1 4 Newton’s Laws Force, net-force, mass & inertia Newton’s Laws of Motion Weight, Contact...
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Transcript of 1 4 Newton’s Laws Force, net-force, mass & inertia Newton’s Laws of Motion Weight, Contact...
1
4 Newton’s Laws
• Force, net-force, mass & inertia
• Newton’s Laws of Motion
• Weight, Contact Forces
• Labeling & Diagramming
• Hk: 37, 49, 53, 57, 59, 61, 65, 67.
66
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
8
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.
9
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
yy
amF
13
velocity
Example: Ball rolls along a smooth level surface
Force Diagram
table force
weight force
xxamF
0
0/0 max
ynyamwFF
wFa ny zero, is Since
15
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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upward (decreasing) velocity
Fnet
acceleration
Ex: Ball rolling up & slowing down (Use PHET Vector Addition for net-force)
17
xy
60cos)270cos(90cos)60cos( FwFFF Nx
wFFwFFF NNy 866.0)270sin(90sin)60sin(
NF
F60w
Ex. m=3kg, F=86N, 60° below horizontal.
18
2/14
360cos86
60cos
sma
a
maFF
x
x
xx
weight) the times3(about
8.103
)8.9(360sin86
060sin
NF
F
wFFF
N
N
Ny
Ex. Continued
20
Ex. Calculate Acceleration of Block on a Frictionless Plane inclined 30°
xx mamgF 30sin
Plane toparallel axis x Choose
mass) ofnt (independe
//9.430sin ssmag x
21
Ex. Calculate Normal Force on a Block on a Plane inclined 30°
)0(30cos
Plane tonormal axisy Choose
mmgFF ny
mass)on depends (value
30cosmgFn
22
Velocity Acceleration Net Force
+ +
– +
+ –
– –
Complete the table below for the sign of the net force. Sketch a motion diagram for each case.
2323
Newton’s 3rd Law of Motion
• equal-sized oppositely-directed forces
• Independent of mass
• Pair-notation
x x
2424
Newton’s 3rd Law Pair Notation
• use “x” marks on forces that are 3rd Law pairs.
• Use “xx” for a different interaction, etc.
2525
Force Diagram each object. Which has greater acceleration when
released?SpringForce
SpringForce
x x
Acceleration= F/m
Acceleration= F/(2m)
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Motion of Ball
Force on Ball Force on Block
Acceleration of BallAcceleration of Block
Newton’s Second and Third Laws in Operation: Ball hits a large block on a smooth level surface.
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Solving Two Body Problems
• Force diagram each object & system (usually with one axis parallel to the acceleration). Use clockwise coordinates for problems with pulleys.
• System has a force-pair that cancels out• Solve simplest diagram first, then use this
information in another diagram• “ma” is not a force• /
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Two Connected Blocks
)1)(100( TFleftx
NT 100
)1)(200( TTFFbothx
)1)(100( TFFrightx
NNTF 200100
NF 2000
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4 Summary
• Zero net-force; constant velocity
• Acceleration = net-force/mass
• All forces are pairs
• Labeling & diagramming
• Solving problems using x, y force template
• Solving two body problems
• /
30
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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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 in?
Its acceleration is parallel to Fnet by Newton’s 2nd Law. So we need to determine the direction of Fnet.
),.(180tan,
,1 IIIIIquadsF
F
xnet
ynet
6.26
4
2tan 1
N
N
We are in Quadrant I since x and y are both +
35
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 x
x //110
10
Example:
37
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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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
41
NF
w F
30
40
NNx FFFFF 5.0766.0120cos)40cos(
wFFwFFF NNy 642.0866.0)40sin(120sin
0 yx aa
05.0766.0 NFF
04.29642.0866.0 FFN
FFN 532.104.29642.0)532.1(866.0 FF
NF
F
FF
0.43
4.29684.0
4.29642.0326.1
NFFN 8.65532.1
xy
42
0
)8.65(5.0)0.43(766.0
?05.0766.0
NFF
Check of Previous Problem:
0
4.29)0.43(642.0)8.65(866.0
?04.29642.0866.0
FFN
xF
yF
43
Q2. 3kg box at rest on frictionless 30° inclined plane. F acts horizontally. Calculate F and Fn.
NF
w F
30
NNx FFFFF 5.030sin
wFwFF NNy 866.030cos
05.0 NFF
0866.0 wFN
NwFN 9.33866.0/4.29866.0/
NFF N 97.165.0
xy
44
3. Three boxes are pushed by force F along a horizontal frictionless surface.
F=26N
3kg5kg
2kg
Force diagram object 1 (left box)
F12, surface reaction force
NF
F
w 3kg
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F21, surface reaction force
NF
w
5kg
F23, surface reaction force
F32, surface reaction force
NF
w
2kg
Diagram object 2:
Diagram object 3:
46
1212 26 FFFFx
Object1: 3kg
NF
w
21F23FObject2: 5kg
2321 FFFx
NF
w
32F
Object3: 2kg
32FFx
Object1+2+3: 3kg+5kg+2kg
)10(2626 32232112 xxx amaFFFFF 2/6.210/26 smax
NF
F
w
12F
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8.7)6.2(326 12 FFx3kg NF 2.1812
)6.2(52.18 232321 FFFFx5kg NF 2.523
)6.2(232 FFx2kg NF 2.532
Summary:
Stimulus=26N Reactions: 18.2N, 5.2N
48
Q3. Recalculate problem3 with order switched to 5kg, 3kg, 2kg.
F=26N3kg
5kg2kg 6.210/26/ mFa xx
)6.2(232 FFx2kg NF 2.532
)6.2(32.5212321 FFFFx3kg NF 1321
49
4. Modified Atwood Machine with frictionless plane
sin11 gmTFCW
TgmFCW 22
TgmgmTFCW 21
21 sin
21
21
21
21sin
mm
gmgm
mm
Fa CW
CW
Let m1 = 1kg, m2 = 2kg, = 30°.
2/9.421
)2(30sin)1(sm
ggaCW
)9.4(2)2(22 TgTgmFCW
NgT 8.98.9)2(
solve for a and T in terms of m1, m2:
sin1 gm
cos1 gm
T T
gm2
50
Q4. Recalculate problem4 with m1 = 6kg m2 = 1kg.
CWCW ammTgmgmTF )(sin 212121
21
21 sin
mm
gmgmaCW
2/8.216
)1(30sin)6(sm
ggaCW
)8.2(1)1(22 TgTgmFCW
NgT 6.12)8.2()1(
Note that T > (m2)g
51
2. Block stays at same place on frictionless wedge.
a) Draw a force diagram for the block with the forces to correct relative scale.
52
b) Use sum of vertical forces to calculate the size of Fn.
c) Use Fn to calculate the size of the acceleration in m/s/s.
53
1. A 0.88 kg block projected up plane. Acceleration is 5.5m/s/s directed down the plane. Sliding friction is present.
Name(s):___________________________________________
a) Draw a force diagram for the block after projection and moving up the plane. Label each force clearly.
54
b) Calculate the kinetic frictional coefficient.
c) The block is projected down the plane. Draw a force diagram for the block after projection and moving down the plane. Label each force clearly.