Transcript T - SeyedAhmad.com

Frictional Forces
Nonequilibriuium Applications
of Newton’s Laws of Motion
Non-equilibrium conditions occur when the object
is accelerating and the forces acting on it are not balanced
so the net force is not zero.
Non-equilibrium: Fx = max and Fy = may
Fx = max and Fy = may
Free Body Diagrams
The 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
y
N1
The force directions are as
indicated in the diagram.
The magnitudes should be
in proportion if possible.
F
T
x
w1
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Tension
This is the force transmitted through a “rope”
from one end to the other.
An ideal cord has zero mass, does not stretch,
and the tension is the same throughout the cord.
MFMcGraw
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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 .
y
T
FBD for the
mass M
x
w
Apply Newton’s 2nd
Law to the mass M.
MFMcGraw
F
PHY 1401- Ch 04b - Revised: 6/9/2010
y
T w 0
T  w  Mg
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Example continued:
Apply Newton’s 2nd Law:
FBD for the pulley:
F
F
y
y
 F sin   T  0

x
T
MFMcGraw
 F cos   T  0
T  F cos  F sin 
F
T
x
This statement is true
only when  = 45 and
F  2 T  2 Mg
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Example
A supertanker of mass m = 1.50 X 108 kg is being towed by two tugboats. The
tensions in the towing cables apply the forces T1 and T2 at equal angles of 30.0o
with respect to the tanker’s axis. In addition, the tanker’s engines produce a
forward drive force D whose magnitude is D = 7.50 X 104 N. Moreover, the
water applies an opposing force R, whose magnitude is R = 4.00 X 104 N. The
tanker moves forward with an acceleration that points along the tanker’s axis and
has a magnitude of 2.00 X10-3 m/s2. Find the magnitudes of T1 and T2.
x component
T1: +T1cos30.0o
T2: +T2cos30.0o
D: +D = 7.50 X104 N
R: -R=-4.00 X104 N
Fx = +T1cos30.0o + T2cos30.0o +D -R = max
T=1.53 X 105 N
y component
T1: +T1sin30.0o
T2: -T2sin30.0o
D: 0
R: 0
Fy = +T1sin30.0o – T2sin30.0o = 0
Example
A flatbed is carrying a crate up a 10.0o hill. the coefficient of
static friction between the truck bed and the crate is s = 0.350.
Find the maximum acceleration that the truck can attain before the
crate begins to slip backward relative to the track.
ax = 1.68 m/s2
Applying Newton’s Second Law
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.
block 2
block 1
F
Assume that the rope stays taut so that both blocks
have the same acceleration.
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FBD for block 2:
FBD for block 1:
y
y
N1
N2
F
T
T
x
x
w1
w2
Apply Newton’s 2nd Law to each block:
F
F
MFMcGraw
x
 T  m2 a
y
 N 2  w2  0
F
F
x
 F  T  m1a
y
 N1  w1  0
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Example continued:
F  T  m1a (1)
These two equations contain
the unknowns: a and T.
T  m2 a (2)
To solve for T, a must be eliminated. Solve for a in (2)
and substitute in (1).
T 
F  T  m1a  m1  
 m2 
T 
 m1 
T
F  m1    T  1 
 m2 
 m2 
T
MFMcGraw
F
10 N

 2.5 N
 m1   3 kg 
1 
 1 

 m2   1 kg 
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Pick Your System Carefully
y
y
N1
N2
T
F
T
x
x
w2
w1
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.
y
FBD for the
person:
N
Apply Newton’s 2nd Law:
x
w
MFMcGraw
F
y
 N  w  ma y
N  mg  ma y
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Apparent Weight
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.
N  mg  a y 
is what the scale reads.
When ay = 0, N = mg. The scale reads your true weight.
When ay  0, N > mg or N < mg.
In free fall ay = -g and N = 0. The person is weightless.
MFMcGraw
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Apparent Weight
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
Apply Newton’s 2nd Law: (1)
N
x
w
MFMcGraw
F
y
 N  w  ma y
N  mg  ma y
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Apparent Weight
Example continued:
Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s2
Unknown: ay
Solving (1) for ay:
N  mg
ay 
 1.8 m/s 2
m
The elevator could be (1) traveling upward with decreasing
speed, or (2) traveling downward with increasing speed.
The change in velocity is DOWNWARD.
MFMcGraw
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Free Body Diagram
N
F
mg
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Hanging Picture
MFMcGraw
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Hanging Picture - Free Body
Diagram
T2
30o
mg
60o
T1
MFMcGraw
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Hanging Picture
T1  mg cos(60 0 )
30o
T2
T2  mg sin(60 0 )
•
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.
mg
60o
T1
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MFMcGraw
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Atwood Machine and
Variations
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Atwood’s Machine
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A 2-Pulley Atwood Machine
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MFMcGraw
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Three Body Problem
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Incline Plane Problems
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Single Incline Plane Problem
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Double Incline Plane Problem
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