Transcript Lecture 02 - Purdue Physics

PHYSICS 220
Lecture 05
Forces and Motion beyond 1 D
Textbook Sections 3.7, 4.1
Lecture 5
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Lecture 5
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Exercise
Fred throws a ball 30 m/s vertically upward. How long does it
take to hit the ground 2 meters below where he released it?
y = y0 + vy0t - 0.5 g t2
y-y0 - vy0t + 0.5g t2 = 0
y = y0 + vy0t + 1/2 gt2
vy = vy0 + gt
vy2 = vy02 + 2g(y-y0)
-2 – 30 t + 0.5*9.8 t2 = 0
ax  bx  c  0
2
b  b2  4ac
x
2a
Lecture 5
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30  30  4 9.8
2
x
9.8
2
t = 6.19 s or -.06 s
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Exercise
Fred throws a ball 30 m/s vertically upward. What is the
maximum height the ball reaches? How long does it take to
reach this height?
v2-vo2 = 2 a Dy
v = v0 + a t
Dy = (v2-vo2 )/ (2 a)
t = (v-v0) / a
= -302 / (2 * (-9.8))
= (0 – 30 m/s )/ (-9.8 m/s2)
= 46 m
= 3.1 seconds
Lecture 5
Purdue University, Physics 220
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Contact Force: Spring
• Force exerted by a spring is directly proportional to
the amount by which it is stretched or compressed.
Fspring = k x always trying to restore its original length
• Example: When a 5 kg mass is suspended from a spring, the
spring stretches 8 cm. Determine the spring constant.
Fspring- Fgravity = 0
k =mg/x
Fspring = Fgravity
= (5 kg) x (9.8 m/s2) /(0.08 m)
kx=mg
= 612 N/m
Fspring
Fgravity
y
x
Lecture 4
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Contact Force: Tension
• Tension
– A force transmitted by a rope, cord,
cable or the like which transmits a
force from one end to an object
attached at the other end
• Ideal string (or cord, rope, etc.):
– Always maintains constant tension
everywhere.
– Has a zero mass.
– Tension is parallel to the string
Lecture 4
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Pulley changes the direction of the force
associated with the tension in the rope.
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Pulley Example
Two boxes are connected by a string over a frictionless
pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg.
Box 2 starts from rest 0.8 meters above the table, how long
does it take to hit the table.
• Compare the acceleration of boxes 1 and 2
A) |a1| > |a2|
B) |a1| = |a2|
1) T - m1 g = m1 a1
2) T - m2 g = -m2 a1
C) |a1| < |a2|
y
using a1 = -a2
x
T
2) T = m2 g -m2 a1
1) m2 g -m2 a1 - m1 g = m1 a1
a1 = (m2 – m1)g / (m1+m2)
Lecture 4
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1
T
2
2
m1g
m2g
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Pulley Example
Two boxes are connected by a string over a frictionless
pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg.
Box 2 starts from rest 0.8 meters above the table, how long
does it take to hit the table.
a1 = (m2 – m1)g / (m1+m2)
a = 2.45 m/s2
Dx = v0t + ½ a t2
Dx = ½ a
t = sqrt(2 Dx/a)
t = 0.81 seconds
y
x
t2
Lecture 4
T
1
T
2
2
m1g
Purdue University, Physics 220
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m2g
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iClicker
• Two boxes are connected by a string over a
frictionless pulley. In equilibrium, box 2 is lower
than box 1. Compare the weight of the two boxes.
A) They have the same weight
B) Box 1 is heavier
C) Box 2 is heavier
SF = 0
1) T - m1 g = 0
2) T – m2 g = 0
=> m1 = m2
Lecture 4
T
1
T
1
m1g
2
2
m2g
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Tension Example
• Determine the force exerted by the hand to
suspend the 45 kg mass as shown in the picture.
1) 220 N
2) 440 N
3) 660 N
4) 880 N
5) 1100 N
SF = 0
T+T–W=0
2T=W
T=mg/2
= (45 kg x (9.8 m/s2)/ 2
= 220 N
y
T T
x
W
Remember the magnitude of the tension is the
same everywhere along the rope!
Lecture 4
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Question
• What is the force on the ceiling?
y
A) 220 N
B) 440 N
D) 880 N
C) 660 N
x
E) 1100 N
SF = 0
Fc
Fc -T - T – T = 0
Fc = 3 T
Fc = 3 x 220 N = 660 N
Lecture 4
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T
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Question
• What does the scale read?
A) 225 N
B) 550 N
C) 1100 N
The sum of the forces here will be 0. Thus, the force
tension will equal the force gravity. Since the force of
gravity is 550N, the force of tension (which is
measured by the scale) will also be 550N.
Lecture 4
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Two blocks one sliding one hanging
•A block of mass m1=3kg rests on
a frictionless horizontal surface. A
second block of mass m2=2kg
hangs from an ideal cord of
negligible mass who runs over an
ideal pulley. Block 2 starts from rest
0.8 meters above the floor, how
long does it take to hit the floor?
a
a
a1 = (m2 )g / (m1+m2)
a = 3.9 m/s2
Dy = v0t + ½ a t2= ½ a t2
t = sqrt(2 Dy/a)
t = 0.64 seconds
Lecture 4
Purdue University, Physics 220
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Friction
• Magnitude of frictional force is
proportional to the normal force.
Fkinetic = mk N
Fstatic  ms N
mk coefficient of kinetic friction
ms coefficient of static friction
Be Careful!
•Static friction , can be
any value up to msN
•Direction always opposes
motion
Lecture 4
F
Ff
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Inclined Plane
Ff
N

N = m g cos
m g sin - Ff = 0
Ff = mN
m g sin - m m g cos = 0

mg
• Special case 1: Start with  at zero and slowly
increase . Just before it slides
ms m g cos = m g sin
tan = ms
• Special case 2: Object is sliding down at constant
velocity, that is a = 0
mk m g cos = m g sin
Lecture 5
tan = mk
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iClicker
What is the normal force of ramp on block?
A) N > mg
B) N = mg
In “y” direction:
SF = ma
N – mg cos  = 0
N = mg cos 
N
T


Lecture 5
C) N < mg
W
N = m g cos 
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W
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Force at Angle Example
• A person is pushing a 15 kg block across a floor with mk= 0.4 at a
constant speed. If she is pushing down at an angle of 25 degrees, what
is the magnitude of her force on the block?
x- direction: SFx = 0
Fpush cos – Ffriction = 0
Fpush cos – m FNormal = 0
FNormal = Fpush cos / m
y- direction: SFy = 0
FNormal –Fweight – FPush sin = 0
FNormal –mg – FPush sin = 0

Combine:
( Fpush cos / m)–mg – FPush sin = 0
Fpush ( cos / m - sin ) = mg
Fpush = m g / ( cos/m – sin)
Fpush = 80 N
Normal
Pushing
y

x
Friction
Weight
Lecture 5
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