Transcript Chapters 5&6

Chapters 5, 6
Force and Laws of Motion
Newtonian mechanics
Sir Isaac Newton
(1643 – 1727)
• Describes motion and interaction of objects
• Applicable for speeds much slower than the speed
of light
• Applicable on scales much greater than the atomic
scale
• Applicable for inertial reference frames – frames
that don’t accelerate themselves
Force
• What is a force?
• Colloquial understanding of a force – a push or a
pull
• Forces can have different nature
• Forces are vectors
• Several forces can act on a single object at a time –
they will add as vectors
Force superposition
• Forces applied to the same object are adding as
vectors – superposition
• The net force – a vector sum of all the forces applied
to the same object
Newton’s First Law
• If the net force on the body is zero, the body’s
acceleration is zero


Fnet  0  a  0
Newton’s Second Law
• If the net force on the body is not zero, the body’s
acceleration is not zero


Fnet  0  a  0
• Acceleration of the body is directly proportional to
the net force on the body
• The coefficient of proportionality is equal to the
mass (the amount of substance) of the object
 
ma  Fnet

 Fnet
a
m
Newton’s Second Law
• SI unit of force kg*m/s2 = N (Newton)
• Newton’s Second Law can be applied to all the
components separately
• To solve problems with Newton’s Second Law we
need to consider a free-body diagram
• If the system consists of more than one body, only
external forces acting on the system have to be
considered
• Forces acting between the bodies of the system are
internal and are not considered
Chapter 5
Problem 14
Three forces acting on an object are given by F1 = (– 2.00^i + 2.00^j) N, F2 =
(5.00^i – 3.00^j) N, and F3 = (–45.0^i) N. The object experiences an acceleration
of magnitude 3.75 m/s2. (a) What is the direction of the acceleration? (b) What
is the mass of the object?
Newton’s Third Law
• When two bodies interact with each other, they exert
forces on each other
• The forces that interacting bodies exert on each
other, are equal in magnitude and opposite in
direction


F12   F21
Forces of different origins
• Gravitational force
• Normal force
• Tension force
• Frictional force (friction)
• Drag force
• Spring force
Gravity force (a bit of Ch. 13)
• Any two (or more) massive bodies attract each other
• Gravitational force (Newton's law of gravitation)

m1m2
F  G 2 rˆ
r
• Gravitational constant G = 6.67*10 –11 N*m2/kg2 =
6.67*10 –11 m3/(kg*s2) – universal constant
Gravity force at the surface of the Earth

mEarthmCrate ˆ
m1m2
FCrate  G 2 rˆ  G
j
2
r
REarth

 GmEarth 
mCrate ˆj  g mCrate ˆj
FCrate   2
 REarth 
g = 9.8 m/s2
Gravity force at the surface of the Earth
• The apple is attracted by the Earth
• According to the Newton’s Third Law, the Earth
should be attracted by the apple with the force of the
same magnitude

mEarthmApple
m1m2
ˆj
FEarth  G 2 rˆ  G
2
r
REarth

a Earth 
G
mEarthm Apple
2
Earth
R
mEarth
m Apple
 GmEarth  mApple ˆ
ˆj
ˆj  



g

j
 R2
m
mEarth
 Earth  Earth
Weight
• Weight (W) of a body is a force that the body exerts
on a support as a result of gravity pull from the Earth
• Weight at the surface of the Earth: W = mg
• While the mass of a body is a constant, the weight
may change under different circumstances
Tension force
• A weightless cord (string, rope, etc.) attached to the
object can pull the object
• The force of the pull is tension ( T )
• The tension is pointing away from the body
Free-body diagrams
Normal force
• When the body presses against the surface
(support), the surface deforms and pushes on the
body with a normal force (n) that is perpendicular to
the surface
• The nature of the normal force – reaction of the
molecules and atoms to the deformation of material
Normal force
• The normal force is not always equal to the
gravitational force of the object
Free-body diagrams
Free-body diagrams
Chapter 5
Problem 28
Two objects are connected by a light string that passes over a frictionless
pulley as shown. Draw free-body diagrams of both objects. Assuming the
incline is frictionless, m1 = 2.00 kg, m2 = 6.00 kg, and θ = 55.0° find (a) the
accelerations of the objects, (b) the tension in the string.
Frictional force
• Friction ( f ) - resistance to the sliding attempt
• Direction of friction – opposite to the direction of
attempted sliding (along the surface)
• The origin of friction – bonding between the sliding
surfaces (microscopic cold-welding)
Static friction and kinetic friction
• Moving an object: static friction vs. kinetic
Friction coefficient
• Experiments show that friction is related to the
magnitude of the normal force
• Coefficient of static friction μs
f s ,max   s n
• Coefficient of kinetic friction μk
f k  k n
• Values of the friction coefficients depend on the
combination of surfaces in contact and their
conditions (experimentally determined)
Free-body diagrams
Free-body diagrams
Chapter 5
Problem 42
Three objects are connected on a table. The rough table has a coefficient of
kinetic friction of 0.350. The objects have masses of 4.00 kg, 1.00 kg, and 2.00
kg, as shown, and the pulleys are frictionless. Draw a free-body diagram for
each object. (a) Determine the acceleration of each object and their directions.
(b) Determine the tensions in the two cords.
Drag force
• Fluid – a substance that can flow (gases, liquids)
• If there is a relative motion between a fluid and a
body in this fluid, the body experiences a resistance
(drag)
• Drag force (R)
R = ½DρAv2
• D - drag coefficient; ρ – fluid density; A – effective
cross-sectional area of the body (area of a crosssection taken perpendicular to the velocity); v - speed
Terminal velocity
• When objects falls in air, the drag force points
upward (resistance to motion)
• According to the Newton’s Second Law
ma = mg – R = mg – ½DρAv2
• As v grows, a decreases. At some point acceleration
becomes zero, and the speed value riches maximum
value – terminal speed
½DρAvt2 = mg
Terminal velocity
• Solving ½DρAvt2 = mg we obtain
vt 
2mg
DA
vt = 300 km/h
vt = 10 km/h
Drag force proportional to speed
• In dense fluids (liquids) a resistance force can be
proportional to speed


R  bv
• b depends on the property of the medium, and on
the shape and dimensions of the object
• The negative sign indicates that the force is in the
opposite direction to motion
Spring force (a bit of Ch. 7)
• Spring in the relaxed state
• Spring force (restoring force) acts to restore the
relaxed state from a deformed state
Hooke’s law
• For relatively small deformations


Fs  kd
Robert Hooke
(1635 – 1703)
• Spring force is proportional to the deformation and
opposite in direction
• k – spring constant
• Spring force is a variable force
• Hooke’s law can be applied not to springs only, but
to all elastic materials and objects
Centripetal force
• For an object in a uniform circular motion, the
centripetal acceleration is
2
v
ac 
R
• According to the Newton’s Second Law, a force
must cause this acceleration – centripetal force
mv
Fc  mac 
R
2
• A centripetal force accelerates a body by changing
the direction of the body’s velocity without changing
the speed
Centripetal force
• Centripetal forces may have different origins
• Gravitation can be a centripetal force
• Tension can be a centripetal force
• Etc.
Centripetal force
• Centripetal forces may have different origins
• Gravitation can be a centripetal force
• Tension can be a centripetal force
• Etc.
Free-body diagram
Chapter 6
Problem 14
A roller-coaster car has a mass of 500 kg when fully loaded with passengers.
(a) If the vehicle has a speed of 20.0 m/s at point A, what is the force exerted by
the track on the car at this point? (b) What is the in maximum speed the vehicle
can have at point B and still remain on the track?
Answers to the even-numbered problems
Chapter 5
Problem 2
(a) 1/3;
(b) 0.750 m/s2
Answers to the even-numbered problems
Chapter 5
Problem 6
(a) 534 N down;
(b) 54.5 kg
Answers to the even-numbered problems
Chapter 5
Problem 18
(b) 1.03 N;
(c) 0.805 N to the right
Answers to the even-numbered problems
Chapter 5
Problem 36
0.306; 0.245
Answers to the even-numbered problems
Chapter 6
Problem 2
215 N horizontally inward
Answers to the even-numbered problems
Chapter 6
Problem 4
(a) 1.65 km/s;
(b) 6.84 × 103 s
Answers to the even-numbered problems
Chapter 6
Problem 12
(a) 1.33 m/s2;
(b) 1.79 m/s2 forward and 48.0° inward
Answers to the even-numbered problems
Chapter 6
Problem 26
(a) 6.27 m/s2;
(b) 784 N up;
(c) 283 N up
Answers to the even-numbered problems
Chapter 6
Problem 40
8.88 N