worked example for engineering mechanics-i

8/17/2019 Worked Example for Engineering Mechanics-I

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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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.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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#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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; >)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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.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

5


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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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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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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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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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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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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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# ) L8!! sin !!i - 8!! cos !! M " ) L2!!.!i - 8*.8 M "

he moment is therefore

12


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*.;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

13


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.: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

14


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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.

15


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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.

16


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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

worked example for engineering mechanics-i

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