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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)%