1 chapter 4: forces and newton’s laws of motion forces newton’s three laws of motion the...
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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.
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Force Concept
Contact Forces
Ex: car on road, ball bounce
Non-Contact
Ex: magnetism, gravity
/
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units
• Force units (SI): newton, N
• 1N ≈ ¼ lb.
• 1N = (1kg)(1m/s/s)
• N/kg = m/s/s
s
sm
kg
N /
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Inertia
• is ‘resistance’ to change in velocity
• Measurement: Mass
• SI Unit: Kilogram (Kg)
• /
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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
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g vs G
• G is universal
• g ~ Mass and Radius
• /
m
Fg g
2r
MG
mrMm
G 2
mgFw g
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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.
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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
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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.
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Force Label Notation
• F = general force
• FN = normal force
• f = frictional force
• w = mg = Fg = weight
• T = tension force
• /
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Net Force = change of motion
yxyxamamFF
vector sum of all forces acting on an object
amFnet
xxamF
yyamF
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constant velocity
Force Diagram
Fnet = 0
a = 0
Example: Net Force = 0, Ball rolls along a smooth level surface
table force
weight force
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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
• /
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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
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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
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Example of a Force Diagram for a Sled
net force equals the mass times its acceleration.
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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.
• /
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Block on Frictionless Incline
• a = wx/m =mgsin/m
• = gsin.
• Fn = wy.
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Newton’s 3rd Law of Motion
• equal-sized oppositely-directed forces
• Independent of mass
• Pair-notation
x x
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Newton’s 3rd Law Pair Notation
• use “x” marks on forces that are 3rd Law pairs.
• Use “xx” for a different interaction, etc.
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Force Diagram each object. Which has greater acceleration when
released?SpringForce
SpringForce
x x
Acceleration= F/m
Acceleration= F/(2m)
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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
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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(
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Applications
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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
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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
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The magnitude of the acceleration is:
ssmkg
N
m
Fa
net//49.1
3
47.4
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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
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Two 1kg Blocks; F = 10N
• a = F/(2m) = 10N/2kg = 5 m/s/s
• T = ma = (1kg)(5m/s/s) = 5N
• /
F
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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)
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3090
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Mg, 300 deg.
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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
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Fnet
acceleration
Ex: Newton’s 2nd Law
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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
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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:
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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)
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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
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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.
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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
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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
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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
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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.
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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: