Engineering mechanics solutions manual

Solutions ManualFor
v v




Engineering v




Mechanics

Manoj Kumar Harbola
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IIT Kanpur
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1

, Chapterv1


1.1 Rotationalvspeedvofvthevearthv isvveryv smallv (aboutv 7v105vradiansvperv second).v Itsveffe
ctvonvparticlevmotionvovervsmallvdistancesvisvthereforevnegligible.v Thisvwillvnotvbevtruevf
orvintercontinentalvmissiles.
1.2 Thevnetvforcevonvthev(belt+person)vsystemvisvzero.vThisvcanvbevseenvasvfollows.vTovpullv
thevropevup,vthevpersonvalsovpushesvthevgroundvandvthereforevthevbeltvonvwhichvhevisvsta
nding.vThisvgivesvzerovnetvforcevonvthevbelt.vForvthevperson,vthevgroundvpushesvhimvupv
onvthevfeetvbutvthevbeltvpullsvhimvdownvwhenvhevpullsvit,vgivingvavzerovnetvforcevonvhim.
1.3 Avvectorv betweenv coordinatesv (x1,v y1,v z1)v andv (x2,y2,z2)v isv givenv by
(x2v vx1v)iˆvv(vy2v v y1v)vˆjvv(z2v vz1v)k .v Thusv(i),v(ii)vandv(iv)varevequal.
ˆ
1.4 Thevvectorsvarev(i)v 2iˆvv3vˆjv v5kˆ (ii) 4iˆvv3vˆjvvkˆ (iii) 2iˆvv9vĵv v5k̂ (iv)
v3iˆvv3vĵv v2k̂

1.5 (i)vThevresultantvvectorsvarev 6iˆvv6kˆv, 4iˆv v 6v ĵvvv10k̂ andv viˆvv3kˆ
(ii) Thevresultantvvectorsvarevv2iˆvv6vĵv v 4k̂ v ,vv2iˆvv6vĵv v4k̂ andv 7iˆvv3kˆ
1.6 1.6

A
⃗ ⃗
⃗ A B
B ⃗
v



B A
v v
B
A


1.7 Onveachvreflection,vthevsignvofvthevvectorvcomponentvperpendicularvtovthevreflectingv
mirrorvchanges.
z




y
1.8 1.8 v

2 x
O

, Thevflyvisvflyingvalongvthevvectorvfromv(2.5,v2,v0)vtov(5,v4,v4).v Thisvvectorvis

2.5iˆvv2vˆjvv4kˆ 2.5iˆvv2vˆjvv4kˆ
2.5iˆvv2vˆjvv4kˆv.v Thevunitvvectorvinvthisvdirectionvis  .
6.25vv4vv16 26.25

Thevvelocityvofvthevflyvisvtherefore

2.5iˆv v 2v ĵv v 4k̂
0.5v vv0.25iˆv v.20v ĵv v 0.39k̂ v .
26.25


⃗vvv ⃗vvvv
1.9 Aftervtimevt,vthevpositionvvectors rA and rB ofvparticlesvAvandvB,vrespectively,vare
⃗v ⃗v
rA vlvsinvtviˆvvlvcostv ˆj rB vlvsinvtv iˆvvlvcostv ˆj
⃗vvvv ⃗
Theirvvelocitiesv v A andv v B arevgivenvbyvdifferentiatingvthesevvectorsvwithvrespectvto
timevtovget
⃗ ⃗vvvv
vA v lvcos tviˆvvlvsinvtvvĵ vB vlvcos tv iˆvvlvsinvtvvĵ
⃗ ⃗ ⃗
Velocity v AB ofvAvwithvrespectvtovBvisvobtainedvbyvsubtracting v B from v A
⃗ ⃗vvvv ⃗vvvv
v AB v v A vv B v2lvcos tviˆ

1.10 Forvrotationvaboutvthevz-axisvbyvanvanglev
vvx'vv v vvxv cosv vvvyv sinv vvyv'vv v vvxv sinvv vvvyvv cos vvzv'vv v vvz
 
Itvisvgivenvthatv vvv30 ◦v.vvTherefore
  vvyvvvvvv 
v   v
1v 1v
vvx' v 3  vvyv vvy' v v
vvzv' v vvz

x x
2 2
3




3

, 1.11 Componentvofvavvectorv Avalongvanvaxisvisvgivenvbyvitsvprojectionvonvthatvaxis.v Thisvisvob
tainedv byv takingv thev dotv productv ofv thev vectorv withv thev unitv vectorv alongv thatv axis.
Thus
⃗v ⃗ ⃗vv ⃗ ⃗vv ⃗
Avv  Avviˆv  Av cos1 ˆ ˆ
x Ayvvvv Avv jv Av cosv2 Azvvvv Avvkv Av cos3
 
Also 
 v A2v v A2v v A
⃗v 2

2
A
x y z



SubstitutingvthevexpressionvforvAx,vAyvandvAzvcompletesvthevproof.
1.12 (i)vDotvproductvofvtwovvectorsvAvvandv Bv is
⃗v ⃗
AvvBv v AxvBxv v Ayv Byvv v AzvBz

Thisv givesv thevdotvproductvofvthevfirstvvectorv ofvproblemv1.4v eachvofvthevothervvectorsvto
vbev4,v2vandv25.
(ii)vCrossvproductvbetweenvtwovvectorsv Avandv Bvis
A vBv v(AyvBzv AzvByv)iˆvv(AzvBxv vAxvBzv)ˆjvv(AxvB vAyvBxv)kˆ
y


Takingv Av tovbevthevfourthvvectorvandv Bv tovbevthevfirst,vsecondvandvthevthirdvvectorvg
ivesvthevcrossvproductsvtovbe
v9iˆv v11vĵvvv3k̂

v 9iˆv v 5v ĵvvv 21k̂

v 33iˆvvv11v ĵvvvv24k̂

1.13 Ifvthevanglevbetweenvtwovvectorsvisv,vthevcosinevofvthisvanglevisvgivenvby

cosv  A
⃗vv⃗Bv .v Thus
Av B

4
Betweenv (i)vandv (ii)v cosv  v0.127vvvvv82.7 ◦
38v 26
 2
Betweenv (i)v andv (iii)v cosv  v0.037vvvvv88.2 ◦
38v 110
v25
Betweenv(i)vandv(iv)vvcos vvv v 0.865vvvvv149.8 ◦
vvvv v


38vv 22
40
Betweenv (ii)vandv(iii)v cosv  v0.748vvvvv 41.6 ◦
26v 110
v5
Betweenv(ii)vandv(iv)vvcos vvv v 0.209vvvvv102.1◦
v v


26vv 22

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