University Physics: Mechanics

University Physics: Mechanics

Ch6. Friction, Drag, and Centripetal ForceLecture 10

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

2012

10/2Erwin Sitompul University Physics: Mechanics

The figure below shows a coin of mass m at rest on a book that has been tilted at an angle θ with the horizontal. By experimenting, you find that when θ is increased to 13°, the coin is on the verge of sliding down the book, which means that even a slight increase beyond 13° produces sliding.What is the coefficient of static friction μs between the coin and the book?Hint: Draw the free-body diagram of the coin first.

Homework 8: Coin On A Book


Forces along the y axis:

Forces along the x axis:

net,y yF ma

net,x xF ma

N g cos 0F F

ssin 0gF f

N g cosF F

s Nsin 0gF F

N cosF mg

ssin cos 0mg mg

ssincos

tan

• Why zero?

• Why zero?

So, the coefficient of static friction is:

s tan13 0.231

hd

Solution of Homework 8: Coin On A Book


An object is kept in rest on an inclined surface. The angle θ is 26°, which is greater than the critical angle θc (μs = tanθc). Upon release, the object directly move and slide down to the bottom. It requires 4.29 s to reach the bottom, which is 18 m away from the initial point.Determine the coefficient of kinetic friction μk between the object and the surface.

Virtual Experiment: Determining μk

θ

18 m


A block of mass m = 3 kg slides along a floor while a force F of magnitude 12 N is applied to it at an upward angle θ. The coefficient of kinetic friction between the block and the floor is μk = 0.4. We can vary θ from 0 to 90° (with the block remains on the floor. What θ gives the maximum value of the block’s acceleration magnitude a?

→

Example: Blue Block


Forces along the y axis: Forces along the x axis:

net,y yF ma

net,x xF ma

N g 0yF F F

N sinF mg F kxF f ma

k NcosF F ma

kcos sinF Fa gm m

• What θ gives the maximum value of a?

• da/dθ = 0

Example: Blue Block


kcos sinF Fa gm m

ksin cosda F Fd m m

0

ktan 1

ktan

then, the derivative of a with respect to θ is

If a is given by

1tan (0.4)

21.80

Example: Blue Block


Block B in the figure below weighs 711 N. The coefficient of static friction between block and table is 0.25; angle θ is 30°. Assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary.

Example: Two Blocks


fs,max

→

Block B

Block A

Knot

TB

→

TA

→

TB

→

TA

→

Wall

TW

→

TW

→

FgA

→

Knot

TW

→

fs,max

→

FgA

→

FNB

→

FgB

→

Example: Two Blocks


Knot

TW

→

fs,max

→

FgA

→

TWx

TWyForces along the y axis:

Forces along the x axis:

net, 0yF

net, 0xF

W gA 0yT F

W s,max 0xT f

W AsinT m g θ

W s NBcosT F

W s BcosT m g

s BA

sin cosm gm g

s BA

sin cosWW

102.624 N

(0.25)(711) tan 30

A s B tanW W

Example: Two Blocks


A block of mass m1 on a rough, horizontal surface is connected to a ball of mass m2 by a lightweight cord over a lightweight, frictionless pulley as shown in the figure below.A force of magnitude F at an angle θ with the horizontal is applied to the block as shown and the block slides to the right. The coefficient of kinetic friction between the block and surface is μk. Find the magnitude of acceleration of the two objects.

Example: Multiple Objects


N g1 yF F F 2 ( )T m g a

k N 1cosF T F m a

fk

→ m1

m2

T→

T→

Fg2

→

FN

→

Fg1

→

F→

Fx

Fy

θ

Forces in m2

net, 2 2y yF m a

g2 2T F m a

Forces in m1

net, 1 1x xF m a

k 1xF T f m a

1 k NcosT F m a F

net, 0yF

N g1 0yF F F

N 1 sinF m g F



N 1 sinF m g F

2 ( )T m g a

1 k NcosT F m a F

2 1 k 1( ) cos ( sin )m g a F m a m g F

1 2 k k 1 2cos sinm a m a F F m g m g

1 2 k k 1 2( ) (cos sin ) ( )m m a F m m g

k k 1 2

1 2

(cos sin ) ( )F m m gam m



When the three blocks in the figure below are released from rest, they accelerate with a magnitude of 0.5 m/s2. Block 1 has mass M, block 2 has 2M, and block 3 has 2M.What is the coefficient of kinetic friction between block 2 and the table?

Example: Trio Blocks


1 ( )T M g a

2 2 ( )T M g a

FN

→

fk

→

T1

→

m1

T1

→

Fg1

→

m2

Fg2

→

T2

→

aa

a

m3

T2

→

Fg3

→

Forces in m1

net, 1 1y yF m a

1 g1T F Ma

Forces in m3

net, 3 3y yF m a

2 g3 2 ( )T F M a

N 2F Mg

Forces in m2

net, 2 2y yF m a

N g2 0F F

2 1 k N 2T T F Ma

net, 2 2x xF m a

2 1 k 2T T f Ma



1 ( )T M g a

2 2 ( )T M g a N 2F Mg2 1 k N 2T T F Ma k2 ( ) ( ) 2 2M g a M g a Mg Ma

k 2 2 ( ) ( ) 2Mg M g a M g a Ma

k

2 ( ) ( ) 22

M g a M g a MaMg

k5

2Mg Ma

Mg

52g ag

(9.8) 5(0.5)2(9.8)

20.372 m s



The Drag Force and Terminal Speed A fluid is anything that can flow – generally a gas or a liquid. When there is a relative velocity between a fluid and a body

(either because the body moves through the fluid or because the fluid moves past the body), the body experiences a drag force D that opposes the relative motion.

→

Here we examine only cases in which air is the fluid, the body is blunt rather than slender, and the relative motion is fast enough so that the air becomes turbulent (breaks up into swirls) behind the body.

In such cases, the magnitude of the drag force is related to the relative speed by an experimentally determined drag coefficient C according to

212D C Av ρ : air specific density

A : effective cross-sectional area of the bodyC : drag coefficient


The Drag Force and Terminal Speed


The Drag Force and Terminal Speed When a blunt body falls from rest through air, the drag force D

is directed upward. This upward force D opposes the downward gravitational force Fg on the body.

→→

gD F ma

212 0t gC Av F

If the body falls long enough, D eventually equals Fg. This means that a = 0, and so the body’s speed no longer increases. The body then falls at a constant speed, called the terminal speed vt.

2 gt

Fv

C A


The Drag Force and Terminal Speed

• Cyclists and downhill speed skiers try to maximize terminal speeds by reducing effective cross-sectional area


If a falling cat reaches a first terminal speed of 97 km/h while it is wrapped up and then stretches out, doubling A, how fast is it falling when it reaches a new terminal speed

Example: Falling Cat

,1

295km hg

t

Fv

C A

,2

22g

t

Fv

C A

95 67.18km h

2


A raindrop with radius R = 1.5 mm falls from a cloud that is at height h = 1200 m above the ground. The drag coefficient C for the drop is 0.6. Assume that the drop is spherical throughout its fall. The density of water ρw is 1000 kg/m3, and the density of air ρa is 1.2 kg/m3.(a) What is the terminal speed of the drop?(b) What would be the drop’s speed just before impact if there

were no drag force?

Example: Raindrop

2 gt

Fv

C A

(a) 2( )w w

a w

v gC A

343

2

2( )( )

w

a

r gC r

2( )w

a w

m gC A

83

w

a

rgC

38 (1000)(1.5 10 )(9.8)

3 (0.6)(1.2)

7.38 m s 26.56 km h

(b) 2 20, 02 ( )y yv v g y y 2(9.8)(0 1200)yv

153.36 m s 552.10 km h


Homework 9AIn the next figure, blocks A and B have weights of 44 N and 22 N, respectively.(a) Determine the minimum weight of block C to keep A from

sliding if μs, between A and the table is 0.20.(b) Block C suddenly is lifted off A. What is the acceleration of

block A if μk between A and the table is 0.15?


Homework 9B1. The figure shows a 1.0-kg University Physics book

connected to a 500-g tea mug. The book is pushed up the slope and reach a speed of 3.0 m/s before being released. The coefficients of friction are μs = 0.50 and μk = 0.20.(a) How far will the book slide upwards?(b) After the book reaches the highest point, will the

book stick to the surface, or will it slide back down?2. In downhill speed skiing, a skier is retarded by both

the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is θ = 40.0°, the snow is dry with a coefficient of kinetic friction μk = 0.04, the mass of the skier and equipment is m = 85.0 kg, the cross-sectional area of the (tucked) skier is A = 1.30 m2, the drag coefficient is C = 0.150, and the air density is 1.20 kg/m3. (a) What is the terminal speed? (b) If a skier can vary C by a slight amount dC by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed?

D

fk

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