Phys 170 Lecture 23 1
Physics 170 Lecture 23
Chapter 13 - Kinetics (Dynamics)
OverviewExample problems
Phys 170 Lecture 23 2
Chapter 13: Kinetics of a Particle
Kinetics means why things move the way they do
Dynamics is another word for kinetics
Particle means we are back to points instead of rigid bodiesand we ignore rotation and moments
The forces no longer add up to zero
The net force causes acceleration
We use kinematics to describe the resulting motion.
Phys 170 Lecture 23 3
13.1: Newton’s Laws
The First Law says that if forces balance, an object stays at restor moves at constant velocity. That’s the basis for statics
The Third Law says that forces come in equal and opposite pairs. We have used that when we have two different free bodies in contact with each other.
The new one is the Second Law: if forces are imbalanced, there is an acceleration in the same direction as the net force, proportional to the net force. In equation form, .
(Newton also invented the inverse-square law of gravity).
F = ma
Phys 170 Lecture 23 4
Mass vs Weight
Weight is a force in Newtons, not a mass in kilograms.(In US units, weight is pounds, mass is slugs = 32.2 pounds).
Your weight on the Moon is 1/6 as much as on Earth, but your mass is the same.
The mass in is inertial mass.The mass in is gravitational mass.
Experimentally, all objects of any composition have the same acceleration under gravity, so inertial and gravitational mass are the same.
W = Mg
F = maF = GMm r2
Phys 170 Lecture 23 5
13.2, 13.4: Equation of Motionand Inertial Reference Frames
If there is more than 1 force, use the vector sum:
is only valid in a coordinate system that is not rotating or accelerating. This is called an inertial coordinate system.
Coordinate systems moving with constant speed in a constant direction are OK.
The rules for translating between inertial coordinate systems leave accelerations unchanged (they just change velocities).
Fi
i∑ = ma
F = ma
Phys 170 Lecture 23 6
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 7
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
N
W
F+
F−
Phys 170 Lecture 23 8
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
F+
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
a
N
W
F+
F−
Phys 170 Lecture 23 9
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
a
N
W
F+
F−
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 10
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
a
N
W
F+
F−
N − mgcosθ − masinθ = 0 + is up (and a bit left)±F + mgsinθ = macosθ + is left (and a bit down)F = µN
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 11
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 12
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 13
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ−µmasinθ + macosθ = µmgcosθ + mgsinθ
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 14
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ−µmasinθ + macosθ = µmgcosθ + mgsinθ−µ sinθ + cosθ( )a = µ cosθ + sinθ( )g
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 15
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ−µmasinθ + macosθ = µmgcosθ + mgsinθ−µ sinθ + cosθ( )a = µ cosθ + sinθ( )g
a = g µ cosθ + sinθ−µ sinθ + cosθ
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 16
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ−µmasinθ + macosθ = µmgcosθ + mgsinθ−µ sinθ + cosθ( )a = µ cosθ + sinθ( )g
a = g µ cosθ + sinθ−µ sinθ + cosθ
= 9.8 0.2 ⋅0.93969 + 0.34202−0.2 ⋅0.34202 + 0.93969
= 5.961 m/s2
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 17
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ−µmasinθ + macosθ = µmgcosθ + mgsinθ−µ sinθ + cosθ( )a = µ cosθ + sinθ( )g
a = g µ cosθ + sinθ−µ sinθ + cosθ
= 9.8 0.2 ⋅0.93969 + 0.34202−0.2 ⋅0.34202 + 0.93969
= 5.961 m/s2
a = v2 ρ→ v = aρ = 5.961⋅100v = 24.41 m/s
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 18
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 19
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ−µmasinθ − macosθ = µmgcosθ − mgsinθ
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 20
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ−µmasinθ − macosθ = µmgcosθ − mgsinθ−µ sinθ − cosθ( )a = µ cosθ − sinθ( )g
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 21
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ−µmasinθ − macosθ = µmgcosθ − mgsinθ−µ sinθ − cosθ( )a = µ cosθ − sinθ( )g
a = g µ cosθ − sinθ−µ sinθ − cosθ
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 22
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ−µmasinθ − macosθ = µmgcosθ − mgsinθ−µ sinθ − cosθ( )a = µ cosθ − sinθ( )g
a = g µ cosθ − sinθ−µ sinθ − cosθ
= 9.8 0.2 ⋅0.93969 − 0.34202−0.2 ⋅0.34202 − 0.93969
= 1.498 m/s2
PROBLEMS 13-53 and 13-54 (page 138, 12th edition)
A 1700 kg sports car travels around a 100 m radius horizontal
circle on a 20o banked track. The coefficient of static friction
between the tires and the road is 0.20.
13-53: Determine the maximum constant speed the car can travel
without sliding up the slope.
13-54: Determine the minimum constant speed the car can travel
without sliding down the slope.
Phys 170 Lecture 23 23
a
N
W
F+
F−
N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ
µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ−µmasinθ − macosθ = µmgcosθ − mgsinθ−µ sinθ − cosθ( )a = µ cosθ − sinθ( )g
a = g µ cosθ − sinθ−µ sinθ − cosθ
= 9.8 0.2 ⋅0.93969 − 0.34202−0.2 ⋅0.34202 − 0.93969
= 1.498 m/s2
a = v2 ρ→ v = aρ = 1.498 ⋅100v = 12.23 m/s
Phys 170 Lecture 23 24
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 25
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
WB
F
N
TB
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 26
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
WA
WB
F
N
TB
TA
TA TA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 27
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
WA
WB
F
N
aB aA
TB
TA
TA TA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 28
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
TB
TA
TA TA
WA
WB
F
N
aB aA
3TA +
WA = mA
aATB +
WB +
F +N = mA
aB
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 29
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
TB
TA
TA TA
WA
WB
F
N
aB aA
3TA +
WA = mA
aATB +
WB +
F +N = mA
aB
YA : 3T − mAg = mAaA′Y B : T − mBgsin60° − F = mBaB′X B : N − mBgcos60° = 0
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 30
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
TB
TA
TA TA
WA
WB
F
N
aB aA
YA : 3T − mAg = mAaA′Y B : T − mBgsin60° − F = mBaB′X B : N − mBgcos60° = 0
Friction: F = µN = 0.4NPulleys: aB = −3aA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 31
TB
TA
TA TA
WA
WB
F
N
aB aA
3T − mAg = mAaAT − mBgsin60° − F = mBaBN − mBgcos60° = 0F = µNaB = −3aA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 32
TB
TA
TA TA
WA
WB
F
N
aB aA
3T − mAg = mAaAT − mBgsin60° − F = mBaB → −3mBaAN − mBgcos60° = 0F = µNaB = −3aA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 33
TB
TA
TA TA
WA
WB
F
N
aB aA
3T − mAg = mAaAT − mBgsin60° − F = mBaB → −3mBaAN − mBgcos60° = 0→ N = mBgcos60°F = µNaB = −3aA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 34
TB
TA
TA TA
WA
WB
F
N
aB aA
3T − mAg = mAaAT − mBgsin60° − F = mBaB → −3mBaAN − mBgcos60° = 0→ N = mBgcos60°F = µN → F = µmBgcos60°aB = −3aA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 35
TB
TA
TA TA
WA
WB
F
N
aB aA
3T − mAg = mAaAT − mBgsin60° − µmBgcos60° = −3mBaA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 36
TB
TA
TA TA
WA
WB
F
N
aB aA
3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 37
TB
TA
TA TA
WA
WB
F
N
aB aA
3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA→ mA g + aA( ) 3− mBgsin60° − µmBgcos60° = −3mBaA
Phys 170 Lecture 23 38
3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA→ mA g + aA( ) 3− mBgsin60° − µmBgcos60° = −3mBaAmA
3+ 3mB
⎛⎝⎜
⎞⎠⎟ aA +
mAg3
− mBg sin60 + µ cos60( ) = 0
Phys 170 Lecture 23 39
3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA→ mA g + aA( ) 3− mBgsin60° − µmBgcos60° = −3mBaAmA
3+ 3mB
⎛⎝⎜
⎞⎠⎟ aA +
mAg3
− mBg sin60 + µ cos60( ) = 0
aA = gmB sin60 + µ cos60( )− mA 3
mA 3+ 3mB
Phys 170 Lecture 23 40
3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA→ mA g + aA( ) 3− mBgsin60° − µmBgcos60° = −3mBaAmA
3+ 3mB
⎛⎝⎜
⎞⎠⎟ aA +
mAg3
− mBg sin60 + µ cos60( ) = 0
aA = gmB sin60 + µ cos60( )− mA 3
mA 3+ 3mB
aA = 9.860 0.8660 + 0.4 ⋅0.5000( )−100 3
100 3+ 3 ⋅60
Phys 170 Lecture 23 41
3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA→ mA g + aA( ) 3− mBgsin60° − µmBgcos60° = −3mBaAmA
3+ 3mB
⎛⎝⎜
⎞⎠⎟ aA +
mAg3
− mBg sin60 + µ cos60( ) = 0
aA = gmB sin60 + µ cos60( )− mA 3
mA 3+ 3mB
aA = 9.860 0.8660 + 0.4 ⋅0.5000( )−100 3
100 3+ 3 ⋅60= 9.8 63.96 − 33.33
213.33
Phys 170 Lecture 23 42
3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA→ mA g + aA( ) 3− mBgsin60° − µmBgcos60° = −3mBaAmA
3+ 3mB
⎛⎝⎜
⎞⎠⎟ aA +
mAg3
− mBg sin60 + µ cos60( ) = 0
aA = gmB sin60 + µ cos60( )− mA 3
mA 3+ 3mB
aA = 9.860 0.8660 + 0.4 ⋅0.5000( )−100 3
100 3+ 3 ⋅60= 9.8 63.96 − 33.33
213.33aA = 1.4069m/s2
Phys 170 Lecture 23 43
3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA→ mA g + aA( ) 3− mBgsin60° − µmBgcos60° = −3mBaAmA
3+ 3mB
⎛⎝⎜
⎞⎠⎟ aA +
mAg3
− mBg sin60 + µ cos60( ) = 0
aA = gmB sin60 + µ cos60( )− mA 3
mA 3+ 3mB
aA = 9.860 0.8660 + 0.4 ⋅0.5000( )−100 3
100 3+ 3 ⋅60= 9.8 63.96 − 33.33
213.33aA = 1.4069m/s2 But this is wrong!
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 44
TB
TA
TA TA
WA
WB
F
N
aB aA
aA = 1.4069 m/s2 → aΒ = −4.2209 m/s2
If B is actually moving down, the sign of kinetic friction flips!
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 45
TB
TA
TA TA
WA
WB
F
N
aB aA
aA = 1.4069 m/s2 → aΒ = −4.2209 m/s2
If B is actually moving down, the sign of kinetic friction flips!3T − mAg = mAaAT − mBgsin60° + F = mBaBN − mBgcos60° = 0F = µNaB = −3aA
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 46
TB
TA
TA TA
WA
WB
F
N
aB aA
aA = 1.4069 m/s2 → aΒ = −4.2209 m/s2
If B is actually moving down, the sign of kinetic friction flips!3T − mAg = mAaAT − mBgsin60° + F = mBaBN − mBgcos60° = 0F = µNaB = −3aA
3T − mAg = mAaAT − mBgsin60° + µmBgcos60° = −3mBaA
aA = gmB sin60 − µ cos60( )− mA 3
mA 3+ 3mB
PROBLEM 13-27 (page 126, 12th edition)
The mass of block A is 100 kg. The mass of block B is 60 kg.
The coefficient of kinetic friction between block B and the inclined
plane is 0.4. A and B are released from rest.
• Determine the acceleration of block A and the tension in the
cord. Neglect the mass of the pulleys and the cord.
Phys 170 Lecture 23 47
TB
TA
TA TA
WA
WB
F
N
aB aA
aA = gmB sin60 − µ cos60( )− mA 3
mA 3+ 3mB
aA = 9.860 0.8660 − 0.4 ⋅0.5000( )−100 3
100 3+ 3 ⋅60= 9.8 39.96 − 33.33
213.33aA = 0.3045 m/s2
T = mA g + aA( ) 3 = 336.82 N
Phys 170 Lecture 23 48
For Next Time
Good Luck on the Midterm!
Make sure you know where to go
Bring ruler etc for free-body diagrams
Bring 1 page both sides handwritten notes, no example problems, and turn it in with exam book
Bring calculator (graphing programmable OK)