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Chapter 2 Force Systems-Worked Examples
1. A man pull with force of !! " on a rope attached to a #uildin$ as shown in fi$% what are the
hori&ontal and 'ertical components of the force exerted #y the rope at point A.
Solution(
tanb )*+,)!.% b=tan-1/!.0)*.,! a)!!-*.,!).1! possi#le to use #oth bor a
Fx)Fcos*.,!
)!!" cos*.,!
) !!sin .1!
Fy) -Fsin*.,! )-!!" sin*.,!)-!!" cos.1!
2.he ca#le A3 pre'ents #ar 4A from rotatin$ clockwise a#out the pi'ot 4. 5f the ca#le tension
!"%determine the n- and t-components of this force actin$ on point A of the #ar.
A32 ) 1.22 6 1.2 -271.271. cos12!! )2.8m
sina/1.2 = sin 1200/2.34 , a=26.30
n)sina
=750sin 26.3
0
=333 N Tt = -Tcosa
=-750cos26.3
0
= -672 N
1
300N
a
6m
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Chapter 2 Force Systems-Worked Examples
.5n the desi$n of a control mechanism% it is determined that rod A3 transmits a 2*!-" force 9 to
the crank 3C.:etermine the x and y scalar components of 9
Solution(
hypotenuse) 2 6122 )1
9x)- 9cos 21!)-2*!/12+10) -28! " 9y)-9sin21
! -2*!/+10) - 1!! "
8.:etermine the resultant ; of the two forces shown #y a0 applyin$ the parallelo$ram rule for
'ector addition #0 summin$ scalar components.
Solution:
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Chapter 2 Force Systems-Worked Examples
#0 ; x)SFx) *!!cos *!! -8!!) -1!! " >
; y)SFy ) *!!sin *!!6!) 2! "
S4 % ;) -1!!i 62! ? " @
.
*.5f the eual tension in the pulley ca#le are 8!! "% express in 'ector notation the force ;
exerted on the pulley #y the two tensions. :etermine the ma$nitude of ;.
Solution(
; @) SF@)8!!68!!cos*!!)*!! "
3
a
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Chapter 2 Force Systems-Worked Examples
; >)SF>)8!!sin*!!)8* "
; ) *!! i 68* ?
;) *!!2 68*2 )* "
.:etermine the resultant of the three forces #elow.
Solution(
∑ F x = 350 cos 25o + 800 cos 70
o - 600 cos 60
o
= 317.2 + 273.6 - 300 = 290.8 N
∑ F y = 350 sin 25o + 800 sin 70o + 600 sin 60o
= 147.9 + 751 + 519.6 = 1419.3 N
i.e. F = 290.8 N i + 1419.3 N j
Resultant, F
F N = + =
= =−
2!, 181 188
181
2!,-, 8
2 2
1 !
. .
tan.
..θ
F = 1449 N 78.4o
,.he two structural mem#ers% one of which is in tension and the other in compression% exert the
indicated forces on ?oint 4. :etermine the ma$nitude of the resultant ; of the two forces and the
an$le which ; makes with the positi'e x-axis./exercise0
4
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Chapter 2 Force Systems-Worked Examples
.A force F of ma$nitude 8!" is applied to the $ear. :etermine the moment of F a#out point 4.
Solution(
1!.A 2!! " force acts on the #racket as shown determine the moment of force a#out BA
( principle of moment0
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Chapter 2 Force Systems-Worked Examples
Given F)2!!" D ) 8o
Reui!e" A )
Solution ;esol'e the force into components F1 am F2
#1) F cosD %#1)2!! cosine 8o =141.42N.
#2) F sin D% #2 ) 2!! sin 8o= 2.46$N.We know that A ) !
A ) moment produce due to component #16 moment produce due to component #2. )#1 x r16 F2 x !2.
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Chapter 2 Force Systems-Worked Examples
Given F1)F2 )!l# F ) F8 ) 12!l#.
Reui!e" oment of couple ) )
Solution he moment of couple can #e determined at any point for example at A% 3 or :.
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Chapter 2 Force Systems-Worked Examples
hus%
Fi$. % MO ) /1!! "0/2 m0 ) 2!! ". m CWFi$ & MO ) /! "0/!. m0 ) . ".m.CW
Fi$ c MO ) /8! l#0/8 ft 6 2 cos !J ft0 ) 22 l#.ft# CW
Fi$. " MO ) /*! l#0/1 sin 8J ft0 ) 82.8 l#.ft CCW
Fi$ e MO ) / k"0/8 m - 1 m0 ) 21.! k".m. CCW
18.:etermine the resultant moment of the four forces actin$ on the rod shown in Fi$. #elow
a#out point 4 .
S')T*'N
Assumin$ that positi'e moments act in the 6+ direction% i.e.% counterclockwise% we ha'e6 / MR0O ) J Fd K
/ MR0O ) -! "/2 m0 6 *! "/!0 6 2! "/ sin !! m0 -8! "/8 m 6 cos !!m0
/ MR0O ) -8 ". m ) 8 ".m CW
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Chapter 2 Force Systems-Worked Examples
For this calculation% note how the moment-arm distances for the 2!-" and 8!-" forces are
esta#lished from the extended /dashed0 lines of action of each of these forces.15.:etermine the moment produced #y the force # in Fi$. tree shown #elow a a#out point O .
Express the result as a Cartesian 'ector.
solution
As shown in Fi$. b % either ! A or ! B can #e used to determine the moment a#out point O . hese
position 'ectors are ! A ) L12+ M m and ! B ) L8i 6 12 M mForce # expressed as a Cartesian 'ector is
*u+
R
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Chapter 2 Force Systems-Worked Examples
N'T: As shown in Fi$. b % O acts perpendicular to the plane that contains #% ! A% and ! B . Nad
this pro#lem #een worked usin$ MO ) Fd % notice the difficulty that would arise in o#tainin$ the
moment arm
1*.wo forces act on the rod shown in Fi$. a . :etermine the resultant moment they create a#outthe flan$e at O . Express the result as a Cartesian 'ector.
solution9osition 'ectors are directed from point O to each force as shown in Fi$. b . hese 'ectors are
1. eermine *e momen o! *e !or"e in Fi(. be$o a abou #oin O .-PN7PL8 F MM8N/
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Chapter 2 Force Systems-Worked Examples
solution i
he moment arm d in Fi$. % can #e found from tri$onometry.
d ) / m0 sin ! ) 2.,, mhus%
M O ) Fd ) / k"0/2.,, m0 ) 18. k". m
Since the force tends to rotate or or#it clockwise a#out point O % the moment is directed into the
pa$e.solution ii
he x and y components of the force are indicated in Fi$. & . Considerin$ counterclockwise
moments as positi'e% and applyin$ the principle of moments% we ha'e 6 MO ) - Fxdy - Fydx
) -/ cos 8!O"0/ sin !!m0 - / sin 8! O"0/ cos !!m0) -18. O". m ) 18. O".m
solution iii
he x and y axes can #e set parallel and perpendicular to the rodPs axis as shown in Fi$. c . Nere
# x produces no moment a#out point O since its line of action passes throu$h this point.
herefore%6 MO ) - Fy dx
) -/ sin !k"0/ m0 ) -18. k".m ) 18. k".m CW1,.Force # acts at the end of the an$le #racket in Fi$ % . :etermine the moment of the force
a#out point/p .m0
solution i (sc%l%! %n%lsis
he force is resol'ed into its x and y components% Fi$. & % then
6 MO ) 8!! sin !! "/!.2 m0 - 8!! cos !! "/!.8 m0 ) -,.* ".m ) ,.* ".mor
O ) L-,.*+ M ".m
solution ii (vecto! %n%lsis
Gsin$ a Cartesian 'ector approach% the force and position 'ectors% Fi$. c % are ! ) L!.8i - !.2 M m
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Chapter 2 Force Systems-Worked Examples
# ) L8!! sin !!i - 8!! cos !! M " ) L2!!.!i - 8*.8 M "
he moment is therefore
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Chapter 2 Force Systems-Worked Examples
*.;eplace the two forces and couple #y a wrench. Find the moment of the wrench and the
coordinates of point 9 in the y-& plane throu$h which the force of the wrench passes.
A"S.
;)SF ) 2!!i 61!? "
Assume positi'e wrench so the direction cosines of m are those of r or !.,% !.*%!
S9 )2!!/!. -Q0? -2!!/!.-y0k 61! & i 61! /!.20k -!i
)/-! 61!Q0i 6/*!-2!!Q0? 6 /-!62!!y0k ".m
Euate the direction cosines of S9 and SF
/-!61!&0+)!.,
/*!-2!!&0+)!.*
/-!62!!y0+)!
Where euals the ma$nitude of S9
Sol'in$ For y)!.1m or y)1!mm
&)!.2*8m or &)2*8mm
++++ )/-!61!/!.2*800+!.,)12 ".m )12/!.,i 6 !.* ?0 ".m
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Chapter 2 Force Systems-Worked Examples
.:etermine and locate the resultant R of the two force and one couple actin$ on the 5-#eam.
>ou saw in the precedin$ lesson that the resultant of two forces may #e determinedeither $raphically or from the tri$onometry of an o#liue trian$le.
. en t!ee o! o!e o!ces %!e involve" , the determination of their resultant
R is #est carried out #y first resol'in$ each force into rectangular components.
wo cases may #e encountered% dependin$ upon the way in which each of the$i'en forces is defined(
%se 1. Te o!ce # is "eine" & its %nitu"e F %n" te %nle a it o!s
8it te x %9is. he x and y components of the force can #e o#tained #y multiplyin$
F #y cos a and sin a% respecti'ely RExample 1.
%se 2. Te o!ce # is "eine" & its %nitu"e F %n" te coo!"in%tes o
t8o oints A %n" B on its line o %ction / Fi$. 2.2 0. he an$le a that # formswith the x axis may first #e determined #y tri$onometry. Nowe'er% the components
of # may also #e o#tained directly from proportions amon$ the 'arious dimensions
in'ol'ed% without actually determinin$ a RExample 2.
;. Rect%nul%! coonents o te !esult%nt. he components R x and R y of the
resultant can #e o#tained #y addin$ al$e#raically the correspondin$ componentsof the $i'en forces RSample 9ro#. 2..>ou can express the resultant in vectorial form usin$ the unit 'ectors i and % which
are directed alon$ the x and y axes% respecti'ely(
R Rxi 1 Ry Alternati'ely% you can determine the magnitude and direction of the resultant #y
sol'in$ the ri$ht trian$le of sides R x and R y for R and for the an$le that R forms
with the x axis.
88888888S8O88.LVING PROBLEMS
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Chapter 2 Force Systems-Worked Examples
ON YO
U8998899999999999999999999999999R .:etermine the x and y components of each of the forces shown.
1!.Onowin$ that the tension in ca#le BC is 2 "% determine the resultant of the three forces
exerted at point B of #eam AB.
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Chapter 2 Force Systems-Worked Examples
11.he forces #1% #2% and #% all of which act on point A of the #racket% are specified in three
different ways. :etermine the x and y scalar components of each of the three forces.
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Chapter 2 Force Systems-Worked Examples
couple
1*.;eplace the 1!-k" force actin$ on the steel column #y an eui'alent forceIcouple system at
point O. his replacement is freuently done in the desi$n of structures.
1