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)

100 90 80 70vel 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 = 0In 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 = 0In 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

FT

Fg

Fnet = Fg – FT

0 = 10N – FT

FT = 10N

1 kg

Ex2) What is the tension in each rope?

1 kg

FT FT Fnet = Fg – 2FT

0 = 10N – 2FT

Ft = 5N

Ex2) What is the tension in each rope?

Fg

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°

Fg Fnet = Fg – 2FTy 0 = 10N – 2(FT ∙ sinθ) 0 = 10N – 2(FT ∙ sin45°) FT = 7N

1 Kg 45°

FTy

FTx

FTy FTy

FT

FTy = FT ∙ sinθ

FTx = FT ∙ cosθ

FT FT

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˚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

Fg

FTyFTyFT

22˚22˚

Fnet = Fg − 2FTy

0 = 168N – 2(FT ∙ sinθ) 0 = 168N – 2(FT ∙ sin22˚) FT = 224N

FT

Ex 5) A 168N sign is held by 2 ropes as shown. What is the tension in rope 2 if FT1 = 86N?

Open

FT1 = 86N

60˚20˚

FT2

Ex 5) A 168N sign is held by 2 ropes as shown. What is the tension in rope 2 if FT1 = 86N?

FTy2

FTy1

Fg

Ch7 HW#1 1 – 5 (Sep paper)

Fnet = Fg − FT1y− FT2y

0 = 168N − FT1∙sinθ1 − FT2∙sinθ2

0 = 168N − 86N∙sin20˚ − FT2∙sin60˚

FT2 = 160N

Open

FT1 = 86N

60˚20˚

FT2

Lab7.1 – Vector Addition

- due in tomorrow

- Ch7 HW#1 due at beginning of period

Ch7 HW#1 1 – 51. What is the tension in each rope?

Fg

20 kg

FT FT

Ch7 HW#1 1 – 51. What is the tension in each rope?

Fnet = Fg – 2FT

0 = 200N – 2FT

Ft = 100N

Fg

20 kg

FT FT

2. What is the tension in each rope?

45°

Fg

20kg

FTy FTyFT FT


Fnet = Fg − 2FTy

0 = 200N – 2(FT ∙ sinθ) 0 = 200N – 2(FT ∙ sin45˚) 45° FT = 141N

Fg

20kg

FTy FTyFT FT


Fg

50kg

FTy FTyFT FT


Fnet = Fg − 2FTy


Fg

50kg

FTy FTyFT FT


45° 60°

Fg

50kg

FTy1 FTy2FT1 =259N FT2


Fnet = Fg − FTy1 – FTy2

0 = 500N – (FT1 ∙ sinθ) – (FT2 ∙ sinθ) 0 = 500N – (259N ∙ sin45°) – (FT2 ∙ sin60°) 45° 60° FT2 = 366N

Fg

50kg



40° 50°

Fg

75kg



Fnet = Fg − FTy1 – FTy2

0 = 750N – (FT1 ∙ sinθ) – (FT2 ∙ sinθ) 0 = 500N – (400N ∙ sin40°) – (FT2 ∙ sin50°) 40° 50° FT2 = 643N

Fg

75kg


Ch7.2 – Inclined Planes

Ch7.2 – Inclined Planes FN

FN = Fg

Fg

θ

Ch7.2 – Inclined Planes FN

FN = Fg

Fg

FN

Fg|| = Fg.sinθ

θ Fg|| Fg ┴ = Fg.cosθ

Fg┴ Fg Fg ┴ = FN

Fg|| θ

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

40°

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

FN

Fg||

Fg┴ Fg

Fg ┴ = Fg.cosθ Fg|| = Fg

.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. Fg|| = Fg.sinθ

= 500N.sin25° FN = 211N

b. FN = Fg ┴

= Fg.cosθ

Fg|| = 500N.cos25° = 453N

Fg┴ Fg

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?

FN

Fg|| Fg┴

Fg

Fnet = Fg||

m.a = Fg.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 = (9.8m/s2).sin42°

a = 6.6 m/s2

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 slipsis 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 slipsis 45°. What is the coefficient of friction between the mass and plane?

FN

Ff,s

Fg|| Fg┴

Fg

Fnet = Fg|| – Ff,s m.a = Fg

.sinθ – μs.FN

0 = Fg.sinθ – μs

. Fg┴ 0 = Fg

.sinθ – μs. Fg

.cosθ 0 = sinθ – μs

. cosθ

Ch7 HW#2 6 – 9

45°

0.145cos

45sin

cos

sin

s

s

s

Ch7 HW#2 6 – 96. A 91N block is placed on a 35˚ inclined plane.

Find Fg ┴ and Fg|| .

Fg


.sinθ

35°

Ch7 HW#2 6 – 96. A 91N block is placed on a 35˚ inclined plane.


Fg


.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. Fg|| = Fg.sinθ

b. FN = Fg┴

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. Fg|| = Fg.sinθ

= 250N.sin50° FN = 192N

b. FN = Fg┴

= Fg.cosθ

Fg|| = 250N.cos50° Fg┴ = 161N

Fg

50°

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

Fg

32°


FN

Fg|| Fg┴

Fg

Fnet = Fg||

m.a = Fg.sinθ

m.a = m.g.sinθ a = g.sinθ a = (9.8m/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 slipsis 35°. What is the coefficient of friction between the mass and plane?

Fg

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 slipsis 35°. What is the coefficient of friction between the mass and plane?

FN Ff,s

Fg|| Fg┴

Fg


.sinθ – μs.FN

0 = Fg.sinθ – μs

. Fg┴ 0 = Fg

.sinθ – μs. Fg

.cosθ 0 = sinθ – μs

. cosθ

35°

7.035cos

35sin

cos

sin

s

s

s

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?

Fg

b. How fast will it be going after 4 sec?

30°


FN

Fnet = Fg||

m.a = Fg.sinθ

m.a = m.g.sinθ a = g.sinθ Fg|| a = (9.8m/s

2).sin30° Fg┴

a = 5 m/s2 Fg

b. How fast will it be going after 4 sec?vi = 0m/s vf = ? t = 4s a = _____

30°


FN

Fnet = Fg||

m.a = Fg.sinθ

m.a = m.g.sinθ a = g.sinθ Fg|| a = (9.8m/s

2).sin30° Fg┴

a = 5 m/s2 Fg

b. How fast will it be going after 4 sec?vi = 0m/s vf = ? t = 4s a = 5 m/s

2

vf = vi + a.t = 0 + (5m/s2)(4s) = 20m/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? FN

Ff,k

Fg|| Fg┴ Fg

30°

c. If the trunk were placed on an incline where the coefficient of kinetic frictionwas 0.15, what would be the acceleration of the trunk?

FN

Fnet = Fg|| – Ff,k Ff,k m.a = Fg

.sinθ – µk∙FN m.a = m.g.sinθ – µk∙Fg┴

m.a = m.g.sinθ – µk∙ m.g.cosθ Fg|| a = (9.8m/s

2).sin30° Fg┴ – (.15)∙(9.8m/s

2).cos30° Fg a = 3m/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 FN

Fg|| Ff,k

Fg┴ Fg

Fnet = F – Fg|| – Ff 0 = F – Fg

.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.


Fg


.sinθ

10°


32°


FN

Fg|| Fg┴

Fg

Fnet = Fg||

m.a = Fg.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?

FN

Fnet = Fg||

m.a = Fg.sinθ

m.a = m.g.sinθ Fg||

Fg┴

a = Fg


vi = 0m/s vf = ? t = 5s a = 6.4m/s2

vf = vi + 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?

FN Ff,k

Fg|| Fg┴

Fg

Fnet = Fg|| – Ff,k m.a = Fg

.sinθ – µk∙FN

m.a = m.g.sinθ – µk∙Fg┴ m.a = m.g.sinθ – µk∙ 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 FN

Fg|| Ff,k

Fg┴ Fg

Fnet = F – Fg|| – Ff 0 = F – Fg

.sinθ – 50N 0 = F – 400N.sin10° – 50N F =

10°

Ch7.3 – Projectile Motion

vi

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


Vertical motion controlled by gravity

vi

t = 1s t = 2s t = 3s t = 4s v1 = vi v2 = vi v3 = vi v4 = vi



Horizontal motion controlled by the initial velocity given to the projectile.

vi


t = 1 s

t = 2 s

t = 3 s

t = 4 s




vi


t = 1 s

t = 2 s

t = 3 s

t = 4 s




Only thing that links the 2 motions is time

vi = 15 m/s

dy = 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?

vi = 15 m/s

dy = 50 m


a. The stone hits the ground at the same time, whether it is thrown horizontallyor dropped. So solve for the dropped stone!

vi = 15 m/s

dy = 50 m



dy = vit + ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

vi = 15 m/s

dy = 50 m



b. If there were no gravity, the ball would travel the same distance away from the cliff, as the actual projectile.

dy = vit + ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

vi = 15 m/s

dy = 50 m




dy = vit + ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

dx = vit + ½at2

= vcont + 0 = 15 m/s(2.9 s) = 43.5m

vi = 15 m/s

dy = 50 m




dy = vit + ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

dx = vit + ½at2

= vcont + 0 = 15 m/s(2.9 s) = 43.5m

c. Find both components of final velocity, and pythag!

vi = 15 m/s

dy = 50 m




dy = vit + ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

dx = vit + ½at2

= vcont + 0 = 15 m/s(2.9 s) = 43.5m

c. Find both components of final velocity, and pythag!

vfx

vfyvf

vfx = 15 m/svfy = viy + at = 0 + (9.8)(2.9) = 28.4 m/s

smv f /1.324.2815 22 Ch7 HW#4 14 – 16

vi = 5 m/s

dy = 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?

vfx

vfyvf

vi = 5 m/s

dy = 78.4 m


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? dy = vit + ½at2

78.4 = 0 + ½(9.8)t2

t = 3.96 sec

vfx

vfyvf

vi = 5 m/s

dy = 78.4 m



78.4 = 0 + ½(9.8)t2

t = 3.96 sec

dx = vit + ½at2

= vcont + 0 = 5 m/s(3.96 s) = 20m

vfx

vfyvf

vi = 5 m/s

dy = 78.4 m



78.4 = 0 + ½(9.8)t2

t = 3.96 sec dx = vit + ½at2

= vcont + 0 = 5 m/s(3.96 s) = 20m

vfx

vfyvf

vfx = 5 m/svfy = viy + at = 0 + (9.8)(3.96) = 39 m/s

smv f /(___)395 22

vi = 10 m/s

dy = 156.8 m


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?

vfx

vfyvf

vi = 10 m/s

dy = 156.8 m



156.8 = 0 + ½(9.8)t2

t = 5.6 sec

vfx

vfyvf

vi = 10 m/s

dy = 156.8 m



156.8 = 0 + ½(9.8)t2

t = 5.6 sec dx = vit + ½at2

= vcont + 0 = 10 m/s(5.6s) = 56m

vfx

vfyvf

vi = 10 m/s

dy = 156.8 m



156.8 = 0 + ½(9.8)t2

t = 5.6 sec dx = vit + ½at2

= vcont + 0 = 10 m/s(5.6s) = 56m

vfx

vfyvf

vfx = 10 m/svfy = viy + at = 0 + (9.8)(5.6) = 55.5 m/s

smv f /4.565.5510 22

vi = ?

dy = 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?

dx = 0.352m

vi = ?

dy = 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?

dy = vit + ½at2

0.950 = 0 + ½(9.8)t2

t = 0.44 sec

dx = 0.352m

vi = ?

dy = 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?

dy = vit + ½at2

0.950 = 0 + ½(9.8)t2

t = 0.44 sec

dx = vit + ½at2

dx = vcont + 00.352m = v.(.44s) v = 0.8m/s

dx = 0.352m

Ch7 Mid Chapter Review1. 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. FN = Fg ┴ c. Fnet = Fg|| – Ff,s

Fg

15°



FN


= Fg.cosθ Ff,k

= 54N.cos15° = 52N Fg|| Fg┴

Fg

15°



FN


= Fg.cosθ Ff,k 0 = Fg

.sinθ – Ff,s = 54N.cos15° Ff,s = m.g.sinθ = 52N Ff,s = 54N.sin15° Fg|| Fg┴

Fg

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.

FN

Ff,k

Fg|| Fg┴

Fg


.sinθ – µk∙FN m.a = m.g.sinθ – µk∙Fg┴ m.a = m.g.sinθ – µk∙ m.g.cosθ a = g.sinθ – µk∙g.cosθ a = (9.8m/s

2).sin60° – (.45)∙(9.8m/s2).cos60°

a = 6.3 m/s2

60°

(Save for HW)3. A block with a mass of 5.0kg is placed on a 30° incline plane,

where µk = 0.14.a. Find acceleration.b. Find velocity after 4s.

Fg

30°

3. A block with a mass of 5.0kg is placed on a 30° incline plane, where µk = 0.14.a. Find acceleration. FN

b. Find velocity after 4s.

a. Fnet = Fg|| – Ff,k m.a = Fg

.sinθ – µk∙FN m.a = m.g.sinθ – µk∙Fg┴ Fg|| m.a = m.g.sinθ – µk∙ m.g.cosθ Fg┴

a = g.sinθ – µk∙g.cosθ Fg

a = (9.8m/s2).sin30° – (.14)∙(9.8m/s

2).cos30° a =


vi = 0m/s vf = ? t = 4s a = 3.7 m/s2

vf = vi + a.t

30°

4. Find the tension force exerted by each cable to support the 625N bag.

Fg

FT

35˚

35˚

FT

4. Find the tension force exerted by each cable to support the 625N bag.

Fg

FTyFTyFT

35˚

35˚

Fnet = Fg − 2FTy


FT

(Save for HW)5. What is the tension in each rope to support the 100kg sign.

60° 60°

Fg

100kg

FT FT

5. What is the tension in each rope to support the 100kg sign.

60° 60°Fnet = Fg – 2FTy

0 = 1000N – 2(FT ∙ sinθ) 0 = 1000N – 2(FT ∙ sin60°) FT =

Fg

100kg

FTy FTyFT FT

vi = 18 m/s

dy = 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? dy = vit + ½at2

dx = vit + ½at2

vi = 18 m/s

dy = 52 m


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? dy = vit + ½at2

52 = 0 + ½(9.8)t2

t = 3.26 sec dx = vit + ½at2

= vcont + 0 = 18m/s(3.26s) = 58.6m

vfx

vfyvf

vfx = 18 m/svfy = viy + at = 0 + (9.8)(3.26) = 31.9m/s

smv f /6.369.3118 22

vi = ?

dy = 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?

dy = vit + ½at2

dx = vcont

dx = 1.20m

Lab7.3 – Projectile Motion

- due tomorrow

- Ch7 Mid Chapter Review due at beginning of period

Ch7.3 cont – Projectiles Launched at an AngleEx1) 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°

4.47

m/s

Ch7.3 cont – Projectiles Launched at an AngleEx1) A ball is launched with an initial velocity of 4.47m/s at an angle of 66°


66°

Step1: Find vi components

4.47

m/s

Ch7.3 cont – Projectiles Launched at an AngleEx1) A ball is launched with an initial velocity of 4.47 m/s at an angle of 66°


66°

Step1: Find vi components vix = vi

.cosθ viy = vi.sinθ

=(4.47m/s)cos66° =(4.47m/s)sin66° = 1.8m/s = 4.1m/s

Step2: Shoot the object straight up at viy.

viy = 4.1m/s

4.47

m/s



66°

Step1: Find vi components Step3: Find time to the top,vix = vi

.cosθ viy = vi.sinθ then double it.

=(4.47m/s)cos66° =(4.47m/s)sin66° = 1.8m/s = 4.1m/s

Step2: Shoot the object straight up at viy.

vfy = 0m/s dy = ? vfy

2 = viy2 + 2ady

02 = 4.12 + 2(-9.8)dy a = –9.8m/s

2

dy = 0.86m

viy = 4.1m/s

4.47

m/s



66°



=(4.47m/s)cos66° =(4.47m/s)sin66° vfy = 0m/s ttop = ? = 1.8m/s = 4.1m/s

Step2: Shoot the object straight up at viy. a = –9.8m/s2

vfy = 0m/s dy = ? vf = vi + a.tvfy

2 = viy2 + 2ady 0 = 4.1 + (-9.8)t

02 = 4.12 + 2(-9.8)dy viy = 4.1m/s ttop = 0.42s a = –9.8m/s

2 ttotal = 0.84sdy = 0.86m Step 4: Range uses vx and ttotal.

viy = 4.1m/s

4.47

m/s



66°



=(4.47m/s)cos66° =(4.47m/s)sin66° vfy = 0m/s ttop = ? = 1.8m/s = 4.1m/s

Step2: Shoot the object straight up at viy. a = –9.8m/s2

vfy = 0m/s dy = ? vf = vi + a.tvfy

2 = viy2 + 2ady 0 = 4.1 + (-9.8)t

02 = 4.12 + 2(-9.8)dy viy = 4.1m/s ttop = 0.42s a = –9.8m/s

2 ttotal = 0.84sdy = 0.86m Step 4: Range uses vx and ttotal.

dx = vx.ttotal

viy = 4.1m/s = (1.8m/s)(0.84s) = 1.5m

4.47

m/s

HW#17) A player kicks a football from ground level with an initial velocity of 27m/s, at 30° above the horizontal. Find: a. Hang time b. Range c. Max Height

27m

/s

30°

Step 1: components: vix = vi

.cosθ

viy = vi.sinθ

HW#17) A player kicks a football from ground level with an initial velocity of 27m/s,at 30° above the horizontal. Find: a. Hang time b. Range c. Max Height

27m

/s

30°


.cosθ =(27m/s)cos30° = 23m/s

viy = vi.sinθ

=(27m/s)sin30° = 13.5m/sStep 2: Hang Time:vfy = 0m/s ttop = ?

vf = vi + a.t a = –9.8m/s

2

viy = 4.1m/s


27m

/s

30°


.cosθ =(27m/s)cos30° = 23m/s

viy = vi.sinθ

=(27m/s)sin30° = 13.5m/sStep 2 (part A): Hang Time:vfy = 0m/s ttop = ?

vf = vi + a.t a = –9.8m/s

2 0 = 13.5 + (-9.8)t ttop = 1.37s viy = 13.5m/s ttotal = 2.75s

Step3 (part B): Range dx = vx

.ttotal


27m

/s

30°


.cosθ =(27m/s)cos30° = 23m/s

viy = vi.sinθ

=(27m/s)sin30° = 13.5m/sStep 2 (part A): Hang Time:vfy = 0m/s ttop = ? dy = ?

vf = vi + a.t a = –9.8m/s



.ttotal = (23m/s)(2.75s)

= 63.8m

Step4 (C): Max Height dy = ? vfy

2 = viy2 + 2ady


27m

/s

30°


.cosθ =(27m/s)cos30° = 23m/s

viy = vi.sinθ

=(27m/s)sin30° = 13.5m/sStep 2 (part A): Hang Time:vfy = 0m/s ttop = ? dy = ?

vf = vi + a.t a = –9.8m/s



.ttotal = (23m/s)(2.75s)

= 63.8m


2 = viy2 + 2ady

02 = 13.52 + 2(-9.8)dy dy = 9.3m


27m

/s

60°

Step 1: components:

Step 2 (part A): Hang Time:vfy = 0m/s

a = –9.8m/s

2

viy = 23m/s

Step3 (part B): Range

Step4 (C): Max Height


27m

/s

60°


.cosθ viy = vi

.sinθ

Step 2 (part A): Hang Time:vfy = 0m/s ttop = ? dy = ?

vf = vi + a.t a = –9.8m/s

2

viy = 23m/s


.ttotal


2 = viy2 + 2ady


27m

/s

60°


.cosθ =(27m/s)cos60° = 13.5m/s

viy = vi.sinθ

=(27m/s)sin60° = 23m/sStep 2 (part A): Hang Time:vfy = 0m/s ttop = ? dy = ?

vf = vi + a.t a = –9.8m/s

2

viy = 23m/s


.ttotal


2 = viy2 + 2ady


27m

/s

60°


.cosθ =(27m/s)cos60° = 13.5m/s

viy = vi.sinθ


vf = vi + a.t a = –9.8m/s

2 0 = 23 + (-9.8)t ttop = 2.4s viy = 23m/s ttotal = 4.8s


.ttotal


2 = viy2 + 2ady


27m

/s

60°


.cosθ =(27m/s)cos60° = 13.5m/s

viy = vi.sinθ


vf = vi + a.t a = –9.8m/s



.ttotal = (13.5m/s)(4.8s)

= 64.8m


2 = viy2 + 2ady


27m

/s

60°


.cosθ =(27m/s)cos60° = 13.5m/s

viy = vi.sinθ


vf = vi + a.t a = –9.8m/s



.ttotal = (13.5m/s)(4.8s)

= 64.8m


2 = viy2 + 2ady

02 = 232 + 2(-9.8)dy dy = 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 20m/s,at 45° above the horizontal. Find: a. Hang time b. Range c. Max Height

27m

/s

45°


.cosθ

viy = vi.sinθ


vf = vi + a.t a = –9.8m/s

2

viy = 14m/s


.ttotal


2 = viy2 + 2ady

Ch7 HW#5 17 – 20


27m

/s

45°


.cosθ =(20m/s)cos45° = 14m/s

viy = vi.sinθ

=(20m/s)sin45° = 14m/s


vf = vi + a.t a = –9.8m/s

2

viy = 14m/s


.ttotal


2 = viy2 + 2ady

Ch7 HW#5 17 – 20


27m

/s

45°


.cosθ =(20m/s)cos45° = 14m/s

viy = vi.sinθ

=(20m/s)sin45° = 14m/s


vf = vi + a.t a = –9.8m/s



.ttotal


2 = viy2 + 2ady

Ch7 HW#5 17 – 20


27m

/s

45°


.cosθ =(20m/s)cos45° = 14m/s

viy = vi.sinθ

=(20m/s)sin45° = 14m/s


vf = vi + a.t a = –9.8m/s



.ttotal = (14m/s)(2.9s)

= 41mStep4 (C): Max Height dy = ?

vfy2 = viy

2 + 2ady

Ch7 HW#5 17 – 20


27m

/s

45°


.cosθ =(20m/s)cos45° = 14m/s

viy = vi.sinθ

=(20m/s)sin45° = 14m/s


vf = vi + a.t a = –9.8m/s



.ttotal = (14m/s)(2.9s)

= 41mStep4 (C): Max Height dy = ?

vfy2 = viy

2 + 2ady

02 = 142 + 2(-9.8)dy dy = 10m

Ch7 HW#5 17 – 20

20. An archershoots an arrow at 10° with an initial velocity of 55m/s,Find: a. Range b. Hang time

55m/s

10°Step 1: components: vix = vi

.cosθ

viy = vi.sinθ

Step 2 (part B): Hang Time:vfy = 0m/s ttop = ? dy = ?

vf = vi + a.t a = –9.8m/s

2

viy = 9.6m/s

Step3 (part A): Range dx = vx

.ttotal


55m/s


.cosθ =(55m/s)cos10° = 54m/s

viy = vi.sinθ

=(55m/s)sin10° = 9.6m/s


vf = vi + a.t a = –9.8m/s

2

viy = 9.6m/s


.ttotal


55m/s


.cosθ =(55m/s)cos10° = 54m/s

viy = vi.sinθ

=(55m/s)sin10° = 9.6m/s


vf = vi + a.t a = –9.8m/s



.ttotal


55m/s


.cosθ =(55m/s)cos10° = 54m/s

viy = vi.sinθ

=(55m/s)sin10° = 9.6m/s


vf = vi + a.t a = –9.8m/s



.ttotal = (54m/s)(1.95s)

= 106m

Ch7.5 – Circular Motion

vi = 0 vf = 25m/s

a = ? t = 2s


vi = 0 vf = 40m/s

a = ? t = 4s vf = vi + a.t a = 10m/s

2

If the car continues at a constant speed of 25m/s as it rounds a corner, is it accelerating?


vi = 0 vf = 25m/s

a = ? t = 2s vf = vi + a.t a = 15m/s

2

If the car continues at a constant speed of 25m/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.


If an object changes direction while maintaining the same speed, its velocity is changing, v=25m/s

so technically it is accelerating. r = 30m

Centripetal Acceleration:

Ex1) A car traveling at a constant speed of 25m/s, rounds a corner with a radius of 30m. What is its centripetal acceleration?

r

vac

2


If an object changes direction while maintaining the same speed, its velocity is changing, v=25m/s

so technically it is accelerating. r = 30m

Centripetal Acceleration:

Ex1) A car traveling at a constant speed of 25m/s, rounds a corner with a radius of 30m. What is its centripetal acceleration?

r

vac

2

2

2

22

2130

625

30

)/25(s

mms

m

m

smac

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

F = m.a


Ex2) A 13g rubber stopper is attached to a 0.93m string. The stopper is swungin a horizontal circle, making 1 revolution in 1.18 sec. Find the tensionforce exerted in the string.

(Top view)

r

mv2cF


Ex2) A 13g rubber stopper is attached to a 0.93m string. The stopper is swungin a horizontal circle, making 1 revolution in 1.18 sec. Find the tensionforce exerted in the string.

(Top view)vel v = d/t

= 2πr/t Fc = 2π(.93)/1.18 = 5m/s

r

mv2cF

Nmr

smkg35.0

93.

)/5)(013.0(F

2

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 4.96m/s. a. What is the centripetal acceleration?b. What is the centripetal force?c. What agent exerts this force?

a.

b.

c.


a.

b.

c.

2

2

5.2693.0

)/96.4(s

mm

smac


a.

b.

c.

2

2

5.2693.0

)/96.4(s

mm

smac

Nm

smkg69.0

93.0

)/96.4)(026.0(F

2

c


a.

b.

c. Tension

2

2

5.2693.0

)/96.4(s

mm

smac

Nm

smkg69.0

93.0

)/96.4)(026.0(F

2

c

HW#23) Racing on a flat track, a 1500kg car going 32m/s, rounds a curve with a 56m radius a. What is the centripetal acceleration? v=32m/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.



r=56m a.

b.

c.

2

2

3.1856

)/32(s

mm

smac



r=56m a.

b. FN

c. Fc

Fg

2

2

3.1856

)/32(s

mm

smac

Nm

smkg428,27

56

)/32)(1500(F

2

c



r=56m a.

b. FN

c. Fc = Ff 27,428N = µ.FN Fc 27,428N = µ.Fg 27,428N = µ.15,000N Fg

µ = 1.80 Ch7 HW#6 21 – 25

2

2

3.1856

)/32(s

mm

smac

Nm

smkg428,27

56

)/32)(1500(F

2

c

Lab7.5 – Circular Motion

- due tomorrow


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

a. ac b. Fc c. Agent?

a.

b.

c.

24. The Moon’s orbital speed is 1026m/s. The distance to the earth is 3.85x108m.The Moon’s mass is 2.35x1022kg. a. ac b. Fc c. Agent?

a.

b.

c.

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

a.

b.

c.

25. If the earth spins at 465m/s at the equator, how come it doesn’t throw a 97kg person off? Earth radius = 6.4x106 m.a. Fc b. Fg c. Apparent weight

a.

b.

c.

26

2

3.3104.6

)/465)(97(s

mmx

smkgFc


a.

b. Fg = m.g = (97kg)(9.8m/s2) = 951N

c.

26

2

3.3104.6

)/465)(97(s

mmx

smkgFc


a.

b. Fg = m.g = (97kg)(9.8m/s2) = 951N

c. Fnet = 951N – 3.3N = 948N

26

2

3.3104.6

)/465)(97(s

mmx

smkgFc

Ch7.6 – Torque

Ch7.6 – Torque Fout

Fin rin

Fin

Torque = F . r

r in

r out

Ch7.6 – Torque Fout

Fin rin

Fin

Torque = F . r

Energy is always conserved.

Workin = Workout

Fin.din = Fout

.dout

F1.r1 = F2

.r2

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 15cmlong, and the fulcrum is located 0.5cm from its tip, and I apply a forceof 50N to the handle, what force is applied to the lid.

Fout

Fin = 50Nrin = 14.5cm

rout = 0.5cm

Fin.din = Fout

.dout (50N)(14.5cm) = Fout

.(0.5cm)

Fout = 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?

F1.r1 = F2

.r2

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?

F1.r1 = F2

.r2

FgB.rB = FgJ

.rJ

750N.rB = 500N.(2m)

rB = 1.3M

rB rJ

FJ

FB

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?

F1.r1 = F2

.r2

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?

F1.r1 = F2

.r2

FgC.rC = FgH

.rH

FgC.2.25m = 550N.(1.5m)

FgC = 367M

rC rH

FH

FC

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 onand sits 4.5m out to make it balance. What is the mass of the backpack?

FgA.rA = FgB

.rB

FgA.rA = (FgB+FgBP).rB

rA rB

FB

FA

Ch7 HW#7 26 – 29

Lab 7.6 – Torque

- due tomorrow


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 whereto make it balance?

FgA.rA = FgB

.rB

FgA.rA = FgB

.rB

rA rB

FB

FA

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 1.75mto make it balance?

FgA.rA = FgB

.rB

rA rB

FB

FA


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?

FgA.rA = FgB

.rB


rA rB m = FB

FA


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?

FgA.rA = FgB

.rB


(800N)(5m) = (650N + FgBP)(4.5m)

FgBP =rA rB

FB

FA

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 fulcrumat the 70cm mark. A heavy rock is just beginning to lift off the ground.What is the mass of the rock?

Fout

Fin

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 fulcrumat the 70cm mark. A heavy rock is just beginning to lift off the ground.What is the mass of the rock?

Fout Fin.din = Fout

.dout (100N)(70cm) = Fout

.(10cm)

Fin

Ch7 Test Review1. A 540N sign is held in equilibrium by 2 wires that both make 35˚

with the horizontal. Find the tension force in each wire.

Fg

FT

35˚35˚

FT

Ch7 Test Review1. A 540N sign is held in equilibrium by 2 wires that both make 35˚

with the horizontal. Find the tension force in each wire.

Fg

FTyFTyFT

35˚35˚

Fnet = Fg − 2FTy


FT

2. A 1120N block is sliding down a 30° incline plane.a. Label all forces.b. Find Fg|| and Fg┴

c. If the coefficient of friction is 0.30, find the friction forced. Find the accl down the incline.

30°


c. If the coefficient of friction is 0.30, find the friction forced. Find the accl down the incline. FN

b. Fg|| Ff

Fg┴

Fg|| Fg┴

Fg

30°



b. Fg|| = m.g.sinθ = 1120N.sin30° = 560N Ff

Fg┴ = m.g.cosθ = 1120N.cos30° = 969N

c. Fg||

Fg┴

Fg

30°




Fg┴ = m.g.cosθ = 1120N.cos30° = 969N

c. Ff,k = µk∙FN = (.30).Fg┴ Fg||

= (.30)(969N) = 291N Fg┴

Fg

30°




Fg┴ = m.g.cosθ = 1120N.cos30° = 969N

c. Ff,k = µk∙FN = (.30).Fg┴ Fg||

= (.30)(969N) = 291N Fg┴

Fg

d. Fnet = Fg|| – Ff,k m.a = 560N – 291N

(112kg)a = 269N a = 2.4 m/s

2

30°

vi = 18 m/s

dy = 52 m


a. How long does it take the stone to hit the water?b. How far from the base of the cliff will it hit?

dy = vit + ½at2 dx = vcont

vi = 18 m/s

dy = 52 m


a. How long does it take the stone to hit the water?b. How far from the base of the cliff will it hit?

dy = vit + ½at2

52 = 0 + ½(9.8)t2

t = 3.26 sec

dx = vit + ½at2

= vcont + 0 = 18m/s(3.26s) = 58.6m

dx = vcont

dy = vit + ½at2

HW#17) A player kicks a football from ground level with an initial velocity of 25m/s,at 45° above the horizontal. Find: a. Find componennts b. Hang time c. Range

25m

/s

45°

a. Components: vix = vi

.cosθ

viy = vi.sinθ

b. Hang Time:vfy = 0m/s ttop = ?

vf = vi + a.t a = –9.8m/s

2

viy = 17.7m/s

c. Range dx = vx

.ttotal


25m

/s

45°


.cosθ =(25m/s)cos45° = 17.7m/s

viy = vi.sinθ

=(25m/s)sin45° = 17.7m/sb. Hang Time:vfy = 0m/s ttop = ?

vf = vi + a.t a = –9.8m/s

2

viy = 17.7m/s

c. Range dx = vx

.ttotal


25m

/s

45°


.cosθ =(25m/s)cos45° = 17.7m/s

viy = vi.sinθ


vf = vi + a.t a = –9.8m/s


c. Range dx = vx

.ttotal


25m

/s

45°


.cosθ =(25m/s)cos45° = 17.7m/s

viy = vi.sinθ


vf = vi + a.t a = –9.8m/s


c. Range dx = vx

.ttotal = (17.7m/s)(3.6s)

= 63.7m

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

b. What is the centripetal force?c. What agent exerts this force.

r=25m a.

b.

c.



r=25m a.

b.

c.

2

2

1625

)/20(s

mm

smac



r=25m a.

b.

c.

2

2

1625

)/20(s

mm

smac

Nm

smkg000,24

25

)/20)(1500(F

2

c



r=25m a.

b.

c. Tension

2

2

1625

)/20(s

mm

smac

Nm

smkg000,24

25

)/20)(1500(F

2

c

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

F1.r1 = F2

.r2

r1 r2

F2

F1

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

F1.r1 = F2

.r2

Fg1.r1 = Fg2

.r2

500N.(3.5m) = (750N).(r2)

r2 = 2.3M

r1 r2

F2

F1

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)

Documents

fn fg fg fg fg

n ft ft

fg fg fn fg

fg fg ch7

fg fnet

ft cos ft ft

n sign

n fg 75kgfty1 fty2ft1