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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). Ch7 – More Forces .

100 90 80 70 vel 60

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- 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)
- ormoving at constant speed (dynamic).

- Ex1) What is the tension in the rope?

1 kg

- 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)
- ormoving at constant speed (dynamic).

- Ex1) What is the tension in the rope?

Fnet = Fg – FT

0 = 10N – FT

FT = 10N

FT

1 kg

Fg

Ex2) What is the tension in each rope?

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

FTyFTy

FTFT

FT

FTy

1 Kg

45°

FTx

FTy= FT ∙ sinθ

FTx= FT ∙ cosθ

Ex 4) A 168 N sign is held by 2 ropes that make 22˚ angles as shown.

What is the tension in each?

22˚

22˚

OPEN

Ex 4) A 168 N sign is held by 2 ropes that make 22˚ angles as shown.

What is the tension in each?

FT

FTy

FTy

FT

Fnet = Fg − 2FTy

0 = 168N – 2(FT ∙ sinθ)

0 = 168N – 2(FT ∙ sin22˚)

FT = 224N

22˚

22˚

OPEN

Fg

Ex 5) A 168N sign is held by 2 ropes as shown.

What is the tension in rope 2 if FT1 = 86N?

FT2

FT1 = 86N

20˚

60˚

Open

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)

FT2

FT1 = 86N

Fnet = Fg− FT1y− FT2y

0 = 168N − FT1∙sinθ1 − FT2∙sinθ2

0 = 168N − 86N∙sin20˚ − FT2∙sin60˚

FT2 = 160N

20˚

60˚

Open

1. What is the tension in each rope?

Fnet = Fg – 2FT

0 = 200N – 2FT

Ft = 100N

Fg

FT

FT

20 kg

2. What is the tension in each rope?

Fnet = Fg − 2FTy

0 = 200N – 2(FT ∙ sinθ)

0 = 200N – 2(FT ∙ sin45˚)45°

FT = 141N

Fg

FTyFTy

FTFT

20kg

3. What is the tension in each rope?

Fnet = Fg − 2FTy

0 = 500N – 2(FT ∙ sinθ)

0 = 500N – 2(FT ∙ sin15˚)

FT = 965N

Fg

FTyFTy

FTFT

50kg

4. What is the tension in each rope?

Fnet = Fg − FTy1 – FTy2

0 = 500N – (FT1 ∙ sinθ) – (FT2 ∙ sinθ)

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

FT2 = 366N

Fg

FTy1 FTy2

FT1 =259N FT2

50kg

4. What is the tension in each rope?

Fnet = Fg − FTy1 – FTy2

0 = 750N – (FT1 ∙ sinθ) – (FT2 ∙ sinθ)

0 = 500N – (400N ∙ sin40°) – (FT2 ∙ sin50°) 40° 50°

FT2 = 643N

Fg

FTy1 FTy2

FT1 =400N FT2

75kg

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.cosθFg|| = 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 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?

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°

Ch7 HW#2 6 – 9 is increased until

A 91N block is placed on a 35˚ inclined plane.

Find Fg ┴ and Fg|| .

Fg

Fg ┴ = Fg.cosθFg|| = Fg.sinθ

35°

Ch7 HW#2 6 – 9 is increased until

A 91N block is placed on a 35˚ inclined plane.

Find Fg ┴ and Fg|| .

Fg

Fg ┴ = Fg.cosθFg|| = 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°

8. A 12kg mass is placed on a frictionless inclined planed angled at 32°.

What is its acceleration down the incline?

FN

Fg||

Fg┴

Fg

Fnet = Fg||

m.a = Fg.sinθ

m.a = m.g.sinθ

a = g.sinθ

a = (9.8m/s2).sin32°

a = 5.3 m/s2

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?

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 slips

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

FNFf,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

35°

Ch7.2 cont – More Incline Planes increased until

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°

Ch7.2 cont – More Incline Planes increased until

Ex5) A trunk weighing 562N is on a 30° 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

a = g.sinθ

Fg||

a = (9.8m/s2).sin30° Fg┴

a = 5 m/s2 Fg

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

vi = 0m/svf = ? t = 4sa = _____

30°

Ch7.2 cont – More Incline Planes increased until

Ex5) A trunk weighing 562N is on a 30° 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

a = g.sinθ

Fg||

a = (9.8m/s2).sin30° Fg┴

a = 5 m/s2 Fg

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

vi = 0m/svf = ? t = 4sa = 5 m/s2

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 friction

was 0.15, what would be the acceleration of the trunk?

FN

Fnet = Fg|| – Ff,k

Ff,km.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/s2).sin30°

Fg┴ – (.15)∙(9.8m/s2).cos30°

Fg a = 3m/s2

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 constant speed.

- due tomorrow

- Ch7 HW#3 10 – 13 due at beginning of period

Ch7 HW#3 10 – 12 constant speed.

A 6kg block is placed on a 10˚ inclined plane.

Find Fg ┴ and Fg|| .

Fg

Fg ┴ = Fg.cosθFg|| = Fg.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?

FN

Fg||

Fg┴

Fg

Fnet = Fg||

m.a = Fg.sinθ

a =

32°

12. A trunk weighing 823N is on a 40° incline plane. angled at 32°.

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. angled at 32°.

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

b. How fast will it be going after 5 sec?

vi = 0m/svf = ? t = 5sa = 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?

FNFf,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 constant speed.

An object shot at an angle

has two motions that are

independent of each other.

vi

An object shot at an angle constant speed.

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

An object shot at an angle constant speed.

has two motions that are

independent of each other.

t = 1s t = 2s t = 3s t = 4s

v1= vi v2= viv3= viv4= vi

vi

Vertical motion controlled by gravity

Horizontal motion controlled

by the initial velocity

given to the projectile.

An object shot at an angle constant speed.

has two motions that are

independent of each other.

t = 1s t = 2s t = 3s t = 4s

v1= vi v2= viv3= viv4= vi

vi

t = 1 s

t = 2 s

t = 3 s

t = 4 s

Vertical motion controlled by gravity

Horizontal motion controlled

by the initial velocity

given to the projectile.

An object shot at an angle constant speed.

has two motions that are

independent of each other.

t = 1s t = 2s t = 3s t = 4s

v1= vi v2= viv3= viv4= vi

vi

t = 1 s

t = 2 s

t = 3 s

t = 4 s

Vertical motion controlled by gravity

Horizontal motion controlled

by the initial velocity

given to the projectile.

An object shot at an angle constant speed.

has two motions that are

independent of each other.

t = 1s t = 2s t = 3s t = 4s

v1= vi v2= viv3= viv4= vi

vi

t = 1 s

t = 2 s

t = 3 s

t = 4 s

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

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= 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?

a. The stone hits the ground at the same time, whether it is thrown horizontally

or dropped. So solve for the dropped stone!

vi = 15 m/s

dy= 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!

vi = 15 m/s

dy= 50 m

dy= vit+ ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

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.

vi = 15 m/s

dy= 50 m

dy= vit+ ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

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.

vi = 15 m/s

dx= vit + ½at2

= vcont + 0

= 15 m/s(2.9 s)

= 43.5m

dy= 50 m

dy= vit+ ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

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.

vi = 15 m/s

dx= vit + ½at2

= vcont + 0

= 15 m/s(2.9 s)

= 43.5m

dy= 50 m

c. Find both components

of final velocity, and pythag!

dy= vit+ ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

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.

vi = 15 m/s

dx= vit + ½at2

= vcont + 0

= 15 m/s(2.9 s)

= 43.5m

dy= 50 m

c. Find both components

of final velocity, and pythag!

vfx

vfx = 15 m/s

vfy = viy + at

= 0 + (9.8)(2.9)

= 28.4 m/s

dy= vit+ ½at2

44 = 0 + ½(9.8)t2

t = 2.9 sec

vfy

vf

Ch7 HW#4 14 – 16

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

vi = 5 m/s

dy= 78.4 m

vfx

vfy

vf

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

dy= vit+ ½at2

78.4 = 0 + ½(9.8)t2

t = 3.96 sec

vi = 5 m/s

dy= 78.4 m

vfx

vfy

vf

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

dy= vit+ ½at2

78.4 = 0 + ½(9.8)t2

t = 3.96 sec

vi = 5 m/s

dx= vit + ½at2

= vcont + 0

= 5 m/s(3.96 s)

= 20m

dy= 78.4 m

vfx

vfy

vf

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

dy= vit+ ½at2

78.4 = 0 + ½(9.8)t2

t = 3.96 sec

vi = 5 m/s

dx= vit + ½at2

= vcont + 0

= 5 m/s(3.96 s)

= 20m

vfx = 5 m/s

vfy = viy + at

= 0 + (9.8)(3.96)

= 39 m/s

dy= 78.4 m

vfx

vfy

vf

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

vi = 10 m/s

dy= 156.8 m

vfx

vfy

vf

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

dy= vit+ ½at2

156.8 = 0 + ½(9.8)t2

t = 5.6 sec

vi = 10 m/s

dy= 156.8 m

vfx

vfy

vf

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

dy= vit+ ½at2

156.8 = 0 + ½(9.8)t2

t = 5.6 sec

vi = 10 m/s

dx= vit + ½at2

= vcont + 0

= 10 m/s(5.6s)

= 56m

dy= 156.8 m

vfx

vfy

vf

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

dy= vit+ ½at2

156.8 = 0 + ½(9.8)t2

t = 5.6 sec

vi = 10 m/s

dx= vit + ½at2

= vcont + 0

= 10 m/s(5.6s)

= 56m

vfx = 10 m/s

vfy = viy + at

= 0 + (9.8)(5.6)

= 55.5 m/s

dy= 156.8 m

vfx

vfy

vf

Ch7 HW#4 14 – 16 of a cliff 44m high.

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?

vi = ?

dy= 0.950m

dx= 0.352m

Ch7 HW#4 14 – 16 of a cliff 44m high.

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

vi = ?

dy= 0.950m

dx= 0.352m

Ch7 HW#4 14 – 16 of a cliff 44m high.

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

vi = ?

dx= vit + ½at2

dx= vcont + 0

0.352m = v.(.44s)

v = 0.8m/s

dy= 0.950m

dx= 0.352m

Ch7 Mid Chapter Review of a cliff 44m high.

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. FN = Fg ┴ c. Fnet = Fg|| – Ff,s

Fg

15°

Ch7 Mid Chapter Review of a cliff 44m high.

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.

FN

b. FN = Fg ┴ c. Fnet = Fg|| – Ff,s

= Fg.cosθFf,k

= 54N.cos15°

= 52N

Fg||

Fg┴

Fg

15°

Ch7 Mid Chapter Review of a cliff 44m high.

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.

FN

b. FN = Fg ┴ c. Fnet = Fg|| – Ff,s

= Fg.cosθFf,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

Fnet = Fg|| – Ff,s

m.a = 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/s2).sin60° – (.45)∙(9.8m/s2).cos60°

a = 6.3 m/s2

60°

(Save for HW) coefficient

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.

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.cosθFg┴

a = g.sinθ – µk∙g.cosθFg

a = (9.8m/s2).sin30° – (.14)∙(9.8m/s2).cos30°

a =

b. How fast will it be going after 5 sec?

vi = 0m/svf = ? t = 4sa = 3.7 m/s2

vf = vi + a.t

30°

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

FTy

FTy

FT

FT

Fnet = Fg − 2FTy

0 = 625N – 2(FT ∙ sinθ)

0 = 625N – 2(FT ∙ sin35˚)

FT = 545N

35˚

35˚

Fg

(Save for HW) the 625N bag.

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

60° 60°

Fg

FTFT

100kg

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

FTFT

FTyFTy

100kg

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

vi = 18 m/s

dx= vit + ½at2

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

52 = 0 + ½(9.8)t2

t = 3.26 sec

vi = 18 m/s

dx= vit + ½at2

= vcont + 0

= 18m/s(3.26s)

= 58.6m

vfx = 18 m/s

vfy = viy + at

= 0 + (9.8)(3.26)

= 31.9m/s

dy= 52 m

vfx

vfy

vf

(Save for HW) horizontally at 18m/s from the top of a cliff 52m high.

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

vi = ?

dx= vcont

dy= 1.00m

dx= 1.20m

Lab7.3 – Projectile Motion horizontally at 18m/s from the top of a cliff 52m high.

- due tomorrow

- Ch7 Mid Chapter Review due at beginning of period

Ch7.3 cont – Projectiles Launched at an Angle horizontally at 18m/s from the top of a cliff 52m high.

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 horizontally at 18m/s from the top of a cliff 52m high.

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 vi components

4.47m/s

Ch7.3 cont – Projectiles Launched at an Angle horizontally at 18m/s from the top of a cliff 52m high.

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 vi components

vix = vi.cosθviy = 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

4.47m/s

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 vi components Step3: Find time to the top,

vix = vi.cosθviy = 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/sdy = ?

vfy2 = viy2+ 2ady

02 = 4.12 + 2(-9.8)dy

a = –9.8m/s2

dy = 0.86m

viy = 4.1m/s

4.47m/s

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 vi components Step3: Find time to the top,

vix = vi.cosθviy = vi.sinθ then double it.

=(4.47m/s)cos66° =(4.47m/s)sin66° vfy = 0m/sttop = ?

= 1.8m/s = 4.1m/s

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

vfy = 0m/sdy = ? vf = vi + a.t

vfy2 = viy2+ 2ady 0 = 4.1 + (-9.8)t

02 = 4.12 + 2(-9.8)dyviy = 4.1m/s ttop = 0.42s

a = –9.8m/s2ttotal = 0.84s

dy = 0.86mStep 4: Range uses vx and ttotal.

viy = 4.1m/s

4.47m/s

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 vi components Step3: Find time to the top,

vix = vi.cosθviy = vi.sinθ then double it.

=(4.47m/s)cos66° =(4.47m/s)sin66° vfy = 0m/sttop = ?

= 1.8m/s = 4.1m/s

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

vfy = 0m/sdy = ? vf = vi + a.t

vfy2 = viy2+ 2ady 0 = 4.1 + (-9.8)t

02 = 4.12 + 2(-9.8)dyviy = 4.1m/s ttop = 0.42s

a = –9.8m/s2ttotal = 0.84s

dy = 0.86mStep 4: Range uses vx and ttotal.

dx = vx.ttotal

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

4.47m/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 timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

viy = vi.sinθ

27m/s

30°

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 timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(27m/s)cos30°

= 23m/s

viy = vi.sinθ

=(27m/s)sin30°

= 13.5m/s

27m/s

30°

Step 2: Hang Time:

vfy = 0m/sttop = ?

vf = vi + a.t

a = –9.8m/s2

viy = 4.1m/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 timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(27m/s)cos30°

= 23m/s

viy = vi.sinθ

=(27m/s)sin30°

= 13.5m/s

27m/s

30°

Step3 (part B): Range

dx = vx.ttotal

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ?

vf = vi + a.t

a = –9.8m/s2 0 = 13.5 + (-9.8)t

ttop = 1.37s

viy = 13.5m/sttotal = 2.75s

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 timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(27m/s)cos30°

= 23m/s

viy = vi.sinθ

=(27m/s)sin30°

= 13.5m/s

27m/s

30°

Step3 (part B): Range

dx = vx.ttotal

= (23m/s)(2.75s)

= 63.8m

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 13.5 + (-9.8)t

ttop = 1.37s

viy = 13.5m/sttotal = 2.75s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

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 timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(27m/s)cos30°

= 23m/s

viy = vi.sinθ

=(27m/s)sin30°

= 13.5m/s

27m/s

30°

Step3 (part B): Range

dx = vx.ttotal

= (23m/s)(2.75s)

= 63.8m

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 13.5 + (-9.8)t

ttop = 1.37s

viy = 13.5m/sttotal = 2.75s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

02 = 13.52 + 2(-9.8)dy

dy = 9.3m

HW#18) A player kicks a football from ground level with an initial velocity of 27m/s,

at 60° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

27m/s

60°

Step3 (part B): Range

Step 2 (part A): Hang Time:

vfy = 0m/s

a = –9.8m/s2

viy = 23m/s

Step4 (C): Max Height

HW#18) A player kicks a football from ground level with an initial velocity of 27m/s,

at 60° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

viy = vi.sinθ

27m/s

60°

Step3 (part B): Range

dx = vx.ttotal

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2

viy = 23m/s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

HW#18) A player kicks a football from ground level with an initial velocity of 27m/s,

at 60° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(27m/s)cos60°

= 13.5m/s

viy = vi.sinθ

=(27m/s)sin60°

= 23m/s

27m/s

60°

Step3 (part B): Range

dx = vx.ttotal

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2

viy = 23m/s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

HW#18) A player kicks a football from ground level with an initial velocity of 27m/s,

at 60° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(27m/s)cos60°

= 13.5m/s

viy = vi.sinθ

=(27m/s)sin60°

= 23m/s

27m/s

60°

Step3 (part B): Range

dx = vx.ttotal

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 23 + (-9.8)t

ttop = 2.4s

viy = 23m/sttotal = 4.8s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

HW#18) A player kicks a football from ground level with an initial velocity of 27m/s,

at 60° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(27m/s)cos60°

= 13.5m/s

viy = vi.sinθ

=(27m/s)sin60°

= 23m/s

27m/s

60°

Step3 (part B): Range

dx = vx.ttotal

= (13.5m/s)(4.8s)

= 64.8m

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 23 + (-9.8)t

ttop = 2.4s

viy = 23m/sttotal = 4.8s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

HW#18) A player kicks a football from ground level with an initial velocity of 27m/s,

at 60° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(27m/s)cos60°

= 13.5m/s

viy = vi.sinθ

=(27m/s)sin60°

= 23m/s

27m/s

60°

Step3 (part B): Range

dx = vx.ttotal

= (13.5m/s)(4.8s)

= 64.8m

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 23 + (-9.8)t

ttop = 2.4s

viy = 23m/sttotal = 4.8s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

02 = 232 + 2(-9.8)dy

dy = 27.9m

Ch7 HW#5 17 – 20

Lab7.4B – Range of a Projectile initial velocity of 27

- due tomorrow

- go over Ch7 HW#5 19,20 @ beginning of period

Ch7 HW#5 17 – 20 initial velocity of 27

19. A player kicks a football from ground level with an initial velocity of 20m/s,

at 45° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

viy = vi.sinθ

27m/s

45°

Step3 (part B): Range

dx = vx.ttotal

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2

viy = 14m/s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

Ch7 HW#5 17 – 20 initial velocity of 27

19. A player kicks a football from ground level with an initial velocity of 20m/s,

at 45° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(20m/s)cos45° = 14m/s

viy = vi.sinθ

=(20m/s)sin45° = 14m/s

27m/s

45°

Step3 (part B): Range

dx = vx.ttotal

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2

viy = 14m/s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

Ch7 HW#5 17 – 20 initial velocity of 27

19. A player kicks a football from ground level with an initial velocity of 20m/s,

at 45° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(20m/s)cos45° = 14m/s

viy = vi.sinθ

=(20m/s)sin45° = 14m/s

27m/s

45°

Step3 (part B): Range

dx = vx.ttotal

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 14 + (-9.8)t

ttop = 1.44s

viy = 14m/sttotal = 2.9s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

Ch7 HW#5 17 – 20 initial velocity of 27

19. A player kicks a football from ground level with an initial velocity of 20m/s,

at 45° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(20m/s)cos45° = 14m/s

viy = vi.sinθ

=(20m/s)sin45° = 14m/s

27m/s

45°

Step3 (part B): Range

dx = vx.ttotal

= (14m/s)(2.9s)

= 41m

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 14 + (-9.8)t

ttop = 1.44s

viy = 14m/sttotal = 2.9s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

Ch7 HW#5 17 – 20 initial velocity of 27

19. A player kicks a football from ground level with an initial velocity of 20m/s,

at 45° above the horizontal. Find:

a. Hang timeb. Range c. Max Height

Step 1: components:

vix = vi.cosθ

=(20m/s)cos45° = 14m/s

viy = vi.sinθ

=(20m/s)sin45° = 14m/s

27m/s

45°

Step3 (part B): Range

dx = vx.ttotal

= (14m/s)(2.9s)

= 41m

Step 2 (part A): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 14 + (-9.8)t

ttop = 1.44s

viy = 14m/sttotal = 2.9s

Step4 (C): Max Height dy = ?

vfy2 = viy2+ 2ady

02 = 142 + 2(-9.8)dy

dy = 10m

20. An initial velocity of 27archershoots an arrow at 10° with an initial velocity of 55m/s,

Find: a. Range b. Hang time

Step 1: components:

vix = vi.cosθ

viy = vi.sinθ

55m/s

10°

Step3 (part A): Range

dx = vx.ttotal

Step 2 (part B): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2

viy = 9.6m/s

20. An initial velocity of 27archershoots an arrow at 10° with an initial velocity of 55m/s,

Find: a. Range b. Hang time

Step 1: components:

vix = vi.cosθ

=(55m/s)cos10° = 54m/s

viy = vi.sinθ

=(55m/s)sin10° = 9.6m/s

55m/s

10°

Step3 (part A): Range

dx = vx.ttotal

Step 2 (part B): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2

viy = 9.6m/s

20. An initial velocity of 27archershoots an arrow at 10° with an initial velocity of 55m/s,

Find: a. Range b. Hang time

Step 1: components:

vix = vi.cosθ

=(55m/s)cos10° = 54m/s

viy = vi.sinθ

=(55m/s)sin10° = 9.6m/s

55m/s

10°

Step3 (part A): Range

dx = vx.ttotal

Step 2 (part B): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 9.6 + (-9.8)t

ttop = 0.97s

viy = 9.6m/sttotal = 1.95s

20. An initial velocity of 27archershoots an arrow at 10° with an initial velocity of 55m/s,

Find: a. Range b. Hang time

Step 1: components:

vix = vi.cosθ

=(55m/s)cos10° = 54m/s

viy = vi.sinθ

=(55m/s)sin10° = 9.6m/s

55m/s

10°

Step3 (part A): Range

dx = vx.ttotal

= (54m/s)(1.95s)

= 106m

Step 2 (part B): Hang Time:

vfy = 0m/sttop = ? dy = ?

vf = vi + a.t

a = –9.8m/s2 0 = 9.6 + (-9.8)t

ttop = 0.97s

viy = 9.6m/sttotal = 1.95s

Ch7.5 – Circular Motion initial velocity of 27

vi = 0vf = 40m/s

a = ? t = 4s

vf = vi + a.ta = 10m/s2

If the car continues at a constant speed

of 25m/s as it rounds a corner, is it accelerating?

Ch7.5 – Circular Motion initial velocity of 27

vi = 0vf = 25m/s

a = ? t = 2s

vf = vi + a.ta = 15m/s2

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.

Ch7.5 – Circular Motion initial velocity of 27

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?

Ch7.5 – Circular Motion initial velocity of 27

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?

If an object is accelerating, then there must be a net force acting on it.

(Newton’s 2nd Law, right?)

F = m.a

If an object is accelerating, then there must be a net force acting on it.

(Newton’s 2nd 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 2nd 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

Fc = 2π(.93)/1.18

= 5m/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 4.96m/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 4.96m/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 4.96m/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 4.96m/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 The stopper is m/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.

HW#23) Racing on a flat track, a 1500kg car going 32 The stopper is m/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.

HW#23) Racing on a flat track, a 1500kg car going 32 The stopper is m/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. FN

c. Fc

Fg

HW#23) Racing on a flat track, a 1500kg car going 32 The stopper is m/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. FN

c. Fc = Ff

27,428N = µ.FNFc

27,428N = µ.Fg

27,428N = µ.15,000N Fg

µ = 1.80

Ch7 HW#6 21 – 25

Ch7 HW#6 21 – 25 The stopper is

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 1026 The stopper is m/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 465 The stopper is m/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 465 The stopper is m/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.

25. If the earth spins at 465 The stopper is m/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. Fg = m.g = (97kg)(9.8m/s2) = 951N

c.

25. If the earth spins at 465 The stopper is m/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. Fg = m.g = (97kg)(9.8m/s2) = 951N

c. Fnet = 951N – 3.3N = 948N

Ch7.6 – Torque The stopper is

Ch7.6 – Torque The stopper is

Fout

Fin

rin

Fin

Torque = F . r

Energy is always conserved.

Workin = Workout

Fin.din = Fout.dout

F1.r1 = F2.r2

rin

rout

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.

Fout

Fin = 50N

rin = 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

rBrJ

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

rCrH

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 on

and sits 4.5m out to make it balance.

What is the mass of the backpack?

FgA.rA = FgB.rB

FgA.rA = (FgB+FgBP).rB

rArB

FB

FA

Ch7 HW#7 26 – 29

Ch7 HW#7 26 – 29 fulcrum of

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?

FgA.rA = FgB.rB

FgA.rA = FgB.rB

rArB

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 1.75m

to make it balance?

FgA.rA = FgB.rB

rArB

FB

FA

Ch7 HW#7 26 – 29 the fulcrum of

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?

FgA.rA = FgB.rB

FgA.rA = (FgB+FgBP).rB

rArB m =

FB

FA

Ch7 HW#7 26 – 29 the fulcrum of

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?

FgA.rA = FgB.rB

FgA.rA = (FgB+FgBP).rB

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

FgBP =

rArB

FB

FA

29. A person applies a the fulcrum of 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?

Fout

Fin

rin

rout

29. A person applies a the fulcrum of 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?

Fout Fin.din = Fout.dout

(100N)(70cm) = Fout.(10cm)

Fin

rin

rout

Ch7 Test Review the fulcrum of

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

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

FT

FT

35˚

35˚

Fg

Ch7 Test Review the fulcrum of

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

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

FT

FTy

FTy

FT

Fnet = Fg − 2FTy

0 = 540N – 2(FT ∙ sinθ)

0 = 540N – 2(FT ∙ sin35˚)

FT = 470N

35˚

35˚

Fg

2. A 1120N block is sliding down a 30° incline plane. the fulcrum of

a. Label all forces.

b. Find Fg|| and Fg┴

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. the fulcrum of

a. Label all forces.

b. Find Fg|| and Fg┴

c. If the coefficient of friction is 0.30, find the friction force

d. Find the accl down the incline. FN

b. Fg|| Ff

Fg┴

Fg||

Fg┴

Fg

30°

2. A 1120N block is sliding down a 30° incline plane. the fulcrum of

a. Label all forces.

b. Find Fg|| and Fg┴

c. If the coefficient of friction is 0.30, find the friction force

d. Find the accl down the incline. FN

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

Fg┴ = m.g.cosθ = 1120N.cos30° = 969N

c.

Fg||

Fg┴

Fg

30°

2. A 1120N block is sliding down a 30° incline plane. the fulcrum of

a. Label all forces.

b. Find Fg|| and Fg┴

c. If the coefficient of friction is 0.30, find the friction force

d. Find the accl down the incline. FN

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

Fg┴ = m.g.cosθ = 1120N.cos30° = 969N

c. Ff,k = µk∙FN

= (.30).Fg┴ Fg||

= (.30)(969N) = 291N Fg┴

Fg

30°

2. A 1120N block is sliding down a 30° incline plane. the fulcrum of

a. Label all forces.

b. Find Fg|| and Fg┴

c. If the coefficient of friction is 0.30, find the friction force

d. Find the accl down the incline. FN

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

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/s2

30°

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?

vi = 18 m/s

dy= 52 m

dx = vcont

dy= vit+ ½at2

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?

dy= vit+ ½at2

52 = 0 + ½(9.8)t2

t = 3.26 sec

vi = 18 m/s

dx= vit + ½at2

= vcont + 0

= 18m/s(3.26s)

= 58.6m

dy= 52 m

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 componenntsb. Hang timec. Range

a. Components:

vix = vi.cosθ

viy = vi.sinθ

25m/s

45°

c. Range

dx = vx.ttotal

b. Hang Time:

vfy = 0m/sttop = ?

vf = vi + a.t

a = –9.8m/s2

viy = 17.7m/s

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 componenntsb. Hang timec. Range

a. Components:

vix = vi.cosθ

=(25m/s)cos45°

= 17.7m/s

viy = vi.sinθ

=(25m/s)sin45°

= 17.7m/s

25m/s

45°

c. Range

dx = vx.ttotal

b. Hang Time:

vfy = 0m/sttop = ?

vf = vi + a.t

a = –9.8m/s2

viy = 17.7m/s

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 componenntsb. Hang timec. Range

a. Components:

vix = vi.cosθ

=(25m/s)cos45°

= 17.7m/s

viy = vi.sinθ

=(25m/s)sin45°

= 17.7m/s

25m/s

45°

c. Range

dx = vx.ttotal

b. Hang Time:

vfy = 0m/sttop = ?

vf = vi + a.t

a = –9.8m/s2 0 = 17.7 + (-9.8)t

ttop = 1.8s

viy = 17.7m/sttotal = 3.6s

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 componenntsb. Hang timec. Range

a. Components:

vix = vi.cosθ

=(25m/s)cos45°

= 17.7m/s

viy = vi.sinθ

=(25m/s)sin45°

= 17.7m/s

25m/s

45°

c. Range

dx = vx.ttotal

= (17.7m/s)(3.6s)

= 63.7m

b. Hang Time:

vfy = 0m/sttop = ?

vf = vi + a.t

a = –9.8m/s2 0 = 17.7 + (-9.8)t

ttop = 1.8s

viy = 17.7m/sttotal = 3.6s

5) Racing on a flat track, a 1500kg car going 20 initial velocity of 25m/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.

5) Racing on a flat track, a 1500kg car going 20 initial velocity of 25m/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.

5) Racing on a flat track, a 1500kg car going 20 initial velocity of 25m/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.

5) Racing on a flat track, a 1500kg car going 20 initial velocity of 25m/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. Tension

6. A 50kg kid sits 3.5m from the fulcrum of a seesaw. initial velocity of 25

Where does his 75kg bro need to sit to balance?

F1.r1 = F2.r2

r1r2

F2

F1

6. A 50kg kid sits 3.5m from the fulcrum of a seesaw. initial velocity of 25

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

r1r2

F2

F1