phys10-chap5-applyingnewtonslaws

7/26/2019 Phys10-Chap5-ApplyingNewtonsLaws

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APPLYING NEWTONS LAWS

->systems in

equilbrium

(

F

= 0

)

-> zero acceleration

Example: A 1kg mass is hanged on a rope and tied

on the ceiling. Find the tension in the rope.

Solution:

x

y

T

w

k1

0 xF

0 wy TFwT

gmT

)/s(1kg)(9.8m 2

N8.9T

3

Example: Find the tension in each rope.

Solution:

FBD for B:

x

y

gwBB

m

1T

FBD for A:

x

y

gwAA

m

2T

0mB21

g-y TTF

m

B

=2kg

Rope 1

Rope 2

m

A

=3kg

2T

0mA2 g-y TF

gA2

mT

)/s(3kg)(9.8m 2

N4.292 T

gB21

mTT

)/s(2kg)(9.8m29.4N 2

N491T

Problem

1:

What

must

be

the

mass

of

A

in

orderfor

the

system

to

be

in

equilibrium?

What

is

thetension

in

the

rope?

B

A

Problem

2:

A

1-

kg

mass

is

hanged

on

the

ceilingusing 3 ropes (see

Figure).

Find the

tension ineach

rope.

1kg

45

o45

o

1 2

3


2/4

->systems with non-

zero acceleration (

F

= ma

)

5

Example: Find the acceleration of the system and

the tension in each rope.

Solution:

FBD for A:

x

y

gwAA

m

1T

FBD for B:

x

y

gwBB

m

2T

FBD for C:

x

y

gwCC

m

C

2T1T

agAA1

mm T

m

A=

1kg m

B=

3kg

Rope 1 Rope 2m

C=

2kg

a

a a

a a

a

agBB2

mm T

aC12

mTT

)1(

)2(

)3(



Solution:

agAA1

mm Tm

A=

1kg

Rope 1 Ropm

C=

2kg

a

a

agBB2

mm T

aC12

mTT

)1(

)2(

)3(

Eliminate

T

1

in

(1)

&

(3)

agAA1

mm T

aC12

mTT

aagCAA2

mmm T )4(

Eliminate

T

2

in

(4)

&

(2)

agBB2

mm T

aagCAA2

mmm T

aaaggBCABA

mmmmm

agCBAAB

mmmmm

a

BA

B

mm

mm

2kg1kg

1k3kg

2s

m27.3a

7



Solution:

m

A=

1kg m

B=

3kg

Rope 1 Rope 2m

C=

2kg

a

a a

To solve for

T

1

we use

(1)

agAA1

mm T

agAA1

mm T

)m/s(1kg)(3.27)/s(1kg)(9.8m 22

N07.131T

To solve for

T

2

we use

(2)

agBB2

mm T

agBB2

mm T

)m/s(3kg)(3.27)/s(3kg)(9.8m

22

N59.192T

Problem

1:

Find

the

acceleration

of

the

system (magnitude

&

direction)

and

the

te

the

rope.

Problem 2:

Black Spider-

Man tries to save RedSpider-

Man

by

pulling his

spider

web

with

a

horizontalforce

of

600

N

but Venom

suddenly pulled

them

down(using

his

weight)

and

they

allmove

with

an

accelerationof

3.

86

m/s

2.

What

must

be

the

mass

of

Venom

in

orderfor

him

to

do

that?

30

o

B

A

5kg

60kg

3.86m/s

2

600N


3/4

9

m

A=

1kg m

B=

3kg

Rope 1 Rope 2m

C=

2kg

a

a a

Example:

Find

the

acceleration

of

the

system.

Thecoefficient of

kinetic

friction

between block

C andthe

table

is

0.

30.

Solution:

FBD for A:

x

y

gwAA

m

1Ta

FBD for B:

x

y

gwBB

m

2T a

FBD for C:

x

y

gwCC

m

C

2T2T

a

f

agAA1

mm T agBB2

mm T afk C12 mTT

)1( )2( )3(

Ckkf

Ckk wf

gf Ckk m

agk CC12 mm TT

m

A=

1kg

Rope 1 Ropm

C=

2kg

a

a

Example:

Find

the

acceleration

of

the

system.

Thecoefficient of

kinetic

friction

between block

C

andthe

table

is

0.

30.

Solution:

agAA1

mm T

ag BB2 mm T

)1(

)2()3(agk CC12 mm TT

ga

CBA

CkAB

mmm

m-mm

2s

m29.2a

The previous answer (frictionless surface) is 3.27 m/s

2.

11

Problem

1:

What

must

be

the

coefficient

of

static

friction

s

in

order

for

the

system

toremain

at

rest

(in

equilibrium)?

30

o

B

A

2.5kg

Problem

2:

Find

the

acceleration

of

thesystem i f a ll b locks r ub aga inst

thesurfaces of the

table.

(The

coefficientof kinetic friction

between the

blocksand

the

table

surface

is

0.

40

)

60kg

5kg

A

45

60

o

1kgC


4/4

13

->

force

exerted

by

a

fluid

on

a

body

moving

through

it

->

friction

due

to

bodies

moving

through

a

fluid

->

Ex.

air

drag

(air

resistance)

2Dvf

For

an

object falling

through

the

air,

the

air

drag

is

mawfF

a

02

mgDvt

D

mgvt Terminal speed

w

f

mamgDv 2

Centripetal

(radial)

acceleration

R

va

rad

2

Centripetal

force

R

vmF

net

2

Beware

misconception

with

centrifugal force

(does

not

exist)

phys10-chap5-applyingnewtonslaws

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