Chapter 6 – Force and Motion IIChapter 6 – Force and Motion II

Fall 2009Fall 2009

I.I. Newton’s first law.Newton’s first law.

II.II. Newton’s second law.Newton’s second law.

III.III. Particular forces:Particular forces: - Gravitational- Gravitational

- Weight- Weight - Normal- Normal

- Tension- Tension

IV. Newton’s third law.IV. Newton’s third law.

Review of chapter 5 – Force Review of chapter 5 – Force and Motion Iand Motion I

In the figure below, mIn the figure below, mblockblock=8.5kg and =8.5kg and θθ=30º. Find (a) Tension in =30º. Find (a) Tension in the cord. the cord.

(b) Normal force acting on the block. (c) If the cord is cut, find the magnitude (b) Normal force acting on the block. (c) If the cord is cut, find the magnitude of the block’s acceleration.of the block’s acceleration.

TN

Fg

y

x

2/9.45.865.410)( smaaNmaFTc gx

NsmkgmgFaa gx 65.415.0)/8.9)(5.8(30sin0)( 2

NmgFNb gy 14.7230sincos)(

Chapter 6 – Force and Motion IIChapter 6 – Force and Motion II

I.I. Frictional forces.Frictional forces.

II.II. Drag forces and terminal speed.Drag forces and terminal speed.

III.III. Uniform circular motion.Uniform circular motion.

I. Frictional forceFrictional force

-Kinetic: Kinetic: (f(fkk)) appears after a large enough external force is appears after a large enough external force is

applied and the body loses its intimate contact with applied and the body loses its intimate contact with the surface, sliding along it.the surface, sliding along it.

No motion

Acceleration

Constant velocity

Counter force that appears when an external force tends to slide a body Counter force that appears when an external force tends to slide a body along a surface. It is along a surface. It is directeddirected parallel parallel to the to the surface surface and and oppositeopposite to the to the sliding motionsliding motion..

FF

(applied(applied force)force)

//Ffs

-Static: Static: (f(fss) compensates the applied force, the body does ) compensates the applied force, the body does

not move.not move.

max,sk ff Nf ss max,

Nf kk

Friction coefficientsFriction coefficientsslidesbodyfFIf s max,//

After the body starts sliding, fAfter the body starts sliding, fkk decreases decreases.

II. Drag force and terminal speedII. Drag force and terminal speed

-Fluid: anything that can flow. Example: gas, liquid.

-Drag force: DDrag force: D

- Appears when there is a relative velocity between a fluid and - Appears when there is a relative velocity between a fluid and a body.a body.- - OpposesOpposes the relative motion of a body in a fluid. the relative motion of a body in a fluid.

- Points in the direction in which the fluid flows.- Points in the direction in which the fluid flows.

2

2

1AvCD

AC

Fv gt

2

-Terminal speed: vTerminal speed: vtt

02

10 2 gg FAvCaifmaFD

AssumptionsAssumptions:: * Fluid = air.* Fluid = air. * Body is baseball.* Body is baseball. * Fast relative motion * Fast relative motion turbulent air. turbulent air.

C = drag coefficient (0.4-1).C = drag coefficient (0.4-1).ρρ = air density (mass/volume). = air density (mass/volume). A= effective body’s cross sectional area A= effective body’s cross sectional area area perpendicular to v area perpendicular to v

- Reached when the acceleration of an object that experiences a - Reached when the acceleration of an object that experiences a vertical movement through the air becomes zero vertical movement through the air becomes zero F Fgg=D=D

III. Uniform circular motionIII. Uniform circular motion

-Centripetal acceleration:Centripetal acceleration:

R

vmF

2

r

va

2

A centripetal force accelerates a body by changing the direction of the A centripetal force accelerates a body by changing the direction of the body’s velocity without changing its speed.body’s velocity without changing its speed.

v, a are constant, but direction changes during motionv, a are constant, but direction changes during motion.

-Centripetal force:Centripetal force:

a, F are directed toward the center of curvaturea, F are directed toward the center of curvatureof the particle’s path.of the particle’s path.

A puck of mass m slides on a frictionless table while attached to a hanging cylinder of mass M by a cord

through a hole in the table. What speed keeps the cylinder at rest?

Mg

mg T

TN

MgTMFor

m

Mgrv

r

vmMg

r

vmTmFor

22

Calculate the drag force on a missile 53 cm in diameter cruising with a speed of 250 m/s at low altitude, where the density of air is 1.2 kg/m3. Assume C=0.75

kNsmmmkgAvCD 2.6/250)2/53.0()/2.1(75.05.02

1 2232

The terminal speed of a ski diver is 160 km/h in the spread eagle position and 310 km/h in the nose-

dive position. Assuming that the diver’s drag coefficient C does not change from one point to

another, find the ratio of the effective cross sectional area A in the slower position to that of the

faster position.

7.32

2

/310

/1602

D

E

E

D

D

g

E

g

gt A

A

A

A

AC

F

AC

F

hkm

hkm

AC

Fv

Two blocks of weights 3.6N and 7.2N, are connected by a massless string and slide down a 30º inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10; that between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string.

A

B

FgA

FgB

NA

NB

T

T

fk,B

fk,A

Movement

Block A Block BNA NB

fkAfkB

TFgxA

FgyA

FgxB

FgyB

T

Light block A leads

Light block A leadsLight block A leads

NT

sma

2.0

/49.3 2

Block B weighs 711N. The coefficient of static friction between the block and the table is 0.25; 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.

FgA

N

FgB

T1

f

T3

T1

T2

T3

NfstationarySystem ss max,

NNTfT

gmNBBlock

s

B

75.17771125.00 1max,1

322

2221

30sin

25.20530cos

75.17730cos

TTT

NN

TTTTKnot

y

x

NNTgmTABlock A 62.10225.2055.030sin23

P

Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μs between A and the table is 0.2. (b) Block C suddenly is lifted of A. What is the acceleration of block A if μk between A and the table is 0.15?

T

T

WB=22N

WA=44N

f

Wc

N

)1(00 max,

max,

NTfTaABlock

Nff

ss

ss

)2(220 NTmgTBBlock

NNT

Ns

1102.0

22)2()1(

NNNWWWNBABlocks CCA 6644110,

(a)

P

Blocks A and B have weights of 44N and 22N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μs between A and the table is 0.2. (b) Block C suddenly is lifted of A. What is the acceleration of block A if μk between A and the table is 0.15?

T

T

WB=22N

WA=44N

f

Wc

N

amgmT

amNT

NgmNdisappearsC

BB

Ak

A

44

(b)

NTaT

smaaT

172.222

/3.25.46.6 2

The two blocks (with m=16kg and m=88kg) shown in the figure below are not attached. The coefficient of static friction between the blocks is: μs=0.38 but the surface beneath the larger block is frictionless.

What is the minimum value of the horizontal force F required to keep the smaller block from slipping

down the larger block?

Mm

FaamF total

)2(00

)1(

mgNmgf

Mm

FmmaNFblockSmall

ss

NM

MmmgFmg

Mm

FM

ss 488)2()1(

Mg

N

N

mg

fN

Movement

FFminmin required to keep m from sliding down? required to keep m from sliding down?

Treat both blocks as a single system sliding across a frictionless floor

mg

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5 kN, and the radius of the circle is 10 m.

What are the magnitude and the direction of the force of the boom on the car at the top of the circle if the car’s speed is (a) 5m/s (b) 12 m/s?

Rg

vWF

R

vmWF BB

22

1

The force of the boom on the car is capable of pointing any direction

FB

W

y

downFsmvb

upNFsmva

B

B

3.2/12)(

7.3/5)(