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Forces & Newton’s Laws of Motion Chapter 4 (angles)
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Transcript of Forces & Newton’s Laws of Motion Chapter 4 (angles)
![Page 1: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/1.jpg)
Forces & Newton’s Laws of Motion
Chapter 4(angles)
![Page 2: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/2.jpg)
Sign Problems
Equilibrium-the state where all forces are balanced and acceleration is zero.
Includes both stationary and constant velocity cases.
![Page 3: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/3.jpg)
Example #15 If the sign has a mass of
15kg, what is the tension in the string? What is the force the beam exerts on the sign?
Assume equilibrium. One component at a time.
![Page 4: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/4.jpg)
Set up equilibrium case
0
0
vertical
horizontal
F
F
cos
sin
Tx
Ty
FF
FF
Fy
Fx
FT
23
Fg
Fbeam
![Page 5: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/5.jpg)
Equilibrium case for vertical
0sin mgFT
0 gy FF
Nmg
FT 2.376sin
0verticalF
![Page 6: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/6.jpg)
Equilibrium case for horizontal
0horizontalF
0 Beamx FF
0cos BeamT FF
BeamT FF cos
N3.34623cos2.376
![Page 7: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/7.jpg)
Example #16 Find the tension in
each rope.
T1T2
![Page 8: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/8.jpg)
Break up Components
111 sinFFy
222 sinFFy
111 cosFFx
222 cosFFx
F1 F2
Fx1 Fx2
Fy1 Fy2
Fg
![Page 9: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/9.jpg)
Horizontal Pieces
Since the sign isn’t accelerating horizontally, we can use: 021 xx FF
0coscos 2211 FF
2211 coscos FF
0 xF
Fx1 Fx2
Fy1 Fy2
Fg
![Page 10: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/10.jpg)
Vertical Pieces
Since the sign isn’t accelerating vertically, we can use: 021 gyy FFF
0sinsin 2211 mgFF
0 F
Fx1 Fx2
Fy1 Fy2
Fg
![Page 11: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/11.jpg)
Putting the two together
0sinsincos
cos221
1
22
mgF
F
0sinsin 2211 mgFF 2211 coscos FF
1
221 cos
cos
F
F
![Page 12: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/12.jpg)
Putting the two together cont.
mgFF
221
1
22 sinsincos
cos
mgF
21
1
22 sinsin
cos
cos
211
2
2
sinsincoscos
mgF
![Page 13: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/13.jpg)
Putting the two together cont.
NF 383
62sin34sin34cos62cos
8.9472
NF
F 6.214cos
cos
1
221
![Page 14: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/14.jpg)
Example #17
What is the tension in the cables that support the sign?
Physics is
Phun!!
m = 1500 kg
40°
![Page 15: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/15.jpg)
Example #18 – Angles with accelerationA sled of mass 30 kg is pulled with a force of 50 N at a 25 degree angle. IF the coefficient of friction between the runners and the snow is 0.1, what is the acceleration of the sled?
![Page 16: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/16.jpg)
Example #19
A 50 kg boy is pushed on a 20 kg scooter with a force shown below at 20 degrees below the horizontal. What is the coefficient of friction that would keep the scooter from accelerating? FA=200 N
![Page 17: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/17.jpg)
Example #20
A 50 kg girl is pushed on a 20 kg scooter with a force shown below at 20-degrees below the horizontal. What is the coefficient of friction that would keep the scooter accelerating at 1.5 m/sec2?
FA=200 N
![Page 18: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/18.jpg)
Example #21 – Ramp Problem
A crate slides down a 37 degree incline. What is the acceleration of the crate?
![Page 19: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/19.jpg)
Example #22
A crate slides down a 35 degree incline. What is the acceleration of the crate if the coefficient of friction between the crate and the incline is 0.2?
![Page 20: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/20.jpg)
Example #23
A skier is accelerating down a mountain that has a 30-degree incline at 4.0 m/sec2.
A) What is the coefficient of friction?
B) How long does it take him to get to the bottom of the 0.5 km long mountain?
C) How fast is he going by the time he reaches the bottom of the mountain assuming that he goes straight down without turns?
![Page 21: Forces & Newton’s Laws of Motion Chapter 4 (angles)](https://reader036.fdocuments.us/reader036/viewer/2022081421/5697bf771a28abf838c81516/html5/thumbnails/21.jpg)
Example #24
A crate is sliding down an incline at constant speed. If the coefficient of friction is 0.45, at what angle is the incline?
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What must be for the second incline?