1 4 topics force and net force inertia and 1 st law acceleration and 2 nd law g notation force pairs...
TRANSCRIPT
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4 Topics
• force and net force
• inertia and 1st law
• acceleration and 2nd law
• g notation
• force pairs and 3rd law
• force diagrams
• equilibrium
• friction
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Force Concept
Force = push or pull
Contact Forces – requires touch
Ex: car on road, ball bounce
Non-Contact – does not require touch
Ex: magnetism, gravity
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Net Force
FFFFnet
21
vector sum of all forces acting on an object
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Example Motion Diagram when Fnet = 0
Newton’s First Law: An object maintains an unchanged constant velocity unless or until it is acted on by a non-zero Net Force.
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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 = 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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Newton’s Second Law: When a Net External Force acts on an object with mass m, the resulting acceleration of the object is parallel to the net external force and has magnitude of
m
Fa
Example Motion Diagrams when Fnet ≠ 0
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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?
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Inertia• The ‘resistance’ to a change in velocity
• Ex: accelerating a ping pong ball• Ex: accelerating a train
• Measurement: Mass
• SI Unit: Kilogram (Kg)
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Newton’s Third Law: Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body
attraction
repulsion
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Motion of Ball
Force on Ball Force on Wall
Acceleration of BallAcceleration of Wall
Newton’s Second and Third Laws in Operation: Ball hits a large block on a smooth level surface.
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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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upward (decreasing) velocity
Fnet
acceleration
Ex: Newton’s 2nd Law
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Velocity Acceleration Net Force
+ +
– +
+ –
– –
Complete the table below for the sign of the net force. Sketch a motion diagram for each case.
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Contact Forces
• Normal Force – perpendicular to surface
• Frictional Force – parallel to surface Static (no sliding) Kinetic (sliding)
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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
41 42 43 44 45
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Static Friction• Objects usually “stick” when at rest
• Must be “budged” to get them moving
• “Budging force” = fs,max.
• fs ranges from 0 to fs,max.
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Kinetic Friction• After objects breaks free, friction
decreases
• fk = force that just keeps object moving at a steady speed
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Coefficients of Friction
N
ss F
f max, dimensionless (no units)
N
kk F
f dimensionless (no units)
Ex. tire on dry road, static coeff. ~ 1.0
Ex. tire on dry road, kinetic coeff. ~ 0.8
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4 Summary
• Newton’s Laws of Motion
• force-diagrams and net force
• contact forces: normal, frictional, other
• equilibrium
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Newton’s 2nd Law Examples
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Block on Frictionless Incline
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Two Connected Blocks
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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 +
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What is the magnitude of the acceleration?
ssmkg
N
m
Fa
net//49.1
3
47.4
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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 x
x //110
10
Example:
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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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2) 3kg box at rest on frictionless 30° inclined plane. F acts 40° below horizontal.
NF
w F
30
40
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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
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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
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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
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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:
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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
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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
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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
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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
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2. Block stays at same place on frictionless wedge.
a) Draw a force diagram for the block with the forces to correct relative scale.
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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.
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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.
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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.
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d) Calculate the net force acting down the plane in newtons.
e) Calculate the acceleration of the block in m/s/s.
f) Is the acceleration i) up the plane, or ii) down the plane?