Download Lecture 03 - Eunil Won

Eunil Won Dept. of Physics, Korea University
1
Ch 03 Force
Movement
of massive
object
Force
Source of the move
Velocity,
acceleration
Eunil Won Dept. of Physics, Korea University
2
1 평(坪) = ?
~
~
3.305
2
m
2
m
1.82 m x 1.82 m
Eunil Won Dept. of Physics, Korea University
Newton’s First Law
Law of Inertia (mass)
Any body will remain at rest of in motion in a
straight line with a constant velocity unless
acted upon by an outside force
If the net force is zero, there is a reference
frame with zero acceleration
3
Eunil Won Dept. of Physics, Korea University
4
Newton’s Second Law
Law of Motion
If a force is applied, the state of motion is changed
dv ⎤
⎡ dp d
= (mv) = m ⎥ = ma newton ⇒ [N = kg ⋅ m/s 2 ]
F=⎢
dt ⎦
⎣ dt dt
Eunil Won Dept. of Physics, Korea University
5
Newton’s Third Law
Action-Reaction Law
Whenever one body exerts a force on a second body,
the second body exerts a force back on the first that
is equal in magnitude and opposite in direction
FAB = −FBA
Eunil Won Dept. of Physics, Korea University
6
Example 3.1
A 440-g can of food on a frictionless level
surface is observed to accelerate at a rate
of 1.5 m/sec2. What is the force on the
can?
Fnet , ext = ma
= (0.440 kg)(1.5 m/s 2 ) = 0.66 kg ⋅ m/s 2
Fnet , ext = 0.66 N
Eunil Won Dept. of Physics, Korea University
7
Example 3.2
An orderly
The orderly exerts a backward
force of 100 N on the floor.
What acceleration is produced,
assuming the friction in the
wheels is negligible?
(question on the system I)
An orderly
mo = 85kg
Gurney
mg = 20kg
Patient
m p = 50kg
Force that
the orderly
exerts
Fa = 100N
Acceleration
a=
=
Ft
Fa
=
mt mo + mg + m p
100
= 0.645m/s 2
85 + 20 + 50
Eunil Won Dept. of Physics, Korea University
8
Example 3.3
Calculate the force the orderly exerts
on the gurney in Example 3.2
(Question on system 2)
An orderly
mo = 85kg
Gurney
mg = 20kg
Patient
m p = 50kg
Force that
the orderly
exerts
Fa = 100N
Force
Fo = ma = (mg + m p )a
= (20 + 50)(0.645)
= 45.2kgm/s 2 = 45.2 N
Eunil Won Dept. of Physics, Korea University
9
Weight and normal force
weight
F = ma
w = mg
Normal force
Ft = 0
N −w=0
⇒
N =w
Eunil Won Dept. of Physics, Korea University
10
Newton’s Universal Law of Gravitation
mM
F=G 2
r
: there is a force of attraction between
any two masses that is proportional to
the product of the masses and inversely
proportional to the distance between
their centers of mass
mM
w = mg ⇔ G 2
r
GM
g= 2
r
Eunil Won Dept. of Physics, Korea University
11
Eunil Won Dept. of Physics, Korea University
12
Friction
: proportional to the normal force
Static Friction
fs = µs N
Kinetic Friction
f (= f k ) = µ k N
µs
f ≤ fs
µs
: coefficient of static friction
µk
: coefficient of kinetic friction
Eunil Won Dept. of Physics, Korea University
13
Friction
Coefficients of friction
Rubber on dry concrete
Rubber on wet concrete
Wood on Wood
Waxed wood on wet snow
Metal on wood
Steel on steel (dry)
Steel on steel (oiled)
Teflon on steel
Bone lubricated with synovial fluid
(활액)
Shoes on wood
Shoes on ice
Ice on ice
Steel on ice
Kinetic
0.7
0.5
0.3
0.1
0.3
0.3
0.03
0.04
Static
1.0
0.7
0.5
0.14
0.5
0.6
0.05
0.04
0.015
0.016
0.7
0.05
0.03
0.02
0.9
0.1
0.1
0.4
Eunil Won Dept. of Physics, Korea University
14
Example 3.4
A 70-kg cross-country skier is on level ground in
wet snow wearing waxed wooden skis. (a) What
is the maximum force that can be exerted on the
skier in a horizontal direction without causing her
to move? (b) Once the skier is moving, what
horizontal force is necessary to keep her moving
at a steady velocity?
w = mg = (70 kg)(9.8 m/s 2 ) = 686 N
f = µs N
= (0.14)(686 N) = 96 N
Fnet , ext = 0 ⇒ F − f = 0
F = f = µk N
= (0.1)(686 N) = 68.6 N
Eunil Won Dept. of Physics, Korea University
15
Tension
A tension is any force carried by a flexible
string, rope, cable, chain etc
Eunil Won Dept. of Physics, Korea University
16
Example 3.5
A child and basket with a total
mass of 10 kg are suspended
from a scale by a cord.
Calculate the tension in the
cord.
Fnet , ext = T − w = 0
T = w = mg
= (10 kg)(9.8 m/s 2 ) = 98 N
Eunil Won Dept. of Physics, Korea University
Tensions of a patient
17
Eunil Won Dept. of Physics, Korea University
18
Vector
1. Vectors have direction &
magnitude
ρ
{a, a}, a = a
2. Two vectors are identical if
both direction and magnitude
a=b
are same
3. Negative vector.
4. Zero vector.
5. Unit vector.
−a
0
eˆ ⇒ eˆ = 1
Eunil Won Dept. of Physics, Korea University
19
Vector analysis
a+b = s
a
b
a+b = b+a
s
a + (b + c) = (a + b) + c
a + ( −a) = 0
a−b = d
a+0=a
−b
d
a
Eunil Won Dept. of Physics, Korea University
20
Components of Vectors
y
a + b = (a x i + a y j) + (bx i + by j)
a
ay
ŷ
x̂
= (a x + bx )i + (a y + by ) j
x
ax
xˆ = i = (1,0,0)
yˆ = j = (0,1,0)
a = ax + ay + az
= a x xˆ + a y yˆ + a z zˆ
a + b = s = sxi + s y j
s x = a x + bx
s y = a y + by
a x = a cosθ , a y = a sin θ
⎛ ay ⎞
θ = tan ⎜⎜ ⎟⎟
⎝ ax ⎠
−1
Eunil Won Dept. of Physics, Korea University
21
Example 3.6
Calculate the total force exerted on a gurney
F1x = F1 cos 20° = 70(0.939) = 65.78
F1 y = F1 sin 20° = 70(0.342) = 23.94
F2 x = F2 cos15° = 60(0.966) = 57.95
F2 y = F2 sin 15° = 60(0.259) = 15.53
Ft = F1 + F2
F1 = F1x i + F1 y j
Ftx = F1x + F2 x = 123.73
Fty = F2 y − F2 y = 8.41
2
2
F2 = F2 x i + F2 y j
Ft = Ftx + Fty = 124.01
Ft = ( F1x + F2 x )i + ( F2 y + F2 y ) j
⎛ Fty ⎞
θ = tan ⎜ F ⎟ = 3.89°
tx ⎠
⎝
−1
Eunil Won Dept. of Physics, Korea University
22
Example 3.7/3.8
Calculate the acceleration (cart+stuff=25 kg, frictional force=2.0 N)
Acceleration of the cart
Fn = Fh − f = 17.3 − 2.0 = 15.3
Fn 15.3
=
= 0.61m/s 2
a=
25
m
Normal Force
Fy = 0
N − w − Fv = 0
N = w + Fv = mg + Fv
Ft = Fh + Fv
Fv + N + mg = 0
= (25)(9.8) + 10 = 255 N
Eunil Won Dept. of Physics, Korea University
Torque and Rotation
23
Eunil Won Dept. of Physics, Korea University
24
Torque
The effectiveness of a force in producing a rotation is
called torque
τ = rFt = rF sin φ = r⊥ F
τ = r × F , τˆ //ω̂
Eunil Won Dept. of Physics, Korea University
25
Equilibrium Condition
∑ F = 0, ∑τ
=0
Eunil Won Dept. of Physics, Korea University
26
Example 3.9
(a) Calculate the force that the biceps muscle exerts to hold the book
(b) Compare the force exerted by the biceps muscle with the weight
of the forearm and book.
(clockwise)
(counterclockwise)
τ cw, net = τ ccw, net
(15 cm)( wa ) + (40 cm)( wt ) = (4.0 cm) FB
(15cm)(ma )( g ) + (40cm)(mt )( g )
4.0cm
(15)(2.5kg )(9.8m/s 2 ) + (40)(4.0kg )(9.8m/s 2 )
=
4.0
367.5 N + 1568 N
=
= 484 N
4.0
FB =
w = wa + wt = ma g + mt g
w = 63.7 N
484 N 〉〉 63.7 N
Eunil Won Dept. of Physics, Korea University
27
Example 3.10
Calculate the force exerted by the extensor muscle in
the upper leg when lifting a weight as shown below
Note: l⊥ = 35cm
l′⊥ = 2.0cm
τ w, net = τ ccw, net
l⊥T = l⊥′ FM ⇒ FM = (l⊥ / l⊥′ )T
FM = (l⊥ / l′⊥ )mg
= (35cm/2.0cm)(98 N)
= 1715 N
Eunil Won Dept. of Physics, Korea University
28
Rotational Motion and Centripetal Force
v2
Fc = mar = m
r
Eunil Won Dept. of Physics, Korea University
29
v p = v px i + v py j
= v cosθ i + v sin θ j
v q = vqx i + vqy j
= v cosθ i − v sin θ j
pq r (2θ )
∆t =
≈
v
v
ax =
ay =
vqx − v px
∆t
vqy − v py
∆t
a = −a y j
=
=
vq cosθ − v p cosθ
∆t
− vq sin θ − v p sin θ
( 2 rθ ) / v
=0
v 2 ⎛ sin θ ⎞
=− ⎜
⎟
r ⎝ θ ⎠
⎛ v2 ⎞
sin θ v 2
a y = lim a y = ⎜⎜ ⎟⎟ lim
=
θ →0
r
⎝ r ⎠ θ →0 θ
Eunil Won Dept. of Physics, Korea University
30
Example 3.11
An ultracentrifuge spins at 50,000 revolutions per minute
and the material being centrifuged is 5.0 cm from the pivot
point. Calculate the centripetal acceleration and express it
as a multiple of the acceleration of gravity (in g’s)
v=
s 50000 × 2π (0.05 m)
=
= 2.62 ×10 2 m/s
60 s
t
v 2 (2.62 × 10 2 m/s) 2
ac =
=
= 1.37 ×106 m/s 2
r
0.05m
ac 1.37 ×106 m/s 2
5
=
=
1
.
40
×
10
g
9.8 m/s 2
ac = (1.40 ×105 ) g