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Physic 231 Lecture 7
•
•
•
Newton’s
N
’ 3d Law:
L
– “When a body exerts a force on another, the second body exerts
an equal oppositely directed force on the first body.”
Frictional forces:
– kinetic friction: f k = μ k N
– static friction
f s < μs N
Examples.
Example


F1 andd F2, actt on the
• Two
T forces,
f
th 5.00
5 00 kg
k block
bl k shown
h
in
i the
th drawing.
d i
The magnitudes of the forces are F1=60 N and F2 = 25 N. What is the
horizontal acceleration (magnitude and direction) of the block?

F1
600

Normal


Normal force cancels the y components of F1 and W
F1,y + Normal y − W = 0

F2

W
F1
60 N
F2
25 N
m
5 kg
θ1
60o
a
?
( )
F1,x = 60 cos 60o N = 30N
Fnet = F1,x − F2 = 30N − 25N = 5N
a = Fnet / m = 5N / 5kg = 1m / s 2
Block accelerates to the right.
conceptual question
•
Consider
C
id a person standing
t di iin an elevator
l
t th
thatt iis accelerating
l ti
upward. The upward normal force N exerted by the elevator
floor on the person is
– a)) llarger th
than
– b) identical to
– c) smaller than
the downward weight W of the person.
Example: with balance scale marked in Newtons, not
the usual scale, but a quite reasonable one.
•
A 100 kkg man stands
t d on a bbathroom
th
scale
l iin an elevator.
l t St
Starting
ti from
f
rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in
0.80 s. It travels with this constant speed for 5.0 s, undergoes a uniform
negative acceleration for 1.5
1 5 s and comes to rest.
rest What does the scale
register (a) before the elevator starts to move? (b) during the first 0.8
s? (c) while the elevator is traveling at constant speed? (d) during the
Scale reading
g =Normal force
negative acceleration?
N
?
m
100 kg
vf
1.2 m/s
Δt
0.8 s
N − mg = ma : scale = N (in Newtons)
a) : a=0; N = mg : scale = N = 980 Newtons
b) : N − mg = ma : scale
l =N
a = Δv / Δt = (1.2m / s) / 0.8s = 1.5m / s 2
N = m(g + a);scale = m(g + a)
scale = (100kg)(9.8m / s 2 + 1.5m / s 2 ) = 1130Newtons
N
vi
1.2 m/s
vf
0
Δt
1.5 s
c) : N = mg : scale = N = 980Newtons
d) : a = (v f − vi ) / Δt = −1.2m / s / (1.5s) = −0.8m / s 2
N = m(g + a);scale = m(g + a) = 900Newtons
No Acceleration: Static Equilibrium
•
All objects
bj t are att restt andd remain
i so. The
Th nett force
f
on any object
bj t mustt
vanish. I.e. on an object:

Fi = 0
i
•
Example: Three ropes are arranged so as to support a 4 kg mass as
shown below. Determine the tension in each rope.
( )
Fi,x = 0 = − T1 + T2 cos 600

T2

T1 A

T3
4 kg
i
0 = − T1 + T2 / 2; T1 = T2 / 2
60o
( )
Fi,y = 0 = T2 sin 600 − T3
i

T3
( )
T2 = T3 / sin 600 = 1.16T3
4 kg
Solve by considering forces at
point A
mg
T3 = mg = 39.1N;T
39 1N;T2 = 45.3N
45 3N
T1 = 22.6N
Atwood’s machine
•
•
Consider the Atwood machine to the
right. The massless string passes
over a massless and frictionless
pulley. It is under tension T, which
we define to be the magnitude of the
tension force. By this definition it is
a positive number.
Choosing up to be positive , what is
the net force on mass 1?
– a) T-m1g
– b) T+m1g
– c) m1g-T
– d) none of the above
m1
m2
Atwood’s machine
•
Choosing
g up
p to be ppositive , what is
the net force on mass 2?
– a) m2g-T
– b) T+m2g
– c) T-m2g
– d) none of the above
m1
m2
Atwood’s machine
•
From Newton’s 2nd law:
T − m1 g = m1a1
 T = m1a1 + m1 g
•
Also
T − m2 g = m2a2
m1
 T = m2a2 + m2 g
•
If m2 exceeds m1, m2 g
goes down and
m1 goes up. If m1 exceeds m2, m1
goes down and m2 goes up. In either
case
Δy2 = − Δy1
v2 = − v1
a2 = −a1
m2
•
Putting it together:
m1a1 + m1 g = m2a2 + m2 g
= −m2a1 + m2 g
m1a1 + m2a1 = m2 g − m1 g
a1
f
(
m2 − m1 )g ⇐ net force
=
pulling m up
1
total mass  m1 + m2 and m down.
2
Reading Quiz
4. An action/reaction pair of forces
A.
B
B.
C.
D.
point in the same direction.
act on the same object.
object
are always long-range forces.
act on two different objects.
Slide 4-13
Newton’s Third Law
• Wh
When a body
b d exerts a force
f
on another,
h the
h secondd body
b d
exerts an equal oppositely directed force on the first body.
– Note: the two forces act on different bodies
body 1
body 2
• Force on body 2
due to body 1:
F21
• Force on body
1 due to body 2:
F12
• 3d law:
l
F21=- F12
Example
•
Two skaters,
T
k t
an 82 kg
k man andd a 48 kg
k woman, are standing
t di on ice.
i
Neglect any friction between the skate blades and the ice. The woman
pushes the man with a force of 45 N due east. Determine the
accelerations (magnitude and direction) of the man and the woman.
woman
x components (West is positive)
−45N
= 0.55m / s 2 east
a man =
82kg
45N
a woman =
= 0.94m / s 2 west
48kg
East
quiz
•
Two skaters,
T
k t
an 100 kg
k man andd a 50 kg
k woman, are standing
t di on ice.
i
Neglect any friction between the skate blades and the ice. By pushing
the man, the woman is accelerated at 2 m/s2 in the direction of due
west What is the corresponding acceleration of the man?
west.
– a) 4 m/s2 due east
– b) 1 m/s2 due east
– c) 2 m/s2 due east
– d) 1.5 m/s2 due east


Fwoman _ on _ man = − Fman _ on _ woman


m man a man = − m woman a woman
East
choosing west to be negative (just to be contrary)
m wo
50
woman
a
a man,x = −
a woman,x = −
( −2m / s 2 ) = 1m / s 2 east
m man
100
Note: the direction would still be east even if you chose west to be positive
Example

F = 60 N
•

N12

N 21
m1=10 kg m2= 5 kg
A bl
blockk with
ith mass 5 kg
k andd a secondd block
bl k with
ith mass 10 kg
k are
supported by a frictionless surface. A force of 60 N is applied to the 10
kg mass. What is the force of the 5 kg block on the 10 kg block?
x components:
t
body 2: N 21,x = m 2a x ; N 21,x = − N12,x from Newtons 3d law
body 1: Fx + N12,x = m1a x ; Fx = − N12,x + m1a x
Fx = N 21,x + m1a x = m 2 a x + m1a x = ( m 2 + m1 ) a x
a x = Fx / ( m 2 + m1 )
N12,x = − N 21,x = − m 2 a x = − m 2 Fx / ( m 2 + m1 )
N12,x = −5kg 60N / (15kg) = −20N
N12,x = 20N to the left
If objects move together, the acceleration is
governed by the total mass
Friction
•
•
Friction
F
i ti impedes
i
d the
th motion
ti off one object
bj t along
l
the
th surfaces
f
off another.
th
It occurs because the surfaces of the two objects temporarily stick
together via “microwelds”. The frictional force can be larger if the two
surfaces are at rest with respect to each other.
other

Experimentally we have two cases:
N


v
fk
– kinetic friction:
f k = μk N
– static friction
f s < μs N

fs

N

F
•
The coefficient of static friction generally exceeds that for kinetic
friction: μ s > μk
•
Frictional
F
i ti l forces
f
always
l
oppose the
th motion
ti off one surface
f
with
ith respectt
to the other.
Example with static friction
•
Consider
C
id the
th figure
fi
below,
b l
with
ith M1=105
105 kkg and
d M2=44.1
44 1 kkg. Wh
Whatt is
i
the minimum static coefficient of friction necessary to keep the block
from slipping.
T = M 2g
If M1 doesn’t move
T = f s ≤ μs N = μs M1g
Putting it together
M 2 g ≤ μs M1g
M2
≤ μs
M1
If this isn’t true, M1 will slip