Forsynthesis,the k in [ ] [ ] dCdtkC =− isequalto. s k Fordegradation, k is. d k Thus
wehave [ ] [] [ ] [] and,sd sd dCdC kCkC dtdt ==−
whichif
⇒=−
dC kCkC dt
⇒=−
dC kkC dt
⇒=
[ ] [ ] [ ] [] [][] [] () [] [] [] sd sd sd o dCdC dC dtdtdt
dC kC dt =+
where o k istheoverallreactionrate.Integratingwithrespecttotime,wehave
dC dtkCdt dt
dCkCtc
⇒=+
⇒−=
⇒==
CkCtc ktCc cc C ktkkt = ⇒=+
[ ] [] [][] [][] () [] [] () 1 1 1 11 1 11, o o o o osd
∫ wherewecouldcalculate1 c ifweweregivenaninitialconcentration.Notethatif
sdsd
sdsd
sdsd
1.17 (xx-Insertfigureshowingset-up)
Letpoint p beapointhalfwaybetweentheappliedforces.Forconvenience,letthepoint belocatedattheoriginofa2-DCartesiancoordinatesystem,withtheappliedforceslocatedat
2 x d = and2, x d =− andtheforcesorientedinthe y directionsuchthat1 ˆ F = Fj and 2 ˆ F =− Fj Computingthemomentsatpoint p,wehave ) 1122, p =×+× ∑ MrFrF whereat p, 1 ˆ 12 ri d = and2
Ifwechooseanarbitrarypoint a locatedbythevector
but ˆˆ , ×= 0 jj leavinguswith
1.18
UsingtheequationsfoundinexampleA1.2,wehave
andtherefore,
0,0,and.xyz T R RRT =−=− ⇒===− RFk
Similarly,forthereactionmoments,wehave ()() ()() ()() 0 00 0000 ,0,and0. xAyAz yAzAx zAxAy xyz MzFyFLLT MxFzFTL
⇒=++=
⇒=−−
⇒=−−
⇒=−− e
(xx-Insert3-Ddiagramshowinganglesalpha,beta,andgammafrom x, y,and z axes, respectively). If ˆˆˆ ˆ coscoscos, α =++βγei j k then ( ) ( ) ( ) 222 2222 122 122 122 ˆ1coscoscos 1coscoscos1 cos1coscos cos1coscos cos1coscos. α βγ αβγ α βγ β αγ γ αβ ==++
Fora2-Dproblem, ˆˆ ˆ coscos, α β =+ eij andthusthecos z AA γ = termmustequalzero, i.e.2. γ π = Thuswehave ( ) ( ) 222 cos01coscos11 2 a π β
==−−=−
wherefor2-D,22ˆ1coscos. α β ==+ e Thustherelationshipbetweenanglesholdsifthe problemisstrictly2-D.
1.20
Becauseapincannotsupportareactionmoment,themomentscreatedby1 F and2 F mustbalance.Therefore,ifweuseastandardCartesiancoordinatesystem,wehave
1.21
(xx-InsertFBDofweightmass mw,gravityforcein–j direction,force T1in j direction)
(xx-InsertFBDofloadcell-force T2in i direction)
(xx-InsertFBDofpulley, T2in–i, T1in–j,reactionforceatanglealpha)
Assumingaweightlesspulleyandsupport,andthatthepulleyisideal(frictionless)i.e.
12, = TT wehavefromthefreebodydiagram(FBD)oftheweight,
Thus219.8ms9.8N, ww Tgmm === where w m isthemassoftheweightinkilogramsandthe gravitationalconstantisapproximatedby9.8m/s2.Therefore,theforceappliedtotheforce
transduceris219.8N. w TTm ==
FromtheFBDofthepulleysupport,wehaveareactionforce
cossin,rr α α =+ rij whichgivesus
Themagnitudeof r isthus2 21 T orientedalongtheaxisofthesupport.
Inadditiontosupportingforces,thebaseofthepulleysupport, r,canalsosustaina reactionmoment,givingus
Notethatif α =45o,cossin, α α = andthusthereisnoreactionmomentatthebaseofthepulley support. 1.22
(xx-insertFBDofblock,withcoordinatesystemorientedonslant)
(xx-insertdecompositionofgravityinto x and y directions)
Theblockisinequilibriumjustbeforeitbeginstoslip.Therefore,usingeq(A1.8),we have
1.23
(xx-InsertFBDofentiretruss,withjointsidentifiedasa(thetoprightjoint),b(thebottomleft joint),c(topleftjoint),andd(bottomrightjoint))
(xx-InsertFBDofeachofthefourpointsofinterest(4panelfigure))
Bygeometry,theangleswithintheequilateraltriangleareall60o,andtheothertriangle isa30o-60o-90otriangle,withsideshavingalengthof L/2,32, L and L.(Usefultrigonometric quantities: ()() sin60cos3032, °=°= ( ) ( ) sin30cos6012. °=°= )Notethattherollercan onlyproduceareactionforceperpendiculartotheplaneoftheroller,andthepinconnections foundthroughoutthetrusscanonlysupportreactionforces,notmoments.
Fortheentiretruss,wehave
Inordertofullydefine ay R and, Rby wemustintroduceanotherequation,inthiscaseasumof moments.Takingthesumofmomentsaboutpoint b,wehave
Bymakingfictitiouscutsbetweenthepointsofinterest,wecandeterminetheforces placedontheindividualpins.Theforcesonthemembersareequalandoppositethatenforcedon thembythepins.Althoughweconsiderthepointsinisolation,theforcesfeltateachpinare relatedsuchthat11, ac FF = 21,ad FF = 13,bc FF = 23,bd FF = and22. cd FF = Summingourforces, wehave
whereweseethat13 bc FF = asitshould.
whereweseethat23, bd FF = and22 cd FF = asexpected.
Finally,thediagonalmembersareincompression,becausetheforcesactingonthepins arenegative(andthustheforcesonthemembersactintothemembers,orarepositive),whilethe verticalmemberandthebottomhorizontalmemberareintension,becausetheforcesactingon thepinsarepositive(andthustheforcesonthemembersactoutofthemembers,orare negative).Thetophorizontalmember,onwhichnoforceacts,isneitherintensionnor compression.
(xx-Insertfigureshowingelementsintensionandcompression)
