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PHYSICS 149: Lecture 7
• Chapter 2
–2
2.8
8 Tension
– 2.9 Fundamental Forces
• Chapter 3
– 3.1 Position and Displacement
Lecture 7
Purdue University, Physics 149
1
ILQ 1
Which statement about frictional forces is not
true?
A) Frictional forces are contact forces parallel to the
contact surface.
surface
B) When frictional forces act to resist motion, two
surfaces slide across each other.
C) Frictional forces are perpendicular to the surface of
contact.
D) Frictional forces always act opposite to the direction
of motion.
y p
proportional
p
to the
E)) The frictional force is always
normal force on an object.
Lecture 7
Purdue University, Physics 149
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Tension
Definition:
Magnitude of Contact Force between different
segments of the string
(or between an end and the object attached there)
Example:
T is the force on the left
portion from the right
portion
p
|T| is the tension at point P
P T
NOTE: T can onlyy p
pull the
other object
Lecture 7
Purdue University, Physics 149
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Tension
•
At any point in the rope (or string
string, cable or chain)
chain), tension is the
pulling force exerted on the rope on one side of the point by the
rope on the other side.
•
At its two ends, tension is the pulling force exerted on the object
attached to its ends by the ropes at the ends.
•
Note that tension can pull but not push.
=T1
•=T4
=T2
=T
T3
Lecture 7
Purdue University, Physics 149
“If” the chain’s weight
is not negligible,
negligible
T1 > T2 > T3 > T4 .
For example,
F
l
T1 = T4 + chain’s weight.
4
Ideal Cord
A id
An
ideall cord
dh
has NO MASS
Consequence:
th tension
the
t
i iis the
th same
at ALL POINTS along
g the cord.
Lecture 7
Purdue University, Physics 149
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Tension with “Ideal Cord”
•
•
•
“Ideal cord”: a cord that has zero mass and thus zero weight
In an ideal cord, (a) the tension has the same value at all points
along the cord, and (b) the tension is equal to the force that the
cord exerts on the objects attached to its ends (as long as there is
no external force on the cord).
Note: In many cases, the weight of a cord is negligibly small
compared to the weight of the objects attached to its ends, and
thus we may assume that it is an ideal cord.
=T1
•=T
T4
=T2
=T3
Lecture 7
Purdue University, Physics 149
“If”
If the chain’s
chain s weight
is negligible (ideal cord),
T1 = T2 = T3 = T4 .
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Ideal Pulley
•
Pulley: A pulley serves to change the direction of a tension force, and
may also (in the case of multiple-pulley systems) change its
magnitude.
•
•
“Ideal pulley”: a pulley that has no mass and no friction.
The tension of an “ideal cord” that runs through an “ideal pulley” is the
same on both sides of the pulley (and at all points along the cord)
cord).
T=
Lecture 7
=T
Purdue University, Physics 149
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ILQ 2
Two blocks with the same mass are connected
by a lightweight cord that runs through an ideal
pulley, as shown. When released, the blocks
will end up
p
A))
B)
C)
D)
Lecture 7
at their
t e current
cu e t heights.
e g ts
at the same height.
with left block on the ground
ground.
with right block on the ground.
Purdue University, Physics 149
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ILQ 3
A heavy ball hangs from a string attached to a
sturdy wooden frame
frame. A second string (same
kind) is attached to the bottom of the ball. You
pull down the lower string
p
g slowlyy and steadily.
y
Which string will break first ?
A) the top one
B)) the bottom one
C) at the same time, because
the tension is the same
D) depends on the weight of the ball
Lecture 7
Purdue University, Physics 149
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Details of ILQ3
Equilibrium Î ΣFy = 0
FBD of ball:
ΣFy = Ttop – Tbottom – W = 0
Î Ttop = Tbottom + W
T2
Thus, Ttop > Tbottom
W
The top one receives
g tension,
stronger
so it will break first.
T1
T1
NOTE: this problem is useful
f CHIP problem
for
bl
with
ith iincline
li
Lecture 7
Purdue University, Physics 149
Therefore
T2 = T1+W>T1
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Example: Tension
• Given conditions:
– “Ideal cord” Æ Tension is same.
– Equilibrium Æ Net force = ΣFi = 0
Lecture 7
Purdue University, Physics 149
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Tension
Determine the tension in the 6 meter rope if it sags 0.12 m in
the center when a gymnast with weight 250 N is standing on it.
∑F = 0
x-direction: ΣF = m a
-TL cosθ + TR cosθ = 0
TL = TR
y
y
y-direction:
ΣF = m a
TL sinθ + TR sinθ - W = 0
2 T sinθ = W
T = W/(2 sinθ) = 3115 Ν
Lecture 7
Purdue University, Physics 149
x
θ
.12 m
3m
0.12
tan θ =
3
θ = 2.3
D
12
Tension
T1
y
y
T2
θ
x
W
Θ = tan-1(0.12/3.00) = 2.291°
tightrope
θ
3.00 m
.12 m
x
T1x = –T1 cosΘ T2x = T2 cosΘ Wx = 0
T1y = T1 sinΘ
T2y = T2 sinΘ Wy = –250 N
x-component: ΣFx = 0
ΣFx= T1x + T2x = –T1cosΘ + T2cosΘ = 0
Î T1 = T2
y-component: ΣFy = 0
ΣFy= T1y + T2y – W = T1sinΘ + T2sinΘ – W = 2⋅T1sinΘ – W = 0
Î T1 = T2 = W / (2⋅sinΘ)
(2 i Θ) = 250 N / [2 ⋅ sin(2.291°)]
i (2 291°)] = 3127.0
3127 0 N
Lecture 7
Purdue University, Physics 149
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Example: A Two-Pulley System
What is the tension of the rope?
– FBD for Pulley L
– Equilibrium
Î ΣFy= Tc + Tc – W = 0
Î Tc = W /2 = 902 N
– Since tension is the same at all
points along the cord C, the
person’s pulling force is equal to
Tc.
– Therefore, the person pulling the
rope only needs to exert a force
equal to half the engine’s weight.
Lecture 7
Purdue University, Physics 149
W=
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Pulley Example
T
How much is T?
T =100 N
Explain why…
Lecture 7
200 N
Purdue University, Physics 149
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ILQ
Whatt can you say about
Wh
b t the
th tensions
t
i
T1 and T2 at the two ends of the cord?
(W is the weight of the cord)
A))
B)
C)
D)
T1
T1 > T2
T2 > T1
T1=T2
1 2
depends
W
NOTE: this is NOT an ideal cord!
Lecture 7
Purdue University, Physics 149
T2
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ILQ
If the weight W=0 then the cord is ideal.
Is it true that T1=T2 ?
A)
B)
C)
D)
no, T1>T2
yes
yes, because of 3rd NL
no, T1<T2
yes, because of 1st NL
NOTE: this IS an ideal cord!
Lecture 7
Purdue University, Physics 149
T1
T2
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Fundamental Forces
• Gravity
–
–
–
–
Acts on p
particles ((and objects)
j
) with mass
Always attractive; recall Newton’s law of universal gravitation
Range: unlimited
The weakest of the four fundamental forces
• Electromagnetism
– Acts on particles with electric charge
– Binds electrons to nuclei to form atoms, and binds atoms in
molecules and solid
– Responsible for contact forces like friction and normal force
– Either attractive or repulsive
– Range: unlimited
– Much stronger than gravity, 2nd strongest of the four fundamental
forces
Lecture 7
Purdue University, Physics 149
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Fundamental Forces
• The Strong Force
– Binds together the protons and neutrons in atomic
nucleus (and also quarks in combinations)
– Very short range: ~10-15 m (about the size of an atomic
nucleus)
– The strongest of the four fundamental forces
• The Weak Force
–R
Responsible
ibl ffor some ttypes off radioactive
di
ti d
decays,
sunlight
– Shortest range:
g ~10-17 m
– 3rd strongest of the four fundamental forces
Lecture 7
Purdue University, Physics 149
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Fundamental Forces
•
•
•
•
Gravity
Strong nuclear force
Weak nuclear force
Electromagnetic force
Lecture 7
Purdue University, Physics 149
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Zero Net Force vs. Nonzero Net Force
• Net Force: the vector sum of all the forces acting
on an object
j
• Zero Net Force (Ch 2)
– When a net force on an object is zero, the velocity
(both direction and magnitude) of the object does not
change.
• Newton
Newton’s
s First Law of Motion
• Nonzero Net Force (from Ch 3)
– When
e a nonzero
o e o net
et force
o ce acts o
on a
an object, tthe
e
velocity of the object changes.
• That is, either the velocity’s direction or magnitude changes, or
both of direction or magnitude change
change.
• Relevant to Newton’s Second Law of Motion
Lecture 7
Purdue University, Physics 149
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Motion in One Dimension
-x
0
+x
The variables are time and distance
t = 0 start of observations at a point x0
t = t end of the observations at a point xf
Objects are in motion and velocity is
(change in distance)/time
V l i can change
Velocity
h
=> acceleration
l
i
(change in velocity)/time
All quantities except time are vectors but the vector “nature”
nature
is contained in whether the quantity is positive or negative
Lecture 7
Purdue University, Physics 149
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Position Vector
• To describe position, we need
– a reference point (origin),
– a distance from the origin, and
– a direction from the origin.
object
at (x,y)
• Position Vector (or Position)
– A vector
t quantity
tit that
th t consists
i t off the
th distance
di t
and
d
direction
– An arrow starting at the “origin”
origin and ending with the
arrowhead on the object
– Position vector is usually denoted by r.
• The x-, y-, and z- component of r are usually written simply
as x, y, and z (instead of rx, ry, and rz).
Lecture 7
Purdue University, Physics 149
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Position
• A vector quantity describing where you are relative
to an “origin”
– Point A is located at x=3, y=1 or (3,1)
– Point B is located at (-1,-2)
• The vector rA indicating
the position of A starts
att the
th origin
i i and
d tterminates
i t
with arrowhead A
• Same for rB and B
-3
y
3
A
3
B
x
-3
Lecture 7
Purdue University, Physics 149
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Distance vs. Displacement
• Distance (scalar)
– Total length
g of p
path traveled
– The path of an object does matter
• Displacement (vector)
– The change of the position vector (∆r), that is, the final
position vector (rf) minus the initial position vector (ri)
= rf + (–ri)
– An arrow starting at the initial position (the tip of the initial
position vector) and ending with the arrowhead at the final
position (the tip of the final position vector)
– The path of an object does not matter. The displacement
d
depends
d only
l on th
the starting
t ti and
d ending
di points.
i t
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Purdue University, Physics 149
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Displacement (m)
• A vector quantity describing a change in position
∆r = rf - ri
– The displacement from A to B is
• We can determine the components y
– x-direction:
3
xf - xi = -1 – 3 = -4
– y-direction:
di ti
A
yf - yi = -2 – 1 = -3
– ∆r = ((-4,, -3)
3)
– |∆r| = sqrt(42 + 32) = 5
-3
• NOTE: The displacement
p
is not the distance traveled
Lecture 7
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Purdue University, Physics 149
B
x
-3
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