01 Torque.pptx

(1)

(2)

(3)

I can . . .

• _{differentiate between}

_torque

_and

_force

_.

• _{differentiate between}

_torque

_and

_work

_.

• _quantify

_torque

_{mathematically.}

• _{explain how the}

_{length of a lever arm}

affects

torque

.

• extend Newton’s 2

nd

Law of Motion and

(4)

TORQUE ≡

A

FORCE

(acting at a DISTANCE from some pivot)

that causes

ROTATION

F

o

rc

e

(5)

The

line of action

of a

force

is an imaginary

line of indefinite length drawn along the

direction of the force.

The

line of action

of a

force

is an imaginary

line of indefinite length drawn along the

direction of the force.

F

₁

F

2 F

₃

line of

(6)

The

line of action

of a

force

is an imaginary

line of indefinite length drawn along the

direction of the force.

The

line of action

of a

force

is an imaginary

line of indefinite length drawn along the

direction of the force.

F

₁

F

2 F

₃

The

lever arm

of a force is the perpendicular

distance from the line of action of a force to

the axis of rotation.

The

lever arm

of a force is the perpendicular

distance from the line of action of a force to

the axis of rotation.

r

₂

(7)

The

line of action

of a

force

is an imaginary

line of indefinite length drawn along the

direction of the force.

The

line of action

of a

force

is an imaginary

line of indefinite length drawn along the

direction of the force.

F

₁

F

2 F

₃

The

lever arm

of a force is the perpendicular

distance from the line of action of a force to

the axis of rotation.

The

lever arm

of a force is the perpendicular

distance from the line of action of a force to

the axis of rotation.

r

₂

r

₁

(8)

(9)

(10)

(11)

TORQUE

≡

Force

· Distance

┴

UNITS

METRIC:

Newton

· Meter

(

N

· m

)

(12)

Force

Distance

Force

Distance

100 N

D

is

ta

nc

_e

2.5 m

10

0 N

TORQUE

=

Force

· Distance

┴

=

100 N

· 2.5 m

=

250 N·m

2.5 m

₁₀₀

N

27 °

τ

=

(

100 ∙

SIN(27)

)∙2.5

27 °

2.5 m

τ

=

100 ∙(2.5∙

SIN(27)

)

τ

=

113 N∙m

(13)

100 N

D

is

ta

nc

_e

27 °

(14)

Force

TORQUE ≡

Force

· Distance

┴

WORK ≡

Force

· Distance

(15)

(16)

“Give me a lever long enough

and a fulcrum on which to

place it, and I can move the

(17)

TORQUE

=

Force

· Distance

┴

70 kg

x

_{10 m}

5.85 x 10^25 N

τ

man

-

τ

earth

>

0 (

(70

· 9.8 N)

· x

) > (

(5.85 x 10^25 N)

· (10 m)

)

(

(686 N)

· x

) > (

(5.85 x 10^25 N)

· (10 m)

)

(

(686 N)

· x

) >

5.85 x 10^26 N m

x

>

8.53 x 10^23 m

τ

man

>

τ

earth `

x

>

5.70 x 10^12 AU

x

>

5.30 x 10^20 mi

(18)

(19)

(20)

(21)

10°

100 N

20 cm

10°

TORQUE

=

Force

· Distance

┴

100∙

SIN(10)

100∙

COS(10)

TORQUE

=

-

100 ∙

SIN(10)

· .2

TORQUE

=

-3.47 N m

TORQUE

=

Force

(22)

(23)

(24)

(25)

(26)

http://www.army-technology.com/projects/chinook/chinook1.html

(27)

http://www.turbosquid.com/Previews/Content_on_9_17_2002_18_40_00%5CSocket_thumbnail1.JPGDB2BD0B0-5C16-4D3C-88D55E8672FE2851.jpgLarge.jpg

TORQUE

≡

Force

· Distance

(28)

http://www.turbosquid.com/Previews/Content_on_9_17_2002_18_40_00%5CSocket_thumbnail1.JPGDB2BD0B0-5C16-4D3C-88D55E8672FE2851.jpgLarge.jpg

TORQUE

≡

Force

· Distance

┴

a) yes

b) no

(29)

(30)

Why are wrenches for bigger bolts also longer?

Draw a clear diagram that shows why.

f

_f

f

_f

(31)

a)

b)

Which door would require

more

torque

to open?

(32)

(33)

(34)

(35)

(36)

(37)

http://www.ne.jp/asahi/ernie/car/wheelr.JPG

Distance

┴

Distance

┴

Torque

Force

282 lbs∙ft

1.04 ft

TORQUE

≡

Force

· Distance

┴

282 ≡

Force

·

1.04 Force

282 ≡

≡

Force

271 lbs

·

1.13 Force

≡

250 lbs

1.13 ft

(38)

http://www.local16.com/web_gallery/pages/convention_40s.htm

(39)

By Convention

Clockwise

Torque is

NEGATIVE ( - )

Counterclockwise

Torque is

POSITIVE ( + )

(40)

-

₊

-+

?

(41)

(42)

(43)

(44)

(45)

(46)

(47)

(48)

An object will continue in its

present state of

LINEAR

MOTION

unless acted upon by a NET

FORCE

.

NET TORQUE = Zero

implies

no change in ROTATIONAL MOTION

An object will continue in its

present state of ROTATIONAL MOTION

unless acted upon by a NET

TORQUE

.

Newton’s FIRST Law of Motion

(49)

(50)

(51)

(52)

(53)

(54)

(55)

100g

200g

50 g

http://www1.iwvisp.com/tronagemclub/ _{http://allenmugs.com/beads.htm}

-

+

= ZERO

(56)

100g

200g

50 g

(57)

(58)

(59)

(60)

(61)

http://www.neon-das.com/481.gif http://www.detroityes.com/news/071014/601pics/AlleyCult-01-05748.jpg

First Condition of Equilibrium

Net Force = Zero

Second Condition of Equilibrium

Net Torque = Zero

x

x x

UP = DOWN

LEFT = RIGHT

(62)

http://store.digitalfaucet.com/detail/1/98/10/1/0/Folding_Ladder.html

?

₁

?

₂

160 lbs.

(63)

?

₁

?

2 6 ft

_{4 ft}

160 lbs.

0 = -160(6) + ?

₂

(10)

?

₂

= 96 lbs

(64)

?

₁

?

2 6 ft

_{4 ft}

160 lbs.

0 = -?

₁

(10) + 160(4)

?

₁

= 64 lbs

(65)

?

₁

?

2 6 ft

_{4 ft}

160 lbs.

?

₁

= 64 lbs

?

₂

= 96 lbs

(66)

43 lbs

25 lbs

(4.5 – x)

x

68 lbs

4.5 ft

0 =

43

·

4.5 -

68

· x

x

=

2.8

0 =

-

25

· x

+

43 (

4.5 –

x

)

0 =

-

25

· x

+

193.5 –

43 x

0 =

- 68

x

+

193.5

68 x

=

193.5

(67)

Lift = ?

C

e

n

te

r

o

f

G

ra

v

it

y

Up = ?

70.0 m

35.7 m

5.70 m

800,000 lbs.

Down = ?

D

(

35.7 -

5.70 ) –

800,000

(

5.70 ) = 0

D

=

152,000 lbs.

L

=

152,000

+

800,000

Center

of Mass

30.0 D

=

(68)

Lift = ?

C

e

n

te

r

o

f

G

ra

v

it

y

Up = ?

70.0 m

35.7 m

3.71 m

800,000 lbs.

Down = ?

D

(

35.7 -

3.71 ) –

800,000

(

3.71 ) = 0

D

=

92,800 lbs.

L

=

92,800

+

800,000

Center

of Mass

31.99 D

=

(69)

First Condition of Equilibrium

, (

ΣF = 0

)

Second Condition of Equilibrium,

(

Στ = 0

)

F

₁

F

₂

W

0.50 m

F

₂

(0.75) -

W

(0.50) = 0

F

₁

+

F

₂

–

W

= 0

F

₂

(0.75) =

W

(0.50)

F

₂

=

W

(0.67)

F

₁

=

W

–

F

₂

(70)

80 .0

N

20 0 N

70

0 N

-

700 (x) - 200(3.0) -

80.0 (6.0) +

900 sin(60.0⁰)6.00 = 0

6.00 m

First Condition of Equilibrium

, (

ΣF =

0 )

Second Condition of Equilibrium,

(

Στ = 0

)

F

₁

F

2 -

700 - 200 -

80.0 +

900 sin(60.0⁰) +

F

₂

= 0

Y

F

₁

-

900 cos(60.0⁰) = 0

X

x

3.00 m

x = 5.18

90

(71)

(72)

(73)

(74)

An

8.0m

,

200N

uniform ladder rests against a smooth

wall. The coefficient of static friction between the

ladder and the ground is

0.60 , and the ladder makes

a

50.0 ⁰

angle with the ground. How far up the ladder

can an

800-N

person climb before the ladder begins

to slip and rotate?

4m

x

200N

800N

F

_N1

F

_N2

F

_f

50 ⁰

8m

First Condition of Equilibrium

, (

ΣF =

0 )

F

_N1

–

F

_f

= 0

F

_N2

–

800 –

200 = 0

Second Condition of Equilibrium,

(

Στ = 0

)

800 (xcos

50 ) +

200 (4cos

50 ) -

F

_N1

(8sin

50 ) = 0

(75)

Center of Gravity

The

center of gravity

of an object is the point

at which all the weight of an object might be

considered as acting for purposes of treating

forces and torques that affect the object.

(76)

m

₁

m

₂

m

₃

m

₁

m

₂

m

₃

Use TORQUE to find the

Center of Gravity

x

₁

_x

2 x

₃

y

₁

_y

2 y

₃

y

x

m

_T

m

_T

· x

=

m

₁

x

₁

+

m

₂

x

₂

+

m

₃

x

₃

m

_T

· y

=

m

₁

y

₁

+

m

₂

y

₂

+

m

₃

y

₃

m

₁

y

₁

+

m

₂

y

₂

+

m

₃

y

₃

m

_T

y

=

m

₁

x

₁

+

m

₂

x

₂

+

m

₃

x

₃

m

_T

· x

=

y

₁

y

₂

_y

(77)

Carpenters Square

x

y

x

₁

m

₁

x

₂

(78)

𝑥

_𝑐𝑚

=

16 𝐻

(

0 )

+

2 ¿ ¿

𝑥

_𝑐𝑚

=

.120363

𝐻

18 𝐻

.0067 nm

A water molecule consists of an oxygen atom with two hydrogen atoms bound

to it, as shown in the figure. The bonds are 0.100 nm in length and the angle

between the two bonds is 106°. Use the x-y axis shown and determine the

location of the center of gravity of the molecule. Consider the mass of an

oxygen atom to be 16 times the mass of a hydrogen.

x

y

(79)

f(x) = 9 – x

2 g(x) = x + 3

(80)

f(x) = 9 – x

2 g(x) = x + 3

(81)

f(x) = 9 – x

2 g(x) = x + 3

(82)

f(x) = 9 – x

2 g(x) = x + 3

(83)

f(x) = 9 – x

2 g(x) = x + 3

Find the center of mass for the shape

defined by the two functions.

f(x)

g(x)

dx

x

Mass

Distance

Mass

y



mass





.

5 M

x

5 



mass

M

y











2

3 )

(

)

(

x

g

x

dx

f

mass



f

x

g

x

_

_f

_x

_g

_x

_

_dx

M

_y

(

)

(

)

2 )

(

)

(

2

3 























2

3 )

(

)

(

x

g

x

dx

f

(84)

I can . . .

• _{differentiate between}

_torque

_and

_force

_.

• _{differentiate between}

_torque

_and

_work

_.

• _quantify

_torque

_{mathematically.}

• _{explain how the}

_{length of a lever arm}

affects

torque

.

• extend Newton’s 2

nd

Law of Motion and