Force And Motion II
Dr. Venkat Kaushik
Phys 211, Lecture 9
Sep 29, 2015
Clicker Question 1 (30 s)
•  CMB as related to physics stands for
A.  Cosmetic Makeup Brand
B.  Coastal Mortar Board
C.  Cell and Molecular Biology
D.  Cosmic Microwave Background
09/29/2015
Lecture 9
2
Clicker Question 2 (30 s)
•  CMB was discovered in what decade?
A.  In the 1940’s
B.  In the 1950’s
C.  In the 1960’s
D.  In 1990’s
09/29/2015
Lecture 9
3
Clicker Question 3 (30 s)
•  What is CMB ?
A.  Microwave radiation of the Earth
B.  Microwave radiation of the Sun
C.  Faint glow (remnants) of the big bang
D.  None of the above
09/29/2015
Lecture 9
4
Clicker Question 4 (30 s)
•  What temperature does CMB correspond to ?
A.  About 273 Kelvin
B.  About 373 K (100 C)
C.  About 10K
D.  About 3 K
09/29/2015
Lecture 9
5
Friction
•  Dissipative Force
§  Friction is a force. It opposes impending (about to happen) motion
§  Direction is opposite to that of impending motion. It is
electromagnetic in nature
•  Static Friction
§  When an object is at rest, it requires a certain minimum external
force to get it moving (sliding) across the contact surface.
§  During this time, the force that opposes the motion is due to static
friction
•  Kinetic Friction
§  When the (sum of external) applied forces causes the object to
move, there can still exist friction opposing motion, this force is
dynamic friction
•  Contact Force
§  Friction force can exist between any two (or more) surfaces in
contact and moving (or at rest) relative to each other.
Chapter 6
6
T î + (N1
(T
(T
1
1
1
m1 g) ĵ = m1 a î
m2 g) ĵFriction
= m2 a ĵ
m g) ĵ = m a ĵ
Formulae
2
m1 g) = 0 , (T 2 m2 g) = m
2a
force
to normal force
g) • = Frictional
0m, g (T m
=ismproportional
mg2 a
2 g) m
2
1 2
= §  The constant
T = of proportionality is called the coefficient of
m21 g+ m2 )
m11 m
(m
(m
+
2mg2 )
friction
T=
(m1 +
(mfriction
) have a value that could be zero or
2)
1 + m2can
§  m
Coefficient
of
greater
Ffr / N
Ffr
fr / N
)
= µ, µ
FN
0
fr
) Fs == µµs N
, µ 0
N
•  Fs (static) and Fk (kinetic)
Friction
Fk = µ k N
Fs = µ s N
Fk = µ k N
Chapter 6
7
T = m1 a , (N1
m1 g) = 0 , (T
)a=
m2 g) = m2 a
Ffr / N
ProblemF 1
m2 g
m1 m2 g
T=
(m1 + m2 )
(m1 + m2 )
)
θ
m
)
Ffr / N
F
#»
r
#»
v
Ffr
= µ, µ
N
F
Fs = µsfrN
fr
= µ, µ
0
N
FBD
ma
Fs =Nµs N
F0k = µk N
0
#»
#» a
#»
r = x î + y ĵ , r 0 =
F sinθ
θ#»
#»
v = v cos ✓ î + v sin
F
cosθ
Fk = µ k N
W = mg
0
0
0
r
#»
a = g ĵ
) #»x î + 1y #»ĵ
#»
#»
r = r(1)
a
0 + v 0t
2
Given m = 4 kg, F = 20 N, θ = 20°
Case 1: Assume object is at rest and find coefficient of static friction µs
(1)
î + y ĵ = v0oft cos
✓ î + v0 t s
Case 2: If applied force increases to F = 25 N and µk = 0.2, find)
thex acceleration
the block
Equations of Motion
(F cos(F✓ cos F
✓ frF)frî )=î =ma
maîî
+ F✓ sin mg)
✓ mg)
(N + F(Nsin
ĵ =ĵ =0 0
#» |F#»fr | = µ N
|F fr | = µ N
Chapter 6
c) #»
vCase
6 0 1 cannot be zero
=
F cos ✓ = Fs
Case 2
x =F v0 t cos ✓
ma = F cos ✓
k
N = mg F sin ✓
1
N =ĵ mg F sin ✓
d) #»
v = (6.0t 4.0t2 ) î + 8.0,
y = v0 t sin ✓
g
Fs = µ s N
F
=
µ
N
k
k
2
2 2 quantity
2
| #»
v | = F(6.0t
4.0t
)
+
(8.0)
=
10
cos ✓
F
µs =
p
a = (cos ✓ + µk sin ✓) µk g
mgquantity
m
1D
2D
6.0F ±sin ✓ 36 4(4.0)(±6.0)
2
#»
=
) µs =t 0.58
) a = 4.34 m/s
r
2(8.0)position
r
position #»
r î
r î8+
Ffr / N
Problem 1 (continued)
F
)
fr
N
F
= µ, µ
0
Fs = µ s N
Fk = µ k N
θ
m
(1)
1) Find normal force and force of friction for Cases 1 and 2
2) For Case 2, what angle (call it θmax) does the block attain maximum acceleration?
Equations of Motion
(F cos ✓
Ffr ) î = ma î
(N + F sin ✓
mg) ĵ = 0
#»
|F fr | = µ N
c) #»
vProblem
6 0 cannot
=
be zero
1
Force
d)
Normal
Friction
Chapter 6
Case 1
#»
v 32.36
= (6.0t
N
| #»
v | = (6.0t
18.79 N
t=
6.0 ±
Problem 2
Case 2
F
(cos ✓ + µk sin ✓) µk g
2
m
4.0t N) î + 8.0,daĵ F
30.65
= (sin ✓ µk cos ✓) = 0
d✓2 =m10
4.0t2 )2 + (8.0)
tan ✓max = µk
6.13 N
p
a=
36 4(4.0)(±6.0)
) ✓max = tan
2(8.0)
1
(0.2) ⇡ 11.31
9
Centripetal Force
•  For an object moving at a constant
speed (v) along a circular arc of
radius (r), the acceleration is
m
v2
ac =
r
2
•  If the object has a
v 2 mass vm, then
ac =Fc = ma =
applying Newton’s
r second
r law gives
Fc
mv 2
Fc = mac =
r
•  The force is directed toward the
center of the circle and is called the
centripetal (center seeking) force.
Chapter 6
10
Example (Vertical Loop)
• 
What is the minimum speed at which an object of mass m traveling
along the circumference of a vertical circle not lose contact at the
highest point?
m
The object will lose contact with the surface when the normal force
becomes zero. The minimum speed at which this happens is v = √gr.
At any larger speed than that, the object will stay put and not lose
contact with the circular path at the highest point.
mv 2
N + mg = ma =
r
2
mvmin
mg ⇡
, N !0
r
p
vmin = rg
Chapter 6
N
W
Fc
11