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Chapter 4: Forces and Newton’s Laws of Motion
• Forces• Newton’s Three Laws of Motion• The Gravitational Force• Contact Forces (normal, friction, tension)• Application of Newton’s Second Law• Apparent Weight• Air Resistance• Fundamental Forces• CQ: 16, 18.• P: 3, 5, 21, 37, 41, 63, 73, 87, 97, 147.
22
Force Concept
Contact Forces
Ex: car on road, ball bounce
Non-Contact
Ex: magnetism, gravity
/
33
units
• Force units (SI): newton, N
• 1N ≈ ¼ lb.
• 1N = (1kg)(1m/s/s)
• N/kg = m/s/s
s
sm
kg
N /
44
Inertia
• is ‘resistance’ to change in velocity
• Measurement: Mass
• SI Unit: Kilogram (Kg)
• /
55
Universal Law of Gravity
• all matter is weakly attracted
• attraction is inverse-square with distance
• G = 6.67x10-11 N·m2/kg2
• Example: Two 100kg persons stand 1.0m apart
221
r
mmGF
NF 72
11 1067.6)1(
)100)(100(1067.6
66
g vs G
• G is universal
• g ~ Mass and Radius
• /
m
Fg g
2r
MG
mrMm
G 2
mgFw g
7
Contact Forces• Surfaces in contact are often under
compression: each surface pushes against the other. The outward push of each object is called the Normal Force.
• If the objects move (even slightly) parallel to their surface the resistance force experienced is called the frictional force.
88
Normal forces are?
1. Always vertically upward.
2. Always vertically downward.
3. Can point in any direction.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
41 42 43 44 45
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Tension & Compression
• Compressed objects push outward away from their center (aka Normal Force).
• Stretched objects pull toward their center. This is called the Tension Force.
1010
Force Label Notation
• F = general force
• FN = normal force
• f = frictional force
• w = mg = Fg = weight
• T = tension force
• /
1111
Net Force = change of motion
yxyxamamFF
vector sum of all forces acting on an object
amFnet
xxamF
yyamF
1212
constant velocity
Force Diagram
Fnet = 0
a = 0
Example: Net Force = 0, Ball rolls along a smooth level surface
table force
weight force
13
Example: Net-force on 0.5kg
• Net-force = 4N: Acceleration = 4N/0.5kg = 8m/s/s
• 5N, Right; 3N Left; Net-force = 2NAcceleration = 2N/0.5kg = 4m/s/s
• Falling; Net-force = mgAcceleration = mg/m = g = 9.8m/s/s
• /
1414
1. An object maintains constant velocity when the Net-Force on it is zero.
3. Forces always occur in pairs equal in size and opposite in direction.
2. An object’s acceleration equals the Net-Force on it divided by its mass.
Newton’s Laws of Motion
1515
Force Diagrams
• Object is drawn as a “point”
• Each force is drawn as a “pulling” vector
• Each force is labeled
• Relevant Angles are shown
• x, y axes are written offset from diagram
• Only forces which act ON the object are shown
NF
w F
30
40
1616
Example of a Force Diagram for a Sled
net force equals the mass times its acceleration.
1717
g’s
• one “g” of acceleration = 9.8m/s/s
• “two g’s” = 19.6m/s/s, etc.
• Example: What is the net force on a 2100kg SUV that is accelerating at 0.75g?
• Net-force = ma = m(0.75g) = 0.75mg = ¾ weight of car.
• /
1818
Block on Frictionless Incline
• a = wx/m =mgsin/m
• = gsin.
• Fn = wy.
1919
Newton’s 3rd Law of Motion
• equal-sized oppositely-directed forces
• Independent of mass
• Pair-notation
x x
2020
Newton’s 3rd Law Pair Notation
• use “x” marks on forces that are 3rd Law pairs.
• Use “xx” for a different interaction, etc.
2121
Force Diagram each object. Which has greater acceleration when
released?SpringForce
SpringForce
x x
Acceleration= F/m
Acceleration= F/(2m)
2222
Friction• Static Friction “sticking force”
• Kinetic Friction “sliding force”
• Coefficients: 0 = min, 1 ~ max
• e.g. teflon around 0.05
• Rubber on concretearound 1.0
2323
Using Coefficients of Friction
• Ex. 10kg block. FN = weight = mg = 98N. Static coef. = 0.50; Kinetic coef. = 0.30.
Nss Ff max, Nkk Ff
N
Nf s49
)98)(50.0(max,
N
Nfk29
)98)(30.0(
24
Applications
2525
A 3kg object sits on a frictionless table. Two horizontal forces act, one is 2N in the y-direction, the other 4N in the x-direction. A top-view diagram will be shown.
Fnet
What is the magnitude of the net-force acting?
4
22
2,
2, )()(|| ynetxnetnet FFF
490cos20cos4, xnetF
290sin20sin4, ynetF
NFnet 47.4)2()4(|| 22
26
What direction does the 3kg mass accelerate?
parallel to Fnet by Newton’s 2nd Law.
),.(180tan,
,1 IIIIIquadsF
F
xnet
ynet
6.26
4
2tan 1
N
N
We are in Quadrant I since x and y are both +
Fnet
4
22
2727
The magnitude of the acceleration is:
ssmkg
N
m
Fa
net//49.1
3
47.4
2828
Two 1kg Blocks; a = 1m/s/s
• Fnet = F = (2m)a = (2kg)(1m/s/s) = 2N
• Fnet = T = ma = (1kg)(1m/s/s) = 1N
• /
F
2929
Two 1kg Blocks; F = 10N
• a = F/(2m) = 10N/2kg = 5 m/s/s
• T = ma = (1kg)(5m/s/s) = 5N
• /
F
3030
4 Summary
• Fnet = ma (Fnet = 0, v = constant)
• forces always occur in pairs of equal size and opposite direction
• various force types (& symbols)
• equilibrium problems (a = 0)
• dynamic problems (a ≠ 0)
31
3090
6030
Mg, 300 deg.
32
Inclined Plane Forces
• Fxnet = FNcos90 + mgcos300 = (0.02)(a)
• = 0 + (0.02)(9.8)(0.5) = (0.02)a
• accel = 4.9 m/s/s
• Fynet = FNsin90 + mgsin300 = (0.02)(0)
• FN + (0.02)(9.8)(-.866) = 0
• FN = 0.17N
3333
Fnet
acceleration
Ex: Newton’s 2nd Law
3434
Coefficients of FrictionEx: Block&Load = 580grams
NkgNkgmgFN 68.5)/8.9)(580.0(
If it takes 2.4N to get it moving and 2.0N to keep it moving
42.068.5
4.2max, N
N
F
f
N
ss
35.068.5
0.2
N
N
F
f
N
ks
3535
1. 3kg box on level frictionless surface. F=86N acts 60° below horizontal.
xy
300cos)270cos(90cos)300cos( FwFFF Nx
wFFwFFF NNy 866.0)270sin(90sin)300sin(
NF
F60w
Example:
3636
xy
0xa
0ya
xx maF
2/14
360cos86
60cos
sma
a
maF
x
x
x
yy maF
NF
F
wFF
N
N
N
8.103
)8.9(360sin86
060sin
1.(cont)
3737
Q1. What are ax and FN if angle is 30?NF
F30w
30cos)90cos(90cos)30cos( FwFFF Nx
wFFwFFF NNy 30sin)90sin()30sin(90sin
2/25
330cos86
30cos
sma
a
maF
x
x
x
NF
F
wFF
N
N
N
4.72
)8.9(330sin86
030sin
3838
Interaction Notation
• Since all forces are ‘pairs’, label as interactions, e.g. 1 on 2, 2 on 1, etc.
• F12 = “force of object 1 on object 2”
• F21 = “force of object 2 on object 1”
• F34 = “force of object 3 on object 4”
• Etc.
3939
Interaction Notation Symbols
• F12 – general force, 1 on 2
• N12 – normal contact force, 1 on 2
• f12 – frictional force, 1 on 2
• W12 – gravitational force, 1 on 2
• T12 – tension force, 1 on 2
• m12 – magnetic force, 1 on 2
• e12 – electrical force, 1 on 2
4040
Gravitational Force
• All masses attract via gravitational force
• Attraction is weak for two small objects
• Ex: Attraction between two bowling balls is so small it is hard to measure.
• Force is proportional to mass product
• Force is inversely proportional to the square of the distance between objects
4141
Example: Net Force = 0. Block on a surface inclined 30° from horizontal. Applied force F acts 40° below horizontal.
NF
w F
30
40
Net Force = 0
velocity = constant
4242
Diagrams with Interaction Notation
• If f21 exists, then f12 also exists, and is opposite in direction to f21.
• f21 and f12 act on different objects.
4343
A 10kg box is being pushed along a horizontal surface by a force of 15N. A frictional force of 5N acts against the motion. We will want to (a) Calculate the net-force acting and (b) calculate the acceleration of the box.
xx maNNNF 10515
0. yy maweightforceNormalF
The net-horizontal force determines its x-acceleration
The y-acceleration is known to be zero because it remains in horizontal motion, thus
The net-force is 10N horizontal (0 vertical)
The x-acceleration is: ssmkg
N
m
Fa xx //1
10
10
Example: