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 .
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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
1 kg
Ch7 – More Forces
Fnet = Fg – FT
0 = 10N – FT
FT = 10N
FT
1 kg
Fg
Ex2) What is the tension in each rope?
1 kg
Ex2) What is the tension in each rope?
Fg
FT
FT
Fnet = Fg – 2FT
0 = 10N – 2FT
Ft = 5N
1 kg
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
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
Lab7.1 – Vector Addition
- due in tomorrow
- Ch7 HW#1 due at beginning of period
Ch7 HW#1 1 – 5
What is the tension in each rope?
Fg
FT
FT
20 kg
Ch7 HW#1 1 – 5
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?
45°
Fg
FTyFTy
FTFT
20kg
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?
Fg
FTyFTy
FTFT
50kg
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?
45° 60°
Fg
FTy1 FTy2
FT1 =259N FT2
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?
40° 50°
Fg
FTy1 FTy2
FT1 =400N FT2
75kg
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
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┴
FgFg ┴ = 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.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
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
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
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
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
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
- due tomorrow
- Ch7 HW#3 10 – 13 due at beginning of period
Ch7 HW#3 10 – 12
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.
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
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
An object shot at an angle
has two motions that are
independent of each other.
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
An object shot at an angle
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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)
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.
FT
FT
35˚
35˚
Fg
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)
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)
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
- 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 vi 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 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
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 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
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 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
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 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
- due tomorrow
- go over Ch7 HW#5 19,20 @ beginning of period
Ch7 HW#5 17 – 20
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
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
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
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
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 archershoots 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 archershoots 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 archershoots 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 archershoots 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
vi = 0vf = 25m/s
a = ? t = 2s
Ch7.5 – Circular Motion
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
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
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
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 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.
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.
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. FN
c. Fc
Fg
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. 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
Lab7.5 – Circular Motion
- due tomorrow
- Ch7 HW#6 due at beginning of period
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.
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. Fg = m.g = (97kg)(9.8m/s2) = 951N
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. Fg = m.g = (97kg)(9.8m/s2) = 951N
c. Fnet = 951N – 3.3N = 948N
Ch7.6 – Torque
Ch7.6 – Torque
Fout
Fin
rin
Fin
Torque = F . r
rin
rout
Ch7.6 – Torque
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
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?
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
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
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 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 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
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
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.
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.
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.
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.
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.
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 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.
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.
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.
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. Tension
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
r1r2
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
r1r2
F2
F1