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revision

EquilibriumVectors & Combining Forces

Force resolutionUnknown force

Moment of a force

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staticsForce is a vector having both magnitude and

direction as well as a point of application

Free !od" #iagramsVector #iagrams

$ind Fw % '

k(

$eight Fsw % )'

k(

'

)'

*cale ) mm )k(

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ForceFrom (ewton+s second law of motion

,n earth the force e-erted b" a mass of ) kg is.

(! ,n earth g varies a bit

2m/1atkg1acceleratetoforcenewton1

onacceleratimassforce

s=

=

N9.81g1force ==

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( and k(/s a (ewton is relativel" small 0the force

e-erted on earth b" a stationar" apple1 wetend to use kilonewtons 0)k( % )''' (1

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equilibrium of forcesEqual in magnitude

,pposite in direction

Co2linear

F)% )' k( F % )' k(

F) % )' k(

F % )' k(

F) % )' k(

F % )' k(

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resultants of forces

F) % )' k(

3esultant and Equilibrant

F% )' k(

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problems'k(

)4k(

)4k(

)'

k('k(

)4k(

'k(

647647

)'k(

8'

k(

)4k(

9'7

- -- -

- -

9'7 9'7

647

)'

k(

'k(

'k(

8'7

- -9'7647

)4k(

8'7

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triangle of forces:hree co2planar forces in equilibrium ma" be

represented b" a triangle

:hree non2parallel forces in equilibrium mustbe concurrent

;''(

m

6m ;''(

3) % < k(

3% 8 k(

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tabular summation:he vertical component of the resultant % the

sum of vertical components

:he hori?ontal component of the resultant %

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contd in a spreadsheet

Force Value 5 Components

V 0k(1 = 0k(1

F) )' 8' 4'' ;99

F )4 64 2)'9) )'9)

F8 ' 9' )>8 )'''

F6 ' 8' )''' 2)>8

:,:/A* )>) ))B4

F) % )'

k(

F6% '

k(

F8 % '

k(

8'7

- -9'7647

F % )4

k(

8'7-

"

positive a-es

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resultant

)>)k(

9)7

3 % 6>;k(

))B4k(

Feedback from past e-am failures D :he h"potenuse must be bigger than the

other sides regardless of signs

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problemCalculate the resultant magnitude and

direction

F) % )'k(

F6% 4k(

F8 % )4

k(

8'7

- -

9'7647

F % '

k(

9'7

8'7

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unknown force

equilibrium

)'' k(

- -9'7647

F)%