Equilibrium

1
Chapter 5Sections 5.5 5.8 Applying Newtons
Laws
Topics

• Equilibrium
• Using Newtons second law
• Mass, weight, and apparent weight
• Static and kinetic friction
• Applying Newtons third law

Sample question
Before his parachute opens, why does this
skydiver fall at a constant speed? And why does
he suddenly slow down when his parachute opens?
Slide 5-1
2
Static Friction
fs max µsn
Slide 5-25
3
Kinetic Friction
fk µkn
Slide 5-26
4
Working with Friction Forces
Slide 5-27
5
Example
A car traveling at 20 m/s stops in a distance of
50 m. Assume that the deceleration is constant.
The coefficients of friction between a passenger
and the seat are µs 0.5 and µk 0.3. Will a 70
kg passenger slide off the seat if not wearing a
seat belt?
Slide 5-28
6
Example
A car traveling at 20 m/s stops in a distance of
50 m. Assume that the deceleration is constant.
The coefficients of friction between a passenger
and the seat are µs 0.5 and µk 0.3. Will a 70
kg passenger slide off the seat if not wearing a
seat belt?
Cars velocity before stopping
n 686 N
W (70 kg)(9.8 m/s2) 686 N
7
Example
A car traveling at 20 m/s stops in a distance of
50 m. Assume that the deceleration is constant.
The coefficients of friction between a passenger
and the seat are µs 0.5 and µk 0.3. Will a 70
kg passenger slide off the seat if not wearing a
seat belt?
Cars velocity before stopping
n 686 N
Cars acceleration during stopping
W (70 kg)(9.8 m/s2) 686 N
8
Example
A car traveling at 20 m/s stops in a distance of
50 m. Assume that the deceleration is constant.
The coefficients of friction between a passenger
and the seat are µs 0.5 and µk 0.3. Will a 70
kg passenger slide off the seat if not wearing a
seat belt?
y
Cars velocity before stopping
n 686 N
persons acceleration during stopping
x
passenger
W (70 kg)(9.8 m/s2) 686 N
9
Example
A car traveling at 20 m/s stops in a distance of
50 m. Assume that the deceleration is constant.
The coefficients of friction between a passenger
and the seat are µs 0.5 and µk 0.3. Will a 70
kg passenger slide off the seat if not wearing a
seat belt?
y
Cars velocity before stopping
n 686 N
persons acceleration during stopping
Friction force
x
passenger
W (70 kg)(9.8 m/s2) 686 N
10
2(acar/person)?x (vx)f2 (vx)i2
(acar/person) (0 m/s)2 (20 m/s)2/(100 m)
-4 m/s2
y
Cars velocity before stopping
n 686 N
persons acceleration during stopping
Friction force
x
passenger
W (70 kg)(9.8 m/s2) 686 N
11
2(acar/person)?x (vx)f2 (vx)i2
(acar/person) (0 m/s)2 (20 m/s)2/(100 m)
-4 m/s2
Is friction big enough to accelerate the person
at this rate???
y
Cars velocity before stopping
n 686 N
persons acceleration during stopping
Friction force
x
passenger
W (70 kg)(9.8 m/s2) 686 N
12
Newton 2 for person (along x axis) aperson
(FNet on person)/Mperson.
Is friction big enough to accelerate the person
at this rate???
y
Cars velocity before stopping
n 686 N
persons acceleration during stopping
Friction force
x
passenger
W (70 kg)(9.8 m/s2) 686 N
13
Newton 2 for person (along x axis) aperson
(FNet on person)/Mperson. The maximum a by fs
Is friction big enough to accelerate the person
at this rate???
y
Cars velocity before stopping
n 686 N
persons acceleration during stopping
Friction force
x
passenger
W (70 kg)(9.8 m/s2) 686 N
14
Newton 2 for person (along x axis) aperson
(FNet on person)/Mperson. The maximum a by fs
(-µsn)/ Mperson (-0.50686 N)/(70 kg) -4.9
m/s2
Is friction big enough to accelerate the person
at this rate???
y
Cars velocity before stopping
n 686 N
persons acceleration during stopping
Friction force
x
passenger
W (70 kg)(9.8 m/s2) 686 N
15
Static friction could accelerate the passenger by
up to -4.9 m/s2. This is less than the required
acceleration of -4 m/s2 so the passenger does not
slide forward (x direction) off the seat!
Is friction big enough to accelerate the person
at this rate???
y
Cars velocity before stopping
n 686 N
persons acceleration during stopping
Friction force
x
passenger
W (70 kg)(9.8 m/s2) 686 N
16
Applying Newtons Third Law Interacting Objects
Acceleration Constraints
Slide 5-29
17
Slide 5-30
18
Example
Block A has a mass of 1 kg block Bs mass is 4
kg. They are pushed with a force of magnitude 10
N. a) What is the acceleration of the blocks? b)
With what force does A push on B? B push on A?
Slide 5-31
19
Example
Block A has a mass of 1 kg block Bs mass is 4
kg. They are pushed with a force of magnitude 10
N. a) What is the acceleration of the blocks? b)
With what force does A push on B? B push on A?
A B accelerate together so FNet on AB
MABaAB
20
B (4 kg)
aAB
10 N
A (1 kg)
21
x
B (4 kg)
aAB
10 N
A (1 kg)
Newton 2 for AB (10 N) (5 kg) aAB So aAB
(10 N)/(5 kg) or 2 m/s2 Both mass A and mass
B accelerate at this rate.
22
b) With what force does A push on B? B push on
A? First, lets consider Newton 2 for mass B
B (4 kg)
aB aAB
FA on B
x
FA on B (4 kg)( 2 m/s2) 8 N
23
b) With what force does A push on B? B push on
A? First, lets consider Newton 2 for mass B
B (4 kg)
aB aAB
FA on B
x
FA on B (4 kg)( 2 m/s2) 8 N We can check
this by writing Newton2 for mass A(now were
doing more than was asked).
24
10 N
A (1 kg)
25
a1 kg 2 m/s2
FB on A - FA on B
10 N
A (1 kg)
-8 N
26
a1 kg 2 m/s2
FB on A - FA on B
10 N
A (1 kg)
-8 N
Newton 2 for 1 kg (10 N) (-8 N) (1 kg)(2
m/s2) 2 N 2 N check!
27
Checking Understanding
Which pair of forces is an action/reaction pair?

• The string tension and the friction force acting
  on A.
• The normal force on A due to B and the weight of
  A.
• The normal force on A due to B and the weight of
  B.
• The friction force acting on A and the friction
  force acting on B.

Slide 5-32
28
Answer
Which pair of forces is an action/reaction pair?
D. The friction force acting on A and the
friction force acting on B.
Slide 5-33
29
Example
What is the acceleration of block B?
Slide 5-34
30
Ropes and Pulleys
Slide 5-35
31
Example
Block A, with mass 4.0 kg, sits on a frictionless
table. Block B, with mass 2.0 kg, hangs from a
rope connected through a pulley to block A. What
is the acceleration of block A?
Slide 5-36
32
nA 39.2 N
T
T
WA 39.2 N
WB 19.6 N
33
nA 39.2 N
aA a
T
T
WA 39.2 N
aB a
WB 19.6 N
34
nA 39.2 N
aA a

T
T
WA 39.2 N
aB a

WB 19.6 N
35
Newton 2 for both masses (along the direction of
movement) For mass A (4 kg),
36
Newton 2 for both masses (along the direction of
movement) For mass A (4 kg), (T) (4 kg)(a)
37
Newton 2 for both masses (along the direction of
movement) For mass A (4 kg), (T) (4
kg)(a) For mass B (2 kg),
38
Newton 2 for both masses (along the direction of
movement) For mass A (4 kg), (T) (4
kg)(a) For mass B (2 kg), (T)(-19.6 N) (2
kg)(-a)
39
Newton 2 for both masses (along the direction of
movement) For mass A (4 kg), (T) (4
kg)(a) For mass B (2 kg), (T)(-19.6 N) (2
kg)(-a) We now have two equations in two unknowns
(a and T). We must solve these simultaneously.
40
Newton 2 for both masses (along the direction of
movement) For mass A (4 kg), (T) (4
kg)(a) For mass B (2 kg), (T)(-19.6 N) (2
kg)(-a) We now have two equations in two unknowns
(a and T). We must solve these simultaneously.
Substitute the first into the second,
41
Newton 2 for both masses (along the direction of
movement) For mass A (4 kg), (T) (4
kg)(a) For mass B (2 kg), (T)(-19.6 N) (2
kg)(-a) We now have two equations in two unknowns
(a and T). We must solve these simultaneously.
Substitute the first into the second, (4 kg)(a)
(-19.6 N) (2 kg)(-a) Solving a (19.6
N)/(6 kg) 3.27 m/s2
42
nA 39.2 N
T
f
T
WA 39.2 N
Now, what if theres friction?
WB 19.6 N
43
nA 39.2 N
T
f
T
WA 39.2 N
Now, what if theres friction? (Static or
kinetic??)
WB 19.6 N