Physics 170 Lecture 23 Chapter 13 - Kinetics (Dynamics)mattison/Courses/Phys170/p170-23.pdf ·...

Phys 170 Lecture 23 1

Physics 170 Lecture 23

Chapter 13 - Kinetics (Dynamics)

OverviewExample problems


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.


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


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


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


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−










F+









a

N

W

F+

F−


















a

N

W

F+

F−


















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










a

N

W

F+

F−

N − mgcosθ − masinθ = 0±µN + mgsinθ = macosθ










a

N

W

F+

F−


µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ










a

N

W

F+

F−


µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ−µmasinθ + macosθ = µmgcosθ + mgsinθ










a

N

W

F+

F−


µN − µmasinθ = µmgcosθµN − macosθ = −mgsinθ−µmasinθ + macosθ = µmgcosθ + mgsinθ−µ sinθ + cosθ( )a = µ cosθ + sinθ( )g










a

N

W

F+

F−



a = g µ cosθ + sinθ−µ sinθ + cosθ










a

N

W

F+

F−




= 9.8 0.2 ⋅0.93969 + 0.34202−0.2 ⋅0.34202 + 0.93969

= 5.961 m/s2










a

N

W

F+

F−




= 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










a

N

W

F+

F−


µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ










a

N

W

F+

F−


µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ−µmasinθ − macosθ = µmgcosθ − mgsinθ










a

N

W

F+

F−


µN − µmasinθ = µmgcosθ−µN − macosθ = −mgsinθ−µmasinθ − macosθ = µmgcosθ − mgsinθ−µ sinθ − cosθ( )a = µ cosθ − sinθ( )g










a

N

W

F+

F−



a = g µ cosθ − sinθ−µ sinθ − cosθ










a

N

W

F+

F−




= 9.8 0.2 ⋅0.93969 − 0.34202−0.2 ⋅0.34202 − 0.93969

= 1.498 m/s2










a

N

W

F+

F−




= 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


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






































WA

WB

F

N

TB

TA

TA TA






































WA

WB

F

N

aB aA

TB

TA

TA TA






































TB

TA

TA TA

WA

WB

F

N

aB aA

3TA +

WA = mA

aATB +

WB +

F +N = mA

aB






































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






































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








TB

TA

TA TA

WA

WB

F

N

aB aA

3T − mAg = mAaAT − mBgsin60° − F = mBaBN − mBgcos60° = 0F = µNaB = −3aA








TB

TA

TA TA

WA

WB

F

N

aB aA

3T − mAg = mAaAT − mBgsin60° − F = mBaB → −3mBaAN − mBgcos60° = 0F = µNaB = −3aA








TB

TA

TA TA

WA

WB

F

N

aB aA

3T − mAg = mAaAT − mBgsin60° − F = mBaB → −3mBaAN − mBgcos60° = 0→ N = mBgcos60°F = µNaB = −3aA








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








TB

TA

TA TA

WA

WB

F

N

aB aA

3T − mAg = mAaAT − mBgsin60° − µmBgcos60° = −3mBaA








TB

TA

TA TA

WA

WB

F

N

aB aA

3T − mAg = mAaA → T = mA g + aA( ) 3T − mBgsin60° − µmBgcos60° = −3mBaA








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


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



3+ 3mB

⎛⎝⎜

⎞⎠⎟ aA +

mAg3

− mBg sin60 + µ cos60( ) = 0

aA = gmB sin60 + µ cos60( )− mA 3

mA 3+ 3mB



3+ 3mB

⎛⎝⎜

⎞⎠⎟ aA +

mAg3

− mBg sin60 + µ cos60( ) = 0


mA 3+ 3mB

aA = 9.860 0.8660 + 0.4 ⋅0.5000( )−100 3

100 3+ 3 ⋅60



3+ 3mB

⎛⎝⎜

⎞⎠⎟ aA +

mAg3

− mBg sin60 + µ cos60( ) = 0


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



3+ 3mB

⎛⎝⎜

⎞⎠⎟ aA +

mAg3

− mBg sin60 + µ cos60( ) = 0


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



3+ 3mB

⎛⎝⎜

⎞⎠⎟ aA +

mAg3

− mBg sin60 + µ cos60( ) = 0


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!








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!








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








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








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


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)

Physics 170 Lecture 23 Chapter 13 - Kinetics (Dynamics)mattison/Courses/Phys170/p170-23.pdf ·...

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