chapter5-svkastro1.panet.utoledo.edu/~khare/teaching/phys2130h...Unit: 1 N = (1kg) .(1m/s2) = 1...
Transcript of chapter5-svkastro1.panet.utoledo.edu/~khare/teaching/phys2130h...Unit: 1 N = (1kg) .(1m/s2) = 1...
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Chapter 5 Force and Motion
• Force F– is the interaction between objects– is a vector– causes acceleration– Net force: vector sum of all the forces on an
object.
am...FFFFFFF 4321
N
1iinettotal
vvvvvvvv=++++==≡ ∑
=
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Force Examples
• Gravitational• Friction• tension• spring• normal
• momentum change• electrostatic• magnetic• nuclear
• etc.......
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Newton’s first law• Newton’s first law: If no force acts on a body,
then the body’s velocity cannot change, that is, the body can not accelerate– rest, still rest– moving, continue moving with same velocity
• Inertia reference frame is one in which Newton’s laws hold
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Mass• Definition: The mass of an object is a measure of its
“resistance” to being accelerated. • Symbol: m • SI Base unit: kg (by the way, the English unit for mass is the “slug”)
• Scalar quantity• Mass is an intrinsic characteristic of an object, however
its value does change at high speeds (Special Relativity) Has to do with the speed of light being c.
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Newton’s Second Law
• Newton’s second law: The net force on a body is equal to the product of the body’s mass and the acceleration of the body
Σ F = m a
Σ F : vector sum of all the forces that act on that bodyΣ Fx = m ax Σ F y = m ay Σ F z = m az
Unit: 1 N = (1kg) .(1m/s2) = 1 kg.m/s2
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• Example: (problem 5-6 in the text book) Three astronauts, propelled by jet backpacks, push and guide a 120 kg asteroid toward a processing dock, exerting forces shown in Fig. What is the asteroid’s acceleration (a) in unit vector notation and as (b)a magnitude and ( c) a direction
x
y
0
F3
F1
F2
Break problem into x- and y-componentsF1 = (41N cos 60o) i + (-41N sin 60o) jF2 = (55N) i + (0) jF3 = (32N cos 30o) i + (32N sin 30o) j
Fnet = (103.2N) i + (-19.51N) j
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• The gravitational force near the surface of a very large object (i.e., the Earth):
Fg = m g
• Weight (gravitational force)W = Fg = m g
– g varies with location– weight and mass are different, e.g. 7.2 kg ball,
same mass on earth and moon, weight 71 N on earth but 12 N on the moon
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• Suppose you are talking by interplanetary telephone to your friend, who lives on the moon. She tells you that he just won a piece of gold weighing one newton in a contest. Excitedly, you tell her that you entered the Earth version of the same contest and also won a newton of gold! Who is richer?
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• The normal force: FN– When a body presses against a surface, the surface
deforms and pushes on the body with a normal force N that is perpendicular to the surface
– FN does not always equal mg
• Friction: f– the resistance force on a body when the body slides or
attempts to slide along a surface
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• Tension: T– When a cord is attached to a body and pulled taut, the cord
pulls on the body with force T
CP 5-5: The body in fig (c) has a weight of 75 N, Is T equal to, greater than, or less than 75 N when the body is moving upward (a) at constant speed(b) at increasing speed (c) at decreasing speed?
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Newton’s Third Law• Why would I feel pain if I hit the wall
with my fist?• Newton’s third law: When two bodies
interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction
FAB = - FBA
– They do not cancel each other since they act on different bodies
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If a sport car collides head-on with a massive truck, (a) which vehicle experiences the greater force?(b) which vehicle experience the greater
acceleration?
Remember that a modern semi-truck has a mass of 25 cars!
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Assume F = 20 N, surface frictionless. What is the acceleration ?
T1 = m1a = (10kg)(1.0 m/s2) = 10N
How about T1?
F = ( m1 + m2 + m3+ m4)a = (20kg)a = 20N==> a = 20N/20kg = 1.0 m/s2
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General scheme for solving Newton’s law problems- Isolate the objects in the problem- For each object of interest, identify all the
external forces on that object and draw a free-body diagram for this object
- establish a convenient coordinate system for each object and find the component of the forces along those axes.
- Apply Newton’s 2nd law in the x and y directions for each object. ( e.g. ΣFx= max, ΣFy = may)
- Solve the resulting set of equations.
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Consider a traffic light of m = 4kg held by one rope which in turn is supported by two other ropes as shown with angles θ1 = 30o θ2 = 45o , Which of the three ropes has the greater tension?
A Quiz
θ1=30o θ2=45o
1 2
3
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Consider a traffic light of m = 4kg held by one rope which in turn is supported by two other ropes as shown with angles θ1 = 30o θ2 = 45o , Which of the three ropes has the greater tension?
A Quiz
θ1=30o θ2=45o
1 23
1) rope 1 2) rope 2 3) rope 3 4) All ropes have the same tension
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Consider a traffic light of m = 4kg held by one rope which in turn is supported by two other ropes as shown with angles θ1 = 30o θ2 = 45o , Which of the three ropes has the greater tension?
A Quiz
θ1=30o θ2=45o
1 23
1) rope 1 2) rope 2 3) rope 3 4) All ropes have the same tension
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Consider a traffic light of m = 4kg held by one rope which in turn is supported by two other ropes as shown with angles θ1 = 30o θ2 = 45o , Which of the three ropes has the greater tension?
A Quiz
θ1=30o θ2=45o
1 2
30F45sinF30sinFF
045cosF30cosFF
3o
2o
1y
o2
o1x
=−+=
=+−=
3o
2o
1
o2
o1
F45sinF30sinF
45cosF30cosF
=+
=
33
32
31
F00.1FF896.0FF732.0F
===
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Newton’s Laws of Motion
• Newton’s first law: If no force acts on a body, then the body’s velocity cannot change, that is, the body can not accelerate.
• Newton’s second law: The net force on a body is equal to the product of the body’s mass and the acceleration of the body ΣF = m a
• Newton’s third law: When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction
FAB = - FBA
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Sample problem 5-5.M = 3.3 kg, m = 2.1 kg, frictionless surfaceH falls as S accelerate to the right(a) What is the acceleration of S?
Fg = mg
T
Acceleration a links the two masses together
mg - Ma = ma
mg - T = ma
T = Ma
Forces on mT = Ma
unbalanced forces on M
gmM
ma+
=
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Consider m1 > m2. What is the acceleration of either mass if the inclined plane is frictionless?
How do you set up this problem?
Which Coordinate system?
x
yHorizontal - Vertical
--or--
slantedxy
Easiest because of direction of motion
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xyFirst set up coordinate system
Consider m1 > m2. What is the acceleration of either mass if the inclined plane is frictionless?
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m1g m2g
xygravity still points down
Consider m1 > m2. What is the acceleration of either mass if the inclined plane is frictionless?
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m1g
N T
m2g
Txygravity still points down
Normal force is perpendicular to the plane
m1g, m2g
Tension is along the rope
Consider m1 > m2. What is the acceleration of either mass if the inclined plane is frictionless?
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xy N TThus, the free bodydiagram becomes:
m2g
T
θ
( ) ( )jcosgmisingmgm
jNi0N
j0iTT
111 −θ+−θ=
+=
+=
v
v
v
Dissect the forces into components m1g
Consider m1 > m2. What is the acceleration of either mass if the inclined plane is frictionless?
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m1g
Thus, the free bodydiagram becomes:
m2g
T
θ
( ) ( )jcosgmisingmgm
jNi0N
j0iTT
111 −θ+−θ=
+=
+=
v
v
v
Disect the forces into components
Consider m1 > m2. What is the acceleration of either mass if the inclined plane is frictionless?
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xy N TThus, the free bodydiagram becomes:
m2g
T
θ
amgmT:m
0cosgmNF:y
amsingmTF:x
222
1iy
11ix
=+−
=θ−=
=θ−=
∑∑
Now apply Newton’s Second Law
a
m1g
Consider m1 > m2. What is the acceleration of either mass if the inclined plane is frictionless?
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xy N TThus, the free bodydiagram becomes:
m2g
T
θ
( )
( ) ( ) ( )θ=
+θ−=⇒
+θ−=
+=θ−
=θ−−=∑
cosgmN:y
imm
singmgmamm
singmgma
ammsingmgmamsingmamgmF:x
1
21
12
21
12
2112
1122ix
v
Now apply Newton’s Second Law m1g
Consider m1 > m2. What is the acceleration of either mass if the inclined plane is frictionless?
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Consider m1 > m2. Is the tension in the rope greater than, less than, or equal to m2g?
m2g
T
A Quiz
1) less than m2g
4) depends on the acceleration
2) greater than m2g 3) equal to m2g
0) none of the above
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Consider m1 > m2. Is the tension in the rope greater than, less than, or equal to m2g? m2g
T
A Quiz
1) less than m2g
4) depends on the acceleration
2) greater than m2g 3) equal to m2g
0) none of the above
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Consider m1 > m2. Is the tension in the rope greater than, less than, or equal to m2g? m2g
T
A Quiz
1) less than m2g
4) depends on the acceleration
2) greater than m2g 3) equal to m2g
0) none of the above
( )agmTamgmT:m 2222 −=⇒=+−
( ) ( ) ( )imm
singmgmamm
singmgma21
12
21
12
+θ−=⇒
+θ−= v