Download - MFMcGrawPHY 1401- Ch 04b - Revised: 6/9/20101 Chapter 4 Forces and Newton’s Laws of Motion
MFMcGraw PHY 1401- Ch 04b - Revised: 6/9/2010 2
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
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The net force is the vector sum of all the forces acting on a body.
321net FFFFF
Net Force
The net force is the resultant of this vector addition.
Bold letters represent vectors. The units of Force are Newtons, or the abbreviation N, which represent the SI units: kg-m/s2
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Free Body DiagramsThe free body diagram (FBD) is a simplified representation of an object, and the forces acting on it. It is called free because the diagram will show the object without its surroundings; i.e. the body is “free” of its environment.
We will consider only the forces acting on our object of interest. The object is depicted as not connected to any other object – it is “free”. Label the forces appropriately. Do not include the forces that this body exerts on any other body.
The best way to explain the free body diagram is to describe the steps required to construct one. Follow the procedure given below.
(1) Isolate the body of interest. Draw a dotted circle around the object that separates our object from its surroundings.
(2) Draw all external force vectors acting on that body.
(3) You may indicate the body’s assumed direction of motion. This does not represent a separate force acting on the body.
(4) Choose a convenient coordinate system.
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Free Body Diagram
T F
w1
N1
x
y
The force directions are as indicated in the diagram. The magnitudes should be in proportion if possible.
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Newton’s First Law of Motion: Inertia and Equilibrium
Newton’s 1st Law (The Law of Inertia):
If no force acts on an object, then the speed and direction of itsmotion do not change.
Inertia is a measure of an object’s resistance to changes in its motion.
It is represented by the inertial mass.
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If the object is at rest, it remains at rest (velocity = 0).
If the object is in motion, it continues to move in a straight line with the same velocity.
No force is required to keep a body in straight line motion when effects such as friction are negligible.
An object is in translational equilibrium if the net force on it is zero and vice versa.
Newton’s First Law of Motion
Translational Equilibrium
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Newton’s Second Law of Motion Net Force, Mass, and Acceleration
Newton’s 2nd Law:
The acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the body’s mass.
Mathematically: aFF
a mm
netnet or
This is the workhorse of mechanics
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An object’s mass is a measure of its inertia. The more mass, the more force is required to obtain a given acceleration.
The net force is just the vector sum of all of the forces acting on the body, often written as F.
Newton’s Second Law of Motion
If a = 0, then F = 0. This body can have:
Velocity = 0 which is called static equilibrium, or
Velocity 0, but constant, which is called dynamic equilibrium.
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Newton’s Third Law of Motion Interaction Pairs
Newton’s 3rd Law:
When 2 bodies interact, the forces on the bodies, due to each other, are always equal in magnitude and opposite in direction.
In other words, forces come in pairs.
Mathematically: .1221 FF
designates the force on object 2 due to object 1.
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Types of Forces
Contact forces: Normal Force & Friction
Tension
Gravitational Force
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Contact Forces
Contact forces: these are forces that arise due
to of an interaction between the atoms in the
surfaces of the bodies in contact.
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Normal force: this force acts in the direction perpendicular to the contact surface.
Normal force of the ramp on the box
N
w
Normal force of the ground on the boxN
w
Normal Forces
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Example: Consider a box on a table.
FBD for box
mgwN
wNFy
that So
0
This just says the magnitude of the normal force equals the magnitude of the weight; they are not Newton’s third law interaction partners.
Apply Newton’s 2nd law
N
w
x
yNormal Forces
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Friction: a contact force parallel to the contact surfaces.
Static friction acts to prevent objects from sliding.
Kinetic friction acts to make sliding objects slow down. Sometimes called Dynamic friction.
Frictional Forces
maxs sf = μ N
d df = μ N
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Tension
An ideal cord has zero mass, does not stretch, and the tension is the same throughout the cord.
This is the force transmitted through a “rope” from one end to the other.
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Example (text problem 4.77): A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is attached to the wall. The rope fastened to the wall makes a right angle with the wall. Neglect the masses of the rope and the pulley.
Find the tension in the rope from which the pulley hangs and the angle .
FBD for the mass M
w
T
x
y
MgwT
wTFy
0Apply Newton’s 2nd Law to the mass M.
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FBD for the pulley:
0sin
0cos
TFF
TFF
y
x
Apply Newton’s 2nd Law:
sincos FFT
This statement is true only when = 45 and
MgTF 2 2
Example continued:
x
y
TF
T
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Gravitational Forces
221
r
MGMF
Gravity is the force between two masses. Gravity is a long-range force. No contact is needed between the bodies. The force of gravity is always attractive!
r is the distance between the two masses M1 and M2 and G = 6.671011 Nm2/kg2.
M2
r
M1F21 F12 .1221 FF
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Let M1 = ME = mass of the Earth. 22M
r
GMF E
Here F = the force the Earth exerts on mass M2. This is the force known as weight, w.
.222gMM
r
GMw
E
E
22
m/s 8.9N/kg 8.9 where E
E
r
GMg
Near the surface of the Earth
km 6400
kg 1098.5
E
24E
r
M
Gravitational Forces
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What is the direction of g?
Note thatm
Fg
is the gravitational force per unit mass. This is called
the gravitational field strength. It is also referred to as
the acceleration due to gravity.
What is the direction of w?
Gravitational Forces
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Example: What is the weight of a 100 kg astronaut on the surface of the Earth (force of the Earth on the astronaut)? How about in low Earth orbit? This is an orbit about 300 km above the surface of the Earth.
On Earth: N 980mgw
In low Earth orbit: N 890)( 2
hR
GMmrmgw
E
Eo
The weight is reduced by about 10%. The astronaut is NOT weightless!
Gravitational Forces
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Applying Newton’s Second Law
aF mThe one equation everyone remembers!
This equation is just the tip of the “iceberg” of the mechanics problem. The student will need to anlyze the forces in the problem and sum the force vector components to build the left hand side of the equation.
Sumof the forces acting on the objects in the system
“m” is the System
Mass
“a” is the System
Response
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Example: A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m1 = 3.00 kg and m2 = 1.00 kg.
Find the tension in the cord connecting the two blocks as shown.
F block 2 block 1
Assume that the rope stays taut so that both blocks have the same acceleration.
Applying Newton’s Second Law
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FBD for block 2:
T F
w1
N1
x
y
x
T
w2
N2
y
FBD for block 1:
011
1
wNF
amTFF
y
x
022
2
wNF
amTF
y
x
Apply Newton’s 2nd Law to each block:
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amTF 1
amT 2
To solve for T, a must be eliminated. Solve for a in (2) and substitute in (1).
(1)
(2)
Example continued:
These two equations contain the unknowns: a and T.
N 5.2
kg 1kg 3
1
N 10
1
1
2
1
2
1
21
211
mm
FT
Tm
mT
m
TmF
m
TmamTF
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Pick Your System Carefully
x
T
w2
N2
y
T F
w1
N1
x
y
Include both objects in the system. Now when you sum the x-components of the forces the tensions cancel. In addition, since there is no friction, y-components do not contribute to the motion.
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Apparent Weight
Stand on a bathroom scale.
FBD for the person: Apply Newton’s 2nd Law:
y
yy
mamgN
mawNF
w
N
x
y
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The normal force is the force the scale exerts on you. By Newton’s 3rd Law this is also the force (magnitude only) you exert on the scale. A scale will read the normal force.
yagmN is what the scale reads.
When ay = 0, N = mg. The scale reads your true weight.
When ay 0, N > mg or N < mg.
Apparent Weight
In free fall ay = -g and N = 0. The person is weightless.
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Example (text problem 4.128):
A woman of mass 51 kg is standing in an elevator. The elevator pushes up on her feet with 408 newtons of force.
What is the acceleration of the elevator?
FBD for woman:
y
yy
mamgN
mawNF
Apply Newton’s 2nd Law: (1)
w
N
x
y
Apparent Weight
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Solving (1) for ay:2m/s 8.1
m
mgNay
The elevator could be (1) traveling upward with decreasing speed, or (2) traveling downward with increasing speed.
The change in velocity is DOWNWARD.
Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s2
Unknown: ay
Example continued:
Apparent Weight
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N
mg
aF m
This is not a methodolgy to solve for the acceleration. It is just graphically demonstrating that the net force is ma
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N
mg
mg
N
Same problem but the applied force is angled up
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mg
N
The normal force, N, is smaller in this case because the upward angled applied force reduces the effective weight of the sled.
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Equilibrium Problem
This is an example of three-vector equilibrium problem. It lends itself to a simple solution because the vector sum of the three vectors closes on itself (equilibrium) and forms a triangle
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Milk Carton
Max static friction force
Non-slip limit on applied force
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Hanging Picture - Free Body Diagram
T1
mg
60o
30o
T2
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Hanging Picture
T 1
m g
6 0 o
3 0 oT 2
01
02
cos(60 )
sin(60 )
T mg
T mg
• Since this turned out to be a right triangle the simple trig functions are that is needed to find a solution.
• If the triangle was not a right triangle then the Law of Sines would have been needed.
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Force SummaryFriction
Opposes motion
Proportional to normal force
Non-conservative
Static & dynamic
Normal Forces
Perpendicular to surface at point of contact.
Magnitude needed to maintain equilibrium
Tension
No mass
No stretching
Pulleys
Massless
No friction (bearing)
Tension in rope continuously changes direction