙ sin45°) – (F T2 ∙ sin60°) 45° 60° F T2 = 366N F g 50kg F Ty1 F Ty2 F T1 =259N F T2

4. What is the tension in each rope? 40° 50° F g 75kg F Ty1 F Ty2 F T1 =400N F T2

4. What is the tension in each rope? F net = F g − F Ty1 – F Ty2 0 = 750N – (F T1 ∙ sinθ) – (F T2 ∙ sinθ) 0 = 500N – (400N ∙ sin40°) – (F T2 ∙ sin50°) 40° 50° F T2 = 643N F g 75kg F Ty1 F Ty2 F T1 =400N F T2

Ch7.2 – Inclined Planes

F N F N = F g F g θ

Ch7.2 – Inclined Planes F N F N = F g F g F N F g|| = F g. sinθ θ F g|| F g ┴ = F g. cosθ F g┴ F g F g ┴ = F N F g|| θ

Ex1) A 100N block is placed on a 40˚ inclined plane. Find F g ┴ and F g||. 40°

Ex1) A 100N block is placed on a 40˚ inclined plane. Find F g ┴ and F g||. F N F g|| F g┴ F g F g ┴ = F g. cosθ F g|| = F g. sinθ = 100N. cos40° = 100N. sin40° = 77N = 64N 40°

Ex2) A 50.0kg mass is placed on a 25° inclined plane. What is the magnitude of the force tending to make it slide down the incline? What is the magnitude of the normal force acting on it? 25°

Ex2) A 50.0kg mass is placed on a 25° inclined plane. What is the magnitude of the force tending to make it slide down the incline? What is the magnitude of the normal force acting on it? a. F g|| = F g. sinθ = 500N. sin25° F N = 211N b. F N = F g ┴ = F g. cosθ F g|| = 500N. cos25° = 453N F g┴ F g 25°

Ex3) A 20.0kg mass is placed on a frictionless inclined planed angled at 42°. What is its acceleration down the incline? 42°

Ex3) A 20.0kg mass is placed on a frictionless inclined planed angled at 42°. What is its acceleration down the incline? F N F g|| F g┴ F g F net = F g|| m. a = F g. sinθ m. a = m. g. sinθ (mass cancels out of the equation, didn’t Galileo already tell us mass doesn’t a = g. sinθ affect accl?) a = (9.8 m / s 2 ). sin42° a = 6.6 m / s 2 42°

Ex4) A 35.0kg mass is placed on an inclined plane. The angle is increased until the mass is just about to slip. The maximum angle reached before it slips is 45°. What is the coefficient of friction between the mass and plane? 45°

Ex4) A 35.0kg mass is placed on an inclined plane. The angle is increased until the mass is just about to slip. The maximum angle reached before it slips is 45°. What is the coefficient of friction between the mass and plane? F N F f,s F g|| F g┴ F g F net = F g|| – F f,s m. a = F g. sinθ – μ s. F N 0 = F g. sinθ – μ s. F g┴ 0 = F g. sinθ – μ s. F g. cosθ 0 = sinθ – μ s. cosθ Ch7 HW#2 6 – 9 45°

Ch7 HW#2 6 – 9 6.A 91N block is placed on a 35˚ inclined plane. Find F g ┴ and F g||. F g F g ┴ = F g. cosθ F g|| = F g. sinθ 35°

Ch7 HW#2 6 – 9 6.A 91N block is placed on a 35˚ inclined plane. Find F g ┴ and F g||. F g F g ┴ = F g. cosθ F g|| = F g. sinθ = 91N. cos35° = 91N. sin35° = 52N = 75N 35°

7) A 25kg mass is placed on a 50° inclined plane. What is the magnitude of the force tending to make it slide down the incline? What is the magnitude of the normal force acting on it? a. F g|| = F g. sinθ b. F N = F g┴ 50°

7) A 25kg mass is placed on a 50° inclined plane. What is the magnitude of the force tending to make it slide down the incline? What is the magnitude of the normal force acting on it? a. F g|| = F g. sinθ = 250N. sin50° F N = 192N b. F N = F g┴ = F g. cosθ F g|| = 250N. cos50° F g┴ = 161N F g 50°

8. A 12kg mass is placed on a frictionless inclined planed angled at 32°. What is its acceleration down the incline? F g 32°

8. A 12kg mass is placed on a frictionless inclined planed angled at 32°. What is its acceleration down the incline? F N F g|| F g┴ F g F net = F g|| m. a = F g. sinθ m. a = m. g. sinθ a = g. sinθ a = (9.8 m / s 2 ). sin32° a = 5.3 m / s 2 32°

9. A 50kg mass is placed on an inclined plane. The angle is increased until the mass is just about to slip. The maximum angle reached before it slips is 35°. What is the coefficient of friction between the mass and plane? F g 35°

9. A 50kg mass is placed on an inclined plane. The angle is increased until the mass is just about to slip. The maximum angle reached before it slips is 35°. What is the coefficient of friction between the mass and plane? F N F f,s F g|| F g┴ F g F net = F g|| – F f,s m. a = F g. sinθ – μ s. F N 0 = F g. sinθ – μ s. F g┴ 0 = F g. sinθ – μ s. F g. cosθ 0 = sinθ – μ s. cosθ 35°

Ch7.2 cont – More Incline Planes Ex5) A trunk weighing 562N is on a 30° incline plane. a. If there’s no friction, what is its accl down the incline? F g b. How fast will it be going after 4 sec? 30°

Ch7.2 cont – More Incline Planes Ex5) A trunk weighing 562N is on a 30° incline plane. a. If there’s no friction, what is its accl down the incline? F N F net = F g|| m. a = F g. sinθ m. a = m. g. sinθ a = g. sinθ F g|| a = (9.8 m / s 2 ). sin30° F g┴ a = 5 m / s 2 F g b. How fast will it be going after 4 sec? v i = 0m/sv f = ? t = 4sa = _____ 30°

Ch7.2 cont – More Incline Planes Ex5) A trunk weighing 562N is on a 30° incline plane. a. If there’s no friction, what is its accl down the incline? F N F net = F g|| m. a = F g. sinθ m. a = m. g. sinθ a = g. sinθ F g|| a = (9.8 m / s 2 ). sin30° F g┴ a = 5 m / s 2 F g b. How fast will it be going after 4 sec? v i = 0m/sv f = ? t = 4sa = 5 m / s 2 v f = v i + a. t= 0 + (5 m / s 2 )(4s)= 20 m / s 30°

c. If the trunk were placed on an incline where the coefficient of kinetic friction was 0.15, what would be the acceleration of the trunk? F N F f,k F g|| F g┴ F g 30°

c. If the trunk were placed on an incline where the coefficient of kinetic friction was 0.15, what would be the acceleration of the trunk? F N F net = F g|| – F f,k F f,k m. a = F g. sinθ – µ k ∙F N m. a = m. g. sinθ – µ k ∙F g┴ m. a = m. g. sinθ – µ k ∙ m. g. cosθ F g|| a = (9.8 m / s 2 ). sin30° F g┴ – (.15)∙(9.8 m / s 2 ). cos30° F g a = 3 m / s 2 30°

Ex6) A 50kg wheel barrow is pushed up a 15° incline at constant speed. If the total friction against the wheel barrow is 100N, what force is required to push it? 15° Ch7 HW #3 10 – 12

Ex6) A 50kg wheel barrow is pushed up a 15° incline at constant speed. If the total friction against the wheel barrow is 100N, what force is required to push it? F F N F g|| F f,k F g┴ F g F net = F – F g|| – F f 0 = F – F g. sinθ – 100N 0 = F – 500N. sin15° – 100N F = ____N 15° Ch7 HW #3 10 – 12

Lab7.2 – Inclined Planes - due tomorrow - Ch7 HW#3 10 – 13 due at beginning of period

Ch7 HW#3 10 – 12 10.A 6kg block is placed on a 10˚ inclined plane. Find F g ┴ and F g||. F g F g ┴ = F g. cosθ F g|| = F g. sinθ 10°

11. A 12kg mass is placed on a frictionless inclined planed angled at 32°. What is its acceleration down the incline? 32°

11. A 12kg mass is placed on a frictionless inclined planed angled at 32°. What is its acceleration down the incline? F N F g|| F g┴ F g F net = F g|| m. a = F g. sinθ a = 32°

12. A trunk weighing 823N is on a 40° incline plane. a. If there’s no friction, what is its accl down the incline? 40°

12. A trunk weighing 823N is on a 40° incline plane. a. If there’s no friction, what is its accl down the incline? F N F net = F g|| m. a = F g. sinθ m. a = m. g. sinθ F g|| F g┴ a = F g b. How fast will it be going after 5 sec? v i = 0m/sv f = ? t = 5sa = 6.4 m / s 2 v f = v i + a. t 40°

c. If the trunk were placed on an incline where the coefficient of kinetic friction was 0.22, what would be the acceleration of the trunk? F N F f,k F g|| F g┴ F g F net = F g|| – F f,k m. a = F g. sinθ – µ k ∙F N m. a = m. g. sinθ – µ k ∙F g┴ m. a = m. g. sinθ – µ k ∙ m. g. cosθ a = 40°

13. A 40kg wheel barrow is pushed up a 10° incline at constant speed. If the total friction against the wheel barrow is 50N, what force is required to push it? 10°

13. A 40kg wheel barrow is pushed up a 10° incline at constant speed. If the total friction against the wheel barrow is 50N, what force is required to push it? F F N F g|| F f,k F g┴ F g F net = F – F g|| – F f 0 = F – F g. sinθ – 50N 0 = F – 400N. sin10° – 50N F = 10°

Ch7.3 – Projectile Motion vivi An object shot at an angle has two motions that are independent of each other.

t = 1 s t = 2 s t = 3 s t = 4 s An object shot at an angle has two motions that are independent of each other. Vertical motion controlled by gravity

vivi t = 1s t = 2s t = 3s t = 4s v 1 = v i v 2 = v i v 3 = v i v 4 = v i An object shot at an angle has two motions that are independent of each other. Vertical motion controlled by gravity Horizontal motion controlled by the initial velocity given to the projectile.

vivi t = 1s t = 2s t = 3s t = 4s v 1 = v i v 2 = v i v 3 = v i v 4 = v i t = 1 s t = 2 s t = 3 s t = 4 s An object shot at an angle has two motions that are independent of each other. Vertical motion controlled by gravity Horizontal motion controlled by the initial velocity given to the projectile.

vivi t = 1s t = 2s t = 3s t = 4s v 1 = v i v 2 = v i v 3 = v i v 4 = v i t = 1 s t = 2 s t = 3 s t = 4 s An object shot at an angle has two motions that are independent of each other. Vertical motion controlled by gravity Horizontal motion controlled by the initial velocity given to the projectile. Only thing that links the 2 motions is time

v i = 15 m/s d y = 44 m Ex1) A stone is thrown horizontally at 15m/s from the top of a cliff 44m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. How fast is it going as it hits the ground?

v i = 15 m/s d y = 50 m Ex1) A stone is thrown horizontally at 15m/s from the top of a cliff 44m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. How fast is it going as it hits the ground? a. The stone hits the ground at the same time, whether it is thrown horizontally or dropped. So solve for the dropped stone!

v i = 15 m/s d y = 50 m Ex1) A stone is thrown horizontally at 15m/s from the top of a cliff 44m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. How fast is it going as it hits the ground? a. The stone hits the ground at the same time, whether it is thrown horizontally or dropped. So solve for the dropped stone! d y = v i t + ½at 2 44 = 0 + ½(9.8)t 2 t = 2.9 sec

v i = 15 m/s d y = 50 m Ex1) A stone is thrown horizontally at 15m/s from the top of a cliff 44m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. How fast is it going as it hits the ground? a. The stone hits the ground at the same time, whether it is thrown horizontally or dropped. So solve for the dropped stone! b. If there were no gravity, the ball would travel the same distance away from the cliff, as the actual projectile. d y = v i t + ½at 2 44 = 0 + ½(9.8)t 2 t = 2.9 sec

v i = 15 m/s d y = 50 m Ex1) A stone is thrown horizontally at 15m/s from the top of a cliff 44m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. How fast is it going as it hits the ground? a. The stone hits the ground at the same time, whether it is thrown horizontally or dropped. So solve for the dropped stone! b. If there were no gravity, the ball would travel the same distance away from the cliff, as the actual projectile. d y = v i t + ½at 2 44 = 0 + ½(9.8)t 2 t = 2.9 sec d x = v i t + ½at 2 = v con t + 0 = 15 m/s(2.9 s) = 43.5m

v i = 15 m/s d y = 50 m Ex1) A stone is thrown horizontally at 15m/s from the top of a cliff 44m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. How fast is it going as it hits the ground? a. The stone hits the ground at the same time, whether it is thrown horizontally or dropped. So solve for the dropped stone! b. If there were no gravity, the ball would travel the same distance away from the cliff, as the actual projectile. d y = v i t + ½at 2 44 = 0 + ½(9.8)t 2 t = 2.9 sec d x = v i t + ½at 2 = v con t + 0 = 15 m/s(2.9 s) = 43.5m c. Find both components of final velocity, and pythag!

v i = 15 m/s d y = 50 m Ex1) A stone is thrown horizontally at 15m/s from the top of a cliff 44m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. How fast is it going as it hits the ground? a. The stone hits the ground at the same time, whether it is thrown horizontally or dropped. So solve for the dropped stone! b. If there were no gravity, the ball would travel the same distance away from the cliff, as the actual projectile. d y = v i t + ½at 2 44 = 0 + ½(9.8)t 2 t = 2.9 sec d x = v i t + ½at 2 = v con t + 0 = 15 m/s(2.9 s) = 43.5m c. Find both components of final velocity, and pythag! v fx v fy vfvf v fx = 15 m/s v fy = v iy + at = 0 + (9.8)(2.9) = 28.4 m/s Ch7 HW#4 14 – 16

v i = 5 m/s d y = 78.4 m Ch7 HW#4 14 – 16 14. A stone is thrown horizontally at 5m/s from the top of a cliff 78.4m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. Find final components of velocity? v fx v fy vfvf

v i = 5 m/s d y = 78.4 m Ch7 HW#4 14 – 16 14. A stone is thrown horizontally at 5m/s from the top of a cliff 78.4m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. Find final components of velocity? d y = v i t + ½at 2 78.4 = 0 + ½(9.8)t 2 t = 3.96 sec v fx v fy vfvf

v i = 5 m/s d y = 78.4 m Ch7 HW#4 14 – 16 14. A stone is thrown horizontally at 5m/s from the top of a cliff 78.4m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. Find final components of velocity? d y = v i t + ½at 2 78.4 = 0 + ½(9.8)t 2 t = 3.96 sec d x = v i t + ½at 2 = v con t + 0 = 5 m/s(3.96 s) = 20m v fx v fy vfvf

v i = 5 m/s d y = 78.4 m Ch7 HW#4 14 – 16 14. A stone is thrown horizontally at 5m/s from the top of a cliff 78.4m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. Find final components of velocity? d y = v i t + ½at 2 78.4 = 0 + ½(9.8)t 2 t = 3.96 sec d x = v i t + ½at 2 = v con t + 0 = 5 m/s(3.96 s) = 20m v fx v fy vfvf v fx = 5 m/s v fy = v iy + at = 0 + (9.8)(3.96) = 39 m/s

v i = 10 m/s d y = 156.8 m Ch7 HW#4 14 – 16 15. A stone is thrown horizontally at 10m/s from the top of a cliff 156.8m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. Find final components of velocity? v fx v fy vfvf

v i = 10 m/s d y = 156.8 m Ch7 HW#4 14 – 16 15. A stone is thrown horizontally at 10m/s from the top of a cliff 156.8m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. Find final components of velocity? d y = v i t + ½at 2 156.8 = 0 + ½(9.8)t 2 t = 5.6 sec v fx v fy vfvf

v i = 10 m/s d y = 156.8 m Ch7 HW#4 14 – 16 15. A stone is thrown horizontally at 10m/s from the top of a cliff 156.8m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. Find final components of velocity? d y = v i t + ½at 2 156.8 = 0 + ½(9.8)t 2 t = 5.6 sec d x = v i t + ½at 2 = v con t + 0 = 10 m/s(5.6s) = 56m v fx v fy vfvf

v i = 10 m/s d y = 156.8 m Ch7 HW#4 14 – 16 15. A stone is thrown horizontally at 10m/s from the top of a cliff 156.8m high. a. How long is it in the air? b. How far from the base of the cliff will it hit? c. Find final components of velocity? d y = v i t + ½at 2 156.8 = 0 + ½(9.8)t 2 t = 5.6 sec d x = v i t + ½at 2 = v con t + 0 = 10 m/s(5.6s) = 56m v fx v fy vfvf v fx = 10 m/s v fy = v iy + at = 0 + (9.8)(5.6) = 55.5 m/s

v i = ? d y = 0.950m Ch7 HW#4 14 – 16 16.A steel marble rolls with a constant velocity across a tabletop 0.950m tall. It rolls off the table and hits the ground 0.352m from the base below the edge. How fast was the ball rolling across the tabletop? d x = 0.352m

v i = ? d y = 0.950m Ch7 HW#4 14 – 16 16. A steel marble rolls with a constant velocity across a tabletop 0.950m tall. It rolls off the table and hits the ground 0.352m from the base below the edge. How fast was the ball rolling across the tabletop? d y = v i t + ½at 2 0.950 = 0 + ½(9.8)t 2 t = 0.44 sec d x = 0.352m

v i = ? d y = 0.950m Ch7 HW#4 14 – 16 16. A steel marble rolls with a constant velocity across a tabletop 0.950m tall. It rolls off the table and hits the ground 0.352m from the base below the edge. How fast was the ball rolling across the tabletop? d y = v i t + ½at 2 0.950 = 0 + ½(9.8)t 2 t = 0.44 sec d x = v i t + ½at 2 d x = v con t + 0 0.352m = v. (.44s) v = 0.8m/s d x = 0.352m

Ch7 Mid Chapter Review 1. A 5.4kg block is in equilibrium on a 15° incline plane. a. Draw and label forces. b. Find the normal force. c. Find the friction force. b. F N = F g ┴ c. F net = F g|| – F f,s F g 15°

Ch7 Mid Chapter Review 1. A 5.4kg block is in equilibrium on a 15° incline plane. a. Draw and label forces. b. Find the normal force. c. Find the friction force. F N b. F N = F g ┴ c. F net = F g|| – F f,s = F g. cosθ F f,k = 54N. cos15° = 52N F g|| F g┴ F g 15°

Ch7 Mid Chapter Review 1. A 5.4kg block is in equilibrium on a 15° incline plane. a. Draw and label forces. b. Find the normal force. c. Find the friction force. F N b. F N = F g ┴ c. F net = F g|| – F f,s = F g. cosθ F f,k 0 = F g. sinθ – F f,s = 54N. cos15° F f,s = m. g. sinθ = 52N F f,s = 54N. sin15° F g|| F g┴ F g 15°

2. A 2.00kg block is placed on a 60° incline plane with a coefficient of kinetic friction of 0.45. Find the acceleration. 60°

2. A 2.00kg block is placed on a 60° incline plane with a coefficient of kinetic friction of 0.45. Find the acceleration. F N F f,k F g|| F g┴ F g F net = F g|| – F f,s m. a = F g. sinθ – µ k ∙F N m. a = m. g. sinθ – µ k ∙F g┴ m. a = m. g. sinθ – µ k ∙ m. g. cosθ a = g. sinθ – µ k ∙g. cosθ a = (9.8 m / s 2 ). sin60° – (.45)∙(9.8 m / s 2 ). cos60° a = 6.3 m / s 2 60°

(Save for HW) 3. A block with a mass of 5.0kg is placed on a 30° incline plane, where µ k = 0.14. a. Find acceleration. b. Find velocity after 4s. F g 30°

3. A block with a mass of 5.0kg is placed on a 30° incline plane, where µ k = 0.14. a. Find acceleration. F N b. Find velocity after 4s. a.F net = F g|| – F f,k m. a = F g. sinθ – µ k ∙F N m. a = m. g. sinθ – µ k ∙F g┴ F g|| m. a = m. g. sinθ – µ k ∙ m. g. cosθ F g┴ a = g. sinθ – µ k ∙g. cosθ F g a = (9.8 m / s 2 ). sin30° – (.14)∙(9.8 m / s 2 ). cos30° a = b. How fast will it be going after 5 sec? v i = 0m/sv f = ? t = 4sa = 3.7 m / s 2 v f = v i + a. t 30°

4. Find the tension force exerted by each cable to support the 625N bag. FgFg FTFT 35 ˚ FTFT

4. Find the tension force exerted by each cable to support the 625N bag. FgFg F Ty FTFT 35 ˚ F net = F g − 2F Ty 0 = 625N – 2(F T ∙ sinθ) 0 = 625N – 2(F T ∙ sin35˚) F T = 545N FTFT

(Save for HW) 5. What is the tension in each rope to support the 100kg sign. 60° F g 100kg F T

5. What is the tension in each rope to support the 100kg sign. 60° 60° F net = F g – 2F Ty 0 = 1000N – 2(F T ∙ sinθ) 0 = 1000N – 2(F T ∙ sin60°) F T = F g 100kg F Ty F T

v i = 18 m/s d y = 52 m 6. A person standing on the edge of a cliff throws a stone horizontally at 18m/s from the top of a cliff 52m high. a. How long does it take the stone to hit the water? b. How far from the base of the cliff will it hit? c. What speed does it the water? d y = v i t + ½at 2 d x = v i t + ½at 2

v i = 18 m/s d y = 52 m 6. A person standing on the edge of a cliff throws a stone horizontally at 18m/s from the top of a cliff 52m high. a. How long does it take the stone to hit the water? b. How far from the base of the cliff will it hit? c. What speed does it the water? d y = v i t + ½at 2 52 = 0 + ½(9.8)t 2 t = 3.26 sec d x = v i t + ½at 2 = v con t + 0 = 18m/s(3.26s) = 58.6m v fx v fy vfvf v fx = 18 m/s v fy = v iy + at = 0 + (9.8)(3.26) = 31.9m/s

v i = ? d y = 1.00m (Save for HW) 7. A ball is projected horizontally from the edge of a table 1.00m tall. It strikes the floor at a point 1.20m from the base of the table. what is the initial speed of the ball? d y = v i t + ½at 2 d x = v con t d x = 1.20m

Lab7.3 – Projectile Motion - due tomorrow - Ch7 Mid Chapter Review due at beginning of period

Ch7.3 cont – Projectiles Launched at an Angle Ex1) A ball is launched with an initial velocity of 4.47m/s at an angle of 66° above the horizontal. a. What is the max height reached? b. How long is it in the air? c. What was the range? 66° 4.47m/s

Ch7.3 cont – Projectiles Launched at an Angle Ex1) A ball is launched with an initial velocity of 4.47m/s at an angle of 66° above the horizontal. a. What is the max height reached? b. How long is it in the air? c. What was the range? 66° Step1: Find v i components 4.47m/s

Ch7.3 cont – Projectiles Launched at an Angle Ex1) A ball is launched with an initial velocity of 4.47 m / s at an angle of 66° above the horizontal. a. What is the max height reached? b. How long is it in the air? c. What was the range? 66° Step1: Find v i components v ix = v i. cosθ v iy = v i. sinθ =(4.47 m / s )cos66° =(4.47 m / s )sin66° = 1.8 m / s = 4.1 m / s Step2: Shoot the object straight up at v iy. v iy = 4.1 m / s 4.47m/s

Ch7.3 cont – Projectiles Launched at an Angle Ex1) A ball is launched with an initial velocity of 4.47 m / s at an angle of 66° above the horizontal. a. What is the max height reached? b. How long is it in the air? c. What was the range? 66° Step1: Find v i components Step3: Find time to the top, v ix = v i. cosθ v iy = v i. sinθ then double it. =(4.47 m / s )cos66° =(4.47 m / s )sin66° = 1.8 m / s = 4.1 m / s Step2: Shoot the object straight up at v iy. v fy = 0 m / s d y = ? v fy 2 = v iy 2 + 2ad y 0 2 = 4.1 2 + 2(-9.8)d y a = –9.8 m / s 2 d y = 0.86m v iy = 4.1 m / s 4.47m/s

Ch7.3 cont – Projectiles Launched at an Angle Ex1) A ball is launched with an initial velocity of 4.47 m / s at an angle of 66° above the horizontal. a. What is the max height reached? b. How long is it in the air? c. What was the range? 66° Step1: Find v i components Step3: Find time to the top, v ix = v i. cosθ v iy = v i. sinθ then double it. =(4.47 m / s )cos66° =(4.47 m / s )sin66° v fy = 0 m / s t top = ? = 1.8 m / s = 4.1 m / s Step2: Shoot the object straight up at v iy. a = –9.8 m / s 2 v fy = 0 m / s d y = ? v f = v i + a. t v fy 2 = v iy 2 + 2ad y 0 = 4.1 + (-9.8)t 0 2 = 4.1 2 + 2(-9.8)d y v iy = 4.1 m / s t top = 0.42s a = –9.8 m / s 2 t total = 0.84s d y = 0.86mStep 4: Range uses v x and t total. v iy = 4.1 m / s 4.47m/s

Ch7.3 cont – Projectiles Launched at an Angle Ex1) A ball is launched with an initial velocity of 4.47 m / s at an angle of 66° above the horizontal. a. What is the max height reached? b. How long is it in the air? c. What was the range? 66° Step1: Find v i components Step3: Find time to the top, v ix = v i. cosθ v iy = v i. sinθ then double it. =(4.47 m / s )cos66° =(4.47 m / s )sin66° v fy = 0 m / s t top = ? = 1.8 m / s = 4.1 m / s Step2: Shoot the object straight up at v iy. a = –9.8 m / s 2 v fy = 0 m / s d y = ? v f = v i + a. t v fy 2 = v iy 2 + 2ad y 0 = 4.1 + (-9.8)t 0 2 = 4.1 2 + 2(-9.8)d y v iy = 4.1 m / s t top = 0.42s a = –9.8 m / s 2 t total = 0.84s d y = 0.86mStep 4: Range uses v x and t total. d x = v x. t total v iy = 4.1 m / s = (1.8 m / s )(0.84s) = 1.5m 4.47m/s

HW#17) A player kicks a football from ground level with an initial velocity of 27 m / s, at 30° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 30° Step 1: components: v ix = v i. cosθ v iy = v i. sinθ

HW#17) A player kicks a football from ground level with an initial velocity of 27 m / s, at 30° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 30° Step 1: components: v ix = v i. cosθ =(27 m / s )cos30° = 23 m / s v iy = v i. sinθ =(27 m / s )sin30° = 13.5 m / s Step 2: Hang Time: v fy = 0 m / s t top = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 4.1 m / s

HW#17) A player kicks a football from ground level with an initial velocity of 27 m / s, at 30° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 30° Step 1: components: v ix = v i. cosθ =(27 m / s )cos30° = 23 m / s v iy = v i. sinθ =(27 m / s )sin30° = 13.5 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? v f = v i + a. t a = –9.8 m / s 2 0 = 13.5 + (-9.8)t t top = 1.37s v iy = 13.5 m / s t total = 2.75s Step3 (part B): Range d x = v x. t total

HW#17) A player kicks a football from ground level with an initial velocity of 27 m / s, at 30° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 30° Step 1: components: v ix = v i. cosθ =(27 m / s )cos30° = 23 m / s v iy = v i. sinθ =(27 m / s )sin30° = 13.5 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 13.5 + (-9.8)t t top = 1.37s v iy = 13.5 m / s t total = 2.75s Step3 (part B): Range d x = v x. t total = (23 m / s )(2.75s) = 63.8m Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y

HW#17) A player kicks a football from ground level with an initial velocity of 27 m / s, at 30° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 30° Step 1: components: v ix = v i. cosθ =(27 m / s )cos30° = 23 m / s v iy = v i. sinθ =(27 m / s )sin30° = 13.5 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 13.5 + (-9.8)t t top = 1.37s v iy = 13.5 m / s t total = 2.75s Step3 (part B): Range d x = v x. t total = (23 m / s )(2.75s) = 63.8m Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y 0 2 = 13.5 2 + 2(-9.8)d y d y = 9.3m

HW#18) A player kicks a football from ground level with an initial velocity of 27 m / s, at 60° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 60° Step 1: components: Step 2 (part A): Hang Time: v fy = 0 m / s a = –9.8 m / s 2 v iy = 23 m / s Step3 (part B): Range Step4 (C): Max Height

HW#18) A player kicks a football from ground level with an initial velocity of 27 m / s, at 60° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 60° Step 1: components: v ix = v i. cosθ v iy = v i. sinθ Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 23 m / s Step3 (part B): Range d x = v x. t total Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y

HW#18) A player kicks a football from ground level with an initial velocity of 27 m / s, at 60° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 60° Step 1: components: v ix = v i. cosθ =(27 m / s )cos60° = 13.5 m / s v iy = v i. sinθ =(27 m / s )sin60° = 23 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 23 m / s Step3 (part B): Range d x = v x. t total Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y

HW#18) A player kicks a football from ground level with an initial velocity of 27 m / s, at 60° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 60° Step 1: components: v ix = v i. cosθ =(27 m / s )cos60° = 13.5 m / s v iy = v i. sinθ =(27 m / s )sin60° = 23 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 23 + (-9.8)t t top = 2.4s v iy = 23 m / s t total = 4.8s Step3 (part B): Range d x = v x. t total Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y

HW#18) A player kicks a football from ground level with an initial velocity of 27 m / s, at 60° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 60° Step 1: components: v ix = v i. cosθ =(27 m / s )cos60° = 13.5 m / s v iy = v i. sinθ =(27 m / s )sin60° = 23 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 23 + (-9.8)t t top = 2.4s v iy = 23 m / s t total = 4.8s Step3 (part B): Range d x = v x. t total = (13.5 m / s )(4.8s) = 64.8m Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y

HW#18) A player kicks a football from ground level with an initial velocity of 27 m / s, at 60° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 60° Step 1: components: v ix = v i. cosθ =(27 m / s )cos60° = 13.5 m / s v iy = v i. sinθ =(27 m / s )sin60° = 23 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 23 + (-9.8)t t top = 2.4s v iy = 23 m / s t total = 4.8s Step3 (part B): Range d x = v x. t total = (13.5 m / s )(4.8s) = 64.8m Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y 0 2 = 23 2 + 2(-9.8)d y d y = 27.9m Ch7 HW#5 17 – 20

Lab7.4B – Range of a Projectile - due tomorrow - go over Ch7 HW#5 19,20 @ beginning of period

19. A player kicks a football from ground level with an initial velocity of 20 m / s, at 45° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 45° Step 1: components: v ix = v i. cosθ v iy = v i. sinθ Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 14 m / s Step3 (part B): Range d x = v x. t total Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y Ch7 HW#5 17 – 20

19. A player kicks a football from ground level with an initial velocity of 20 m / s, at 45° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 45° Step 1: components: v ix = v i. cosθ =(20 m / s )cos45° = 14 m / s v iy = v i. sinθ =(20 m / s )sin45° = 14 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 14 m / s Step3 (part B): Range d x = v x. t total Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y Ch7 HW#5 17 – 20

19. A player kicks a football from ground level with an initial velocity of 20 m / s, at 45° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 45° Step 1: components: v ix = v i. cosθ =(20 m / s )cos45° = 14 m / s v iy = v i. sinθ =(20 m / s )sin45° = 14 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 14 + (-9.8)t t top = 1.44s v iy = 14 m / s t total = 2.9s Step3 (part B): Range d x = v x. t total Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y Ch7 HW#5 17 – 20

19. A player kicks a football from ground level with an initial velocity of 20 m / s, at 45° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 45° Step 1: components: v ix = v i. cosθ =(20 m / s )cos45° = 14 m / s v iy = v i. sinθ =(20 m / s )sin45° = 14 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 14 + (-9.8)t t top = 1.44s v iy = 14 m / s t total = 2.9s Step3 (part B): Range d x = v x. t total = (14 m / s )(2.9s) = 41m Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y Ch7 HW#5 17 – 20

19. A player kicks a football from ground level with an initial velocity of 20 m / s, at 45° above the horizontal. Find: a. Hang timeb. Range c. Max Height 27m/s 45° Step 1: components: v ix = v i. cosθ =(20 m / s )cos45° = 14 m / s v iy = v i. sinθ =(20 m / s )sin45° = 14 m / s Step 2 (part A): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 14 + (-9.8)t t top = 1.44s v iy = 14 m / s t total = 2.9s Step3 (part B): Range d x = v x. t total = (14 m / s )(2.9s) = 41m Step4 (C): Max Height d y = ? v fy 2 = v iy 2 + 2ad y 0 2 = 14 2 + 2(-9.8)d y d y = 10m Ch7 HW#5 17 – 20

20. An archershoots an arrow at 10° with an initial velocity of 55 m / s, Find: a. Range b. Hang time 55m/s 10° Step 1: components: v ix = v i. cosθ v iy = v i. sinθ Step 2 (part B): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 9.6 m / s Step3 (part A): Range d x = v x. t total

20. An archershoots an arrow at 10° with an initial velocity of 55 m / s, Find: a. Range b. Hang time 55m/s 10° Step 1: components: v ix = v i. cosθ =(55 m / s )cos10° = 54 m / s v iy = v i. sinθ =(55 m / s )sin10° = 9.6 m / s Step 2 (part B): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 9.6 m / s Step3 (part A): Range d x = v x. t total

20. An archershoots an arrow at 10° with an initial velocity of 55 m / s, Find: a. Range b. Hang time 55m/s 10° Step 1: components: v ix = v i. cosθ =(55 m / s )cos10° = 54 m / s v iy = v i. sinθ =(55 m / s )sin10° = 9.6 m / s Step 2 (part B): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 9.6 + (-9.8)t t top = 0.97s v iy = 9.6 m / s t total = 1.95s Step3 (part A): Range d x = v x. t total

20. An archershoots an arrow at 10° with an initial velocity of 55 m / s, Find: a. Range b. Hang time 55m/s 10° Step 1: components: v ix = v i. cosθ =(55 m / s )cos10° = 54 m / s v iy = v i. sinθ =(55 m / s )sin10° = 9.6 m / s Step 2 (part B): Hang Time: v fy = 0 m / s t top = ? d y = ? v f = v i + a. t a = –9.8 m / s 2 0 = 9.6 + (-9.8)t t top = 0.97s v iy = 9.6 m / s t total = 1.95s Step3 (part A): Range d x = v x. t total = (54 m / s )(1.95s) = 106m

Ch7.5 – Circular Motion v i = 0v f = 25 m / s a = ? t = 2s

Ch7.5 – Circular Motion v i = 0v f = 40 m / s a = ? t = 4s v f = v i + a. ta = 10 m / s 2 If the car continues at a constant speed of 25 m / s as it rounds a corner, is it accelerating?

Ch7.5 – Circular Motion v i = 0v f = 25 m / s a = ? t = 2s v f = v i + a. ta = 15 m / s 2 If the car continues at a constant speed of 25 m / s as it rounds a corner, is it accelerating? If an object changes direction while maintaining the same speed, its velocity is changing, so technically it is accelerating.

Ch7.5 – Circular Motion If an object changes direction while maintaining the same speed, its velocity is changing, v=25 m / s so technically it is accelerating. r = 30m Centripetal Acceleration: Ex1) A car traveling at a constant speed of 25 m / s, rounds a corner with a radius of 30m. What is its centripetal acceleration?

If an object is accelerating, then there must be a net force acting on it. (Newton’s 2 nd Law, right?) F = m. a

If an object is accelerating, then there must be a net force acting on it. (Newton’s 2 nd Law, right?) Ex2) A 13g rubber stopper is attached to a 0.93m string. The stopper is swung in a horizontal circle, making 1 revolution in 1.18 sec. Find the tension force exerted in the string. (Top view)

If an object is accelerating, then there must be a net force acting on it. (Newton’s 2 nd Law, right?) Ex2) A 13g rubber stopper is attached to a 0.93m string. The stopper is swung in a horizontal circle, making 1 revolution in 1.18 sec. Find the tension force exerted in the string. (Top view) velv = d/t = 2πr/t F c = 2π(.93)/1.18 = 5 m / s

HW#21) A 26g rubber stopper is attached to a 0.93m string. The stopper is swung in a horizontal circle, with a velocity of 4.96 m / s. a. What is the centripetal acceleration? b. What is the centripetal force? c. What agent exerts this force? a. b. c.

HW#21) A 26g rubber stopper is attached to a 0.93m string. The stopper is swung in a horizontal circle, with a velocity of 4.96 m / s. a. What is the centripetal acceleration? b. What is the centripetal force? c. What agent exerts this force? a. b. c. Tension

HW#23) Racing on a flat track, a 1500kg car going 32 m / s, rounds a curve with a 56m radius a. What is the centripetal acceleration?v=32 m / s b. What is the centripetal force? c. What minimum coefficient of friction is necessary for the car not to slip? r=56m a. b. c.

HW#23) Racing on a flat track, a 1500kg car going 32 m / s, rounds a curve with a 56m radius a. What is the centripetal acceleration? v=32 m / s b. What is the centripetal force? c. What minimum coefficient of friction is necessary for the car not to slip? r=56m a. b. c.

HW#23) Racing on a flat track, a 1500kg car going 32 m / s, rounds a curve with a 56m radius a. What is the centripetal acceleration? v=32 m / s b. What is the centripetal force? c. What minimum coefficient of friction is necessary for the car not to slip? r=56m a. b. F N c. F c F g

HW#23) Racing on a flat track, a 1500kg car going 32 m / s, rounds a curve with a 56m radius a. What is the centripetal acceleration? v=32 m / s b. What is the centripetal force? c. What minimum coefficient of friction is necessary for the car not to slip? r=56m a. b. F N c. F c = F f 27,428N = µ. F N F c 27,428N = µ. F g 27,428N = µ. 15,000N F g µ = 1.80 Ch7 HW#6 21 – 25

Lab7.5 – Circular Motion - due tomorrow - Ch7 HW#6 due at beginning of period

Ch7 HW#6 21 – 25 22. A 50kg runner @ 8.8 m / s, rounds a curve with a radius of 25m. a. a c b. F c c. Agent? a. b. c.

24. The Moon’s orbital speed is 1026 m / s. The distance to the earth is 3.85x10 8 m. The Moon’s mass is 2.35x10 22 kg. a. a c b. F c c. Agent? a. b. c.

25. If the earth spins at 465 m / s at the equator, how come it doesn’t throw a 97kg person off? a. F c b. F g c. Apparent weight a. b. c.

25. If the earth spins at 465 m / s at the equator, how come it doesn’t throw a 97kg person off? Earth radius = 6.4x10 6 m. a. F c b. F g c. Apparent weight a. b. c.

25. If the earth spins at 465 m / s at the equator, how come it doesn’t throw a 97kg person off? Earth radius = 6.4x10 6 m. a. F c b. F g c. Apparent weight a. b. F g = m. g = (97kg)(9.8 m / s 2 ) = 951N c.

25. If the earth spins at 465 m / s at the equator, how come it doesn’t throw a 97kg person off? Earth radius = 6.4x10 6 m. a. F c b. F g c. Apparent weight a. b. F g = m. g = (97kg)(9.8 m / s 2 ) = 951N c. F net = 951N – 3.3N = 948N

Ch7.6 – Torque

F out F in r in F in Torque = F. r r in r out

Ch7.6 – Torque F out F in r in F in Torque = F. r Energy is always conserved. Work in = Work out F in. d in = F out. d out F 1. r 1 = F 2. r 2 r in r out

Ex1) A screwdriver is used to pry open a paint can lid. If the screwdriver is 15cm long, and the fulcrum is located 0.5cm from its tip, and I apply a force of 50N to the handle, what force is applied to the lid.

Ex1) A screwdriver is used to pry open a paint can lid. If the screwdriver is 15cm long, and the fulcrum is located 0.5cm from its tip, and I apply a force of 50N to the handle, what force is applied to the lid. F out F in = 50N r in = 14.5cm r out = 0.5cm F in. d in = F out. d out (50N)(14.5cm) = F out. (0.5cm) F out = 1450N

Ex2) Jane weighs 500N and sits 2m from the fulcrum of a seesaw. If Bob weighs 750N, where does he need to sit to balance? F 1. r 1 = F 2. r 2

Ex2) Jane weighs 500N and sits 2m from the fulcrum of a seesaw. If Bob weighs 750N, where does he need to sit to balance? F 1. r 1 = F 2. r 2 F gB. r B = F gJ. r J 750N. r B = 500N. (2m) r B = 1.3M r B r J F J F B

Ex3) Holly has a mass of 55kg and sits 1.5m from the fulcrum of a seesaw. If Chad sits 2.25m from the fulcrum, what is his weight? F 1. r 1 = F 2. r 2

Ex3) Holly has a mass of 55kg and sits 1.5m from the fulcrum of a seesaw. If Chad sits 2.25m from the fulcrum, what is his weight? F 1. r 1 = F 2. r 2 F gC. r C = F gH. r H F gC. 2.25m = 550N. (1.5m) F gC = 367M r C r H F H F C

HW#28 Hint) Person A weighing 800N, sits 5 meters from the fulcrum of a very long seesaw. Person B weighing 650N has a backpack on and sits 4.5m out to make it balance. What is the mass of the backpack? F gA. r A = F gB. r B F gA. r A = (F gB +F gBP ). r B r A r B F B F A Ch7 HW#7 26 – 29

Lab 7.6 – Torque - due tomorrow - Ch7 HW#7 due at beginning of period

Ch7 HW#7 26 – 29 HW#26) Person A weighing 800N, sits 5 meters from the fulcrum of a very long seesaw. Person B weighing 650N sits where to make it balance? F gA. r A = F gB. r B r A r B F B F A

HW#27) Person A with a mass of 62kg, sits 2.5 meters from the fulcrum of a seesaw. What is the weight of person B who sits at 1.75m to make it balance? F gA. r A = F gB. r B r A r B F B F A

Ch7 HW#7 26 – 29 HW#28) Person A weighing 800N, sits 5 meters from the fulcrum of a very long seesaw. Person B weighing 650N has a backpack on and sits 4.5m out to make it balance. What is the mass of the backpack? F gA. r A = F gB. r B F gA. r A = (F gB +F gBP ). r B r A r B m = F B F A

Ch7 HW#7 26 – 29 HW#28) Person A weighing 800N, sits 5 meters from the fulcrum of a very long seesaw. Person B weighing 650N has a backpack on and sits 4.5m out to make it balance. What is the mass of the backpack? F gA. r A = F gB. r B F gA. r A = (F gB +F gBP ). r B (800N)(5m) = (650N + F gBP )(4.5m) F gBP = r A r B F B F A

r in r out 29. A person applies a froce of 100N to one end of a prybar, 80cm long. The pry bar is set against a triangular block of wood, making a fulcrum at the 70cm mark. A heavy rock is just beginning to lift off the ground. What is the mass of the rock? F out F in

r in r out 29. A person applies a froce of 100N to one end of a prybar, 80cm long. The pry bar is set against a triangular block of wood, making a fulcrum at the 70cm mark. A heavy rock is just beginning to lift off the ground. What is the mass of the rock? F out F in. d in = F out. d out (100N)(70cm) = F out. (10cm) F in

Ch7 Test Review 1.A 540N sign is held in equilibrium by 2 wires that both make 35˚ with the horizontal. Find the tension force in each wire. FgFg FTFT 35˚ FTFT

Ch7 Test Review 1.A 540N sign is held in equilibrium by 2 wires that both make 35˚ with the horizontal. Find the tension force in each wire. FgFg F Ty FTFT 35˚ F net = F g − 2F Ty 0 = 540N – 2(F T ∙ sinθ) 0 = 540N – 2(F T ∙ sin35˚) F T = 470N FTFT

2. A 1120N block is sliding down a 30° incline plane. a. Label all forces. b. Find F g|| and F g┴ c. If the coefficient of friction is 0.30, find the friction force d. Find the accl down the incline. 30°

2. A 1120N block is sliding down a 30° incline plane. a. Label all forces. b. Find F g|| and F g┴ c. If the coefficient of friction is 0.30, find the friction force d. Find the accl down the incline. F N b. F g|| F f F g┴ F g|| F g┴ F g 30°

2. A 1120N block is sliding down a 30° incline plane. a. Label all forces. b. Find F g|| and F g┴ c. If the coefficient of friction is 0.30, find the friction force d. Find the accl down the incline. F N b. F g|| = m. g. sinθ = 1120N. sin30° = 560N F f F g┴ = m. g. cosθ = 1120N. cos30° = 969N c. F g|| F g┴ F g 30°

2. A 1120N block is sliding down a 30° incline plane. a. Label all forces. b. Find F g|| and F g┴ c. If the coefficient of friction is 0.30, find the friction force d. Find the accl down the incline. F N b. F g|| = m. g. sinθ = 1120N. sin30° = 560N F f F g┴ = m. g. cosθ = 1120N. cos30° = 969N c. F f,k = µ k ∙F N = (.30). F g┴ F g|| = (.30)(969N) = 291N F g┴ F g 30°

2. A 1120N block is sliding down a 30° incline plane. a. Label all forces. b. Find F g|| and F g┴ c. If the coefficient of friction is 0.30, find the friction force d. Find the accl down the incline. F N b. F g|| = m. g. sinθ = 1120N. sin30° = 560N F f F g┴ = m. g. cosθ = 1120N. cos30° = 969N c. F f,k = µ k ∙F N = (.30). F g┴ F g|| = (.30)(969N) = 291N F g┴ F g d. F net = F g|| – F f,k m. a = 560N – 291N (112kg)a = 269N a = 2.4 m / s 2 30°

v i = 18 m/s d y = 52 m 3. A person standing on the edge of a cliff throws a stone horizontally at 18m/s from the top of a cliff 52m high. a. How long does it take the stone to hit the water? b. How far from the base of the cliff will it hit? d y = v i t + ½at 2 d x = v con t

v i = 18 m/s d y = 52 m 3. A person standing on the edge of a cliff throws a stone horizontally at 18m/s from the top of a cliff 52m high. a. How long does it take the stone to hit the water? b. How far from the base of the cliff will it hit? d y = v i t + ½at 2 52 = 0 + ½(9.8)t 2 t = 3.26 sec d x = v i t + ½at 2 = v con t + 0 = 18m/s(3.26s) = 58.6m d x = v con t d y = v i t + ½at 2

HW#17) A player kicks a football from ground level with an initial velocity of 25 m / s, at 45° above the horizontal. Find: a. Find componenntsb. Hang timec. Range 25m/s 45° a. Components: v ix = v i. cosθ v iy = v i. sinθ b. Hang Time: v fy = 0 m / s t top = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 17.7 m / s c. Range d x = v x. t total

HW#17) A player kicks a football from ground level with an initial velocity of 25 m / s, at 45° above the horizontal. Find: a. Find componenntsb. Hang timec. Range 25m/s 45° a. Components: v ix = v i. cosθ =(25 m / s )cos45° = 17.7 m / s v iy = v i. sinθ =(25 m / s )sin45° = 17.7 m / s b. Hang Time: v fy = 0 m / s t top = ? v f = v i + a. t a = –9.8 m / s 2 v iy = 17.7 m / s c. Range d x = v x. t total

HW#17) A player kicks a football from ground level with an initial velocity of 25 m / s, at 45° above the horizontal. Find: a. Find componenntsb. Hang timec. Range 25m/s 45° a. Components: v ix = v i. cosθ =(25 m / s )cos45° = 17.7 m / s v iy = v i. sinθ =(25 m / s )sin45° = 17.7 m / s b. Hang Time: v fy = 0 m / s t top = ? v f = v i + a. t a = –9.8 m / s 2 0 = 17.7 + (-9.8)t t top = 1.8s v iy = 17.7 m / s t total = 3.6s c. Range d x = v x. t total

HW#17) A player kicks a football from ground level with an initial velocity of 25 m / s, at 45° above the horizontal. Find: a. Find componenntsb. Hang timec. Range 25m/s 45° a. Components: v ix = v i. cosθ =(25 m / s )cos45° = 17.7 m / s v iy = v i. sinθ =(25 m / s )sin45° = 17.7 m / s b. Hang Time: v fy = 0 m / s t top = ? v f = v i + a. t a = –9.8 m / s 2 0 = 17.7 + (-9.8)t t top = 1.8s v iy = 17.7 m / s t total = 3.6s c. Range d x = v x. t total = (17.7 m / s )(3.6s) = 63.7m

5) Racing on a flat track, a 1500kg car going 20 m / s, rounds a curve with a 25m radius a. What is the centripetal acceleration? V=20 m / s b. What is the centripetal force? c. What agent exerts this force. r=25m a. b. c.

5) Racing on a flat track, a 1500kg car going 20 m / s, rounds a curve with a 25m radius a. What is the centripetal acceleration? V=20 m / s b. What is the centripetal force? c. What agent exerts this force. r=25m a. b. c. Tension

6. A 50kg kid sits 3.5m from the fulcrum of a seesaw. Where does his 75kg bro need to sit to balance? F 1. r 1 = F 2. r 2 r 1 r 2 F 2 F 1

6. A 50kg kid sits 3.5m from the fulcrum of a seesaw. Where does his 75kg bro need to sit to balance? F 1. r 1 = F 2. r 2 F g1. r 1 = F g2. r 2 500N. (3.5m) = (750N). (r 2 ) r 2 = 2.3M r 1 r 2 F 2 F 1

100 90 80 70 vel 60 (m/s) 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 time (sec)

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100 90 80 70 vel 60 (m/s) 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 time (sec)

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Presentation on theme: "100 90 80 70 vel 60 (m/s) 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 time (sec)"— Presentation transcript:

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