chap two worked example engineering mecha-i

8/17/2019 CHAP TWO Worked Example Engineering Mecha-I

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1. A man pull with force of 300 N on a rope attached to a building as shown in fig, what are thehorizontal and vertical components of the force exerted by the rope at point A.

Solution

tanb !"#$!0.%&, b= tan '( )0.%&*!3".$%0 a !+0 0'3".$% 0!&3.(3 0 possible to use both b or a

x! cos3".$% 0!300N cos3".$% 0! 300sin &3.(3 0

y! ' sin3".$% 0 !'300N sin3".$% 0!'300N cos&3.(3 0

-. he cable A/ prevents bar A from rotating cloc1wise about the pivot . 2f the cable tension%&0N, determine the n' and t'components of this force acting on point A of the bar.

A/ - ! (.- - (.&- '-4(.-4(.& cos(-0 0 !-.35m

sin a /1.2 = sin 120 0/2.34 , a =26.3 0

n! sin a =750sin 26.30

=333 N T t = -Tcos a =-750cos26.30

= -672 N

300N

a

6m


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3.2n the design of a control mechanism, it is determined that rod A/ transmits a -"0'N force 6 tothe cran1 /7.8etermine the x and y scalar components of 6

Solution

9ypotenuse! & - (- - !(3

6 x!' 6 cos -( 0!'-"0)(-#(3*! '-50 N 6 y!'6sin-( 0 '-"0)(3*! ' (00 N

5.8etermine the resultant : of the two forces shown by a* applying the parallelogram rule forvector addition b* summing scalar components.

Solution:

;aws of cosines: - !"00 - 500 - '-4"004500cos "0 0! &-+N

;aws of sines

&-+#sin"00 !"00#sin a < , a= %+.(0

P=260 N


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b* : x! S x! "00cos "0 0 '500! '(00 N =

: y! S y ! "00sin "0 0 0! &-0 N

S , :! '(00i &-0 > N ?

&.

".2f the e@ual tension in the pulley cable are 500 N, express in vector notation the force :exerted on the pulley by the two tensions. 8etermine the magnitude of :.

Solution

: ? ! S ? !500 500cos"0 0!"00 N

a


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: = ! S = !500sin"0 0!35" N

: ! "00 i 35" >

:! "00 - 35" - !"+3 N

%. 8etermine the resultant of the three forces below.

Solution

∑ F x = 350 cos 25 o + 800 cos 70 o - 600 cos 60 o

= 317.2 + 273.6 - 300 = 290.8 N

∑ F y = 350 sin 25 o + 800 sin 70 o + 600 sin 60 o

= 147.9 + 751 + 519.6 = 1419.3 N

i.e. F = 290.8 N i + 1419.3 N j

Resultant, F

F N = + =

= =−

-+0 $ (5(+ 3 (55+

(5(+ 3

-+0 $%$ 5

- -

( 0

. .

tan .

..θ

F = 1449 N 78.4 o

$. he two structural members, one of which is in tension and the other in compression, exert theindicated forces on >oint . 8etermine the magnitude of the resultant : of the two forces and theangle which : ma1es with the positive x'axis.)exercise*


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+.A force of magnitude 50N is applied to the gear. 8etermine the moment of about point .

Solution

(0.A -00 N force acts on the brac1et as shown determine the moment of force about AB ( principle of moment*


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Given !-00N C ! 5& o

Re ui!e" DA !ESolution :esolve the force into components ( am -#1 ! cosC , #1 !-00 cosine 5& o =141.42N.#2 ! sin C, #2 ! -00 sin 5& o= 2.46$N.Fe 1now that DA ! 0DA ! moment produce due to component #1 moment produce due to component #2 . ! #1 x r( - x !2 .;et us consider that cloc1 wise moment is ve. DA ! ( x r( - x r- ! ' (5(.5- x 0.( -.5"$ x )0.( 0.(* ! ' (3."5$ N ! (3 ."5$ N anti cloc1 wise.((. 8etermine the moment of couple acting on the moment shown )7 G6;H*

GivenF1=200 N, L 1=4m F 2=200 N , L 2 = 2m.Required Moment o !ou"#e = M =$%or&in' Formu#a M = F ( r.Solution

Put t)e va#ue* in +or&in' ormu#aM= -F1(L1 / -F2 (L2 = -200N (4m / -200N(2m =- 00N.m -400N.m =-1200N.m = 1200N.m %12. 8etermine the moment of couple acting on the moment shown .


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Given ( ! - !+0lb 3 ! 5 ! (-0lb.Re ui!e" Doment of couple ! D!ESolution he moment of couple can be determined at any point for example at A, / or 8.;et us ta1e the moment about point / D / ! I :. ! ' ( x r ( J - x r -. ! ' +0)3* J (-0 )(* ! ' 3+0 lb ft:esult D / ! D A!D 8 !3+0 lb .ft counter cloc1 wise.Doment of couple )D* ! 3+0 lb.ft counter cloc1wise )77F*13. or each case illustrated in ig. below, determine the moment of the force about point O.Solution (scalar analysis)

he line of action of each force is extended as a dashed line in order to establish the moment armd. Also illustrated is the tendency of rotation of the member as caused by the force. urthermore,

the orbit of the force about O is shown as a colored curl. Assuming cloc1wise positive


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hus,ig.% D ! )(00 N*)- m* ! -00 N. m 7Fig & D ! )&0 N*)0.%& m* ! 3%.& N.m.7Fig c D ! )50 lb*)5 ft - cos 30ft* ! --+ lb.ftb 7Fig." D ! )"0 lb*)( sin 5&ft* ! 5-.5 lb.ft 77Fig e D ! )% 1N*)5 m ' ( m* ! -(.0 1N.m . 77F

(5. 8etermine the resultant moment of the four forces acting on the rod shown in ig. belowabout point .


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S' )T*'NAssuming that positive moments act in the + direction, i.e., countercloc1wise, we have

) MR*O ! K Fd L) MR*O ! '&0 N)- m* "0 N)0* -0 N)3 sin 30 0 m* '50 N)5 m 3 cos 30 0m*) MR*O ! '335 N. m ! 335 N.m 7F

or this calculation, note how the moment'arm distances for the -0'N and 50'N forces areestablished from the extended )dashed* lines of action of each of these forces.1 . 8etermine the moment produced by the force # in ig. tree shown below a about point O .Hxpress the result as a 7artesian vector.SolutionAs shown in ig. b , either ! A or ! B can be used to determine the moment about point O . hese

position vectors are ! A ! M(- + m and ! B ! M5i (- morce # expressed as a 7artesian vector is

)u*


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R

N'T : As shown in ig. b , O acts perpendicular to the plane that contains # , ! A, and ! B . 9adthis problem been wor1ed using MO ! Fd , notice the difficulty that would arise in obtaining themoment arm(". wo forces act on the rod shown in ig. a . 8etermine the resultant moment they create aboutthe flange at O . Hxpress the result as a 7artesian vector.

Solution6osition vectors are directed from point O to each force as shown in ig. b . hese vectors are

(%. 8etermine the moment of the force in ig. below a about point O .)62N726;H D DHN *


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Solution ihe moment arm d in ig. % can be found from trigonometry.

d ! )3 m* sin %&0 ! -.$+$ mhus,


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M O ! Fd ! )& 1N*)-.$+$ m* ! (5.& 1N. mSince the force tends to rotate or orbit cloc1wise about point O , the moment is directed into the

page.Solution ii

he x and y components of the force are indicated in ig. & . 7onsidering countercloc1wise

moments as positive, and applying the principle of moments, we have MO ! ' Fxdy ' Fydx! ')& cos 5& 0ON*)3 sin 30 0m* ' )& sin 5&0 ON*)3 cos 30 0m*! '(5.& ON. m ! (5.& ON.mSolution iii

he x and y axes can be set parallel and perpendicular to the rodPs axis as shown in ig. c . 9ere# x produces no moment about point O since its line of action passes through this point.

herefore, MO ! ' Fy dx

! ')& sin %&01N*)3 m* ! '(5.& 1N.m ! (5.& 1N.m 7F($. orce # acts at the end of the angle brac1et in ig % . 8etermine the moment of the forceabout point)p .m*

Solution i (sc%l%! %n%l sishe force is resolved into its x and y components, ig. & , then MO ! 500 sin 30 0 N)0.- m* ' 500 cos 30 0 N)0.5 m* ! '+$." N.m ! +$." N.m

or O ! M'+$." + N.m

Solution ii (vecto! %n%l sisGsing a 7artesian vector approach, the force and position vectors, ig. c , are ! ! M0.5i ' 0.- m# ! M500 sin 30 0i ' 500 cos 30 0 N ! M-00.0 i ' 35".5 N

he moment is therefore

chap two worked example engineering mecha-i

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