S L Loney statics Dynamics with Solutions

396 Pages·17.92 MB·English

SOLUTIONS OF THE EXAMPLES
THE ELEMENTS
OF
STATICS AND DYNAMICS
BY
LONEY
S. L.
CAMBRIDGE:
AT THE UNIVERSITY PRESS
.
Published July 1893
,
Second Edition, 3899
Third Edition with additions 1902
, ,
Fourth Edition additions 1906
, ,
Bepiinted 1910, 1913, 1917, 1923, 1927, 193J 1940
PRINTED IN GREAT BRITAIN
PREFACE.
For the following Solutions of the Examples in my
Elements of Statics and Dynamics I am almost entirely
indebted to a friend, to whom my best thanks are due
Ho has also carefully revised the whole of the proof-
sheets.
I hope these solutions will be found usoful to
teachers and private students.
LONEY.
S. L.
Royal Holloway College,
Egham Surrey.
July 3, 18(J3.
The Fourth Edition has been altered so that the
solutions correspond with the Tenth Edition of the
Elements of Statics and Dynamics.
October 15, 1906.

.
ELEMENTS OF STATICS.—SOLUTIONS.
EXAMPLES. I. (Pages 15, 16.)
1. (i) R=,v/(24)a+7J=n/C26=25.
(ii) Q=J(14)2-(13)s=^27=8^3.
(iii) ii=,/7»+82+277.8cos60°=^169=13.
(iv) R=js>+‘>+2.5.9cos'120"=Jbl.
2. (v) 7a=3a+5,+2.3.5cosa,
whence cosa=g, i.e. a=60°.
3,
(vi) COBa=Jl^aWa= \~(i|)*=±^‘.
" (13)2+(14)S±12-13-14-^=-v/sOS; or 15.
(vii) 72=5,+Qa+2(5.<J)cos60°,
whence Qa+5Q-24=0, and so Q=3.
The resultant of two forces is greatest or least according as
theyactinthe same straightlinein thesame direction orin opposite
directions. Hence [cf. Art. 23], the greatest resultant of forces of
r1e2sulblst.anwtt.=a(n12d-88l)bsl.bsw.tw.t=.=(142+lbs8.)wltb.s. wt.=20lbs. wt.; andtheleast
TheforcesofSlbs. wt. and4lbs.wt. actin thesamestraight
lineinoppositedirections,andare, therefore, equivalent toaforceof
1 lb.wt. in the direction of the foroe of 4lbs.wt., i.e south. The
forcesof6lbs. wt.and6 lbs. wt. also actinthe samestraightlinein
opposite directions, andare, therefore, equivalent to a force of 1lb.
wt. in the direction of the force of 6lbs. wt., i.e. west. Hence the
fourforcesareequivalenttotwoforcesof 1lb.wt.each, actingatright
angles; therefore the resultants*yiHl*=^2lb. wt., and evidently
actsalongalinebiseotingtheanglebetween thelinesofaotionofthe
forcesof4lbs. wt.and6lbs. wt.,i.e. inadirectionsouth-west.
L. S. K, 1
V
P
,
2 ELEMENTS OF STATICS Exs.
4. The resultant= =s/4'2025=206lbs. wt.
6. Here R=JP>+(PJ2—)»+2.P.Pj2eos135°=P lbs. wt. at
an angle tan-1P+P^oos135°, i.e. tan-loo, i.e. at right angles to
the direction of the first component.
0. IfQbetherequiredforce, wehave
8. ( 3)*=2a4 Qa4-2.2. Qcos60°,
whence Qfl+2Q-8=0, and therefore Q=21bs. wt.
79.. If tana=1o2 then cosa lo.
P=/v/(13)’+(ll)J+2.13,ll.A=N/400=201bs. wt.
Iftana=^, then cosa=?.
o 5
R=eJ(10)2+9J+2.10.9 *= V289=171bs.wt.
Ifthe forces be each equal to P, a be the angle between them,
andRbe theirresultant, we have
i22=3P8, so that 3P2= 2(2+2cosa)
whence cosa=(j, i.e. a=60°.
10. IfPand Qbe therequiredforces, we have
U/10)»=P*+Q», and (n/13)j=P4+Q2+2P(i)cob60o,
i.e. P»+g3=10, and Fi+Q‘*+PQ=13.
Solvingthese equations, wehave
P=3lbs. wt., and <2=1 lb. wt.
11. (i) p=pV*u+cosa), i.e. 1=2+2cosa;
whence cosa=-i, i.e. a=120°.
(2) ^=P*J2(1+cosa), i.e. ^=2+2cosa;
whence cosa=- ue. a=cos-1f-x)=151°3'.
£
~
I SOLUTIONS 3
12. Here, ifabetherequiredangle, wehave
( + 2)a=[A+1])*+{A-13)*+2(1+B)(.A-B)coso,
swohetnhcaet ooso=da+A2*B(+aA=1&2-(Bdya+iB.es.)+a=2c(o4ss-B8)cl2osdAa*a,+-Biiaa\)‘
thi.r1d3.gi_veFninfdortch^e;rtehseulgtraenatte(sRt)roefsutlhteanttwoRgainvden8focracnesh;al\eot Sisbiei+thSe
mwuhsetnatchteyinacttheinditrheectsiaonmeofdiRr.ection in thesame straightline; i.t, S
14. Takethe figureof Art. 27.
(i) Make CM =5 ins., lA0B~%1°, and cut off 0B=7| ins.;
complete the parallelogram OACB; then OC is Ji.
[2unitsofforce=oneinch.]
(ii) Make 0A=4^ ins., lAOB 133° and cut off 0B=3^ ins.;
complete the parallelogram OACB; then OC=ll.
radi(iii5i)anMda2k^eiOnsA.=to3imeients.i;nwCi;thcocemnptlreetse0thaendpaAradlelseclroigbreamciOrcAleCsBof
then zAOB=a. ;
(iv) Make 0A=3-05 ins. and zdOB=65°; draw AG parallel to
OB; with centre 0and radius 4*35 describe acircle tocutACin C;
complete the parallelogram OACB then OB is Q.
;
1—8
P
,
ELEMENTS OF STATICS Exb.
EXAMPLES. II. (Pages 19, 20.)
1. Theresolvedpartsare 10oos80° and 10sin30°, respectively
i.e. 5^/3lbs. wt. and 6 lbs. wt.
2. (1) Pcos45°, i.e. ^P^/2. (2) Pcos^cos— i.e. 1|p.
3. Therequired force=100cos60°=50lbs. wt.
4. Iftherequired forcesbeeachequal toP, wehave
(100)2= 2(2+2cos60°);
whence 8P2=(100)a and P=i5|V?=67-785lbs. wt.
,
5. If*an<Jy be the required forcesrespectively, wehave
ainx45°_“siny60°"sin51005°^"'cob6016°“_cos(465°0-30°)
=^x=60 8-1>-
Hence *=60(iv/3-l)=36‘603lbs. wt.,
and y=25(-t/18-^6)=44*83 lbs. wt., nearly.
6. If*andybethe requiredcomponentsrespectively, wehave
sin*45°_"siny80°"_EtPT0”_sin(45P°+80°).“-^P/jX+V12 ^n f-
Hence x=P(*y3-l), and (^6-^/2).
7. Therequired forced
8. IfPand Qbetherequiredforcesrespectively, wehave
P=Ptan60°-P,y3, and Q=Pseo60°=2P.
=
’ . ;
n
SOLUTIONS 5
and9,P bIefaatfroirgchetFanbgelesrestoolFvedanidnteoqutawlotcooimtpionnmeangtniftourcdees,PthaenndtQh,e
otheranglesareeach45°,andQ=Pyf2 F,J2. Alsotheanglebetween
thecomponentforcesis135°.
'10. Draw OB vertical and equal to 20 units of length, and OA
horizontal and equal to 10 units of length. Completetheparallelo-
gram OABC. Then OCrepresents theotherforce.
ClearlyOC=AB=J(20)1+(10)a=10*/5=22*86lbs. wt.
AlsotanCOB=tanOBA=|, sothat the inclination tothe vertical
=tan-i|=26°84'.
11. InFig.Art. 84make
OC=3Jins., LCOA=98° and LCOB=iO°.
EXAMPLES. ILL (Pages 25, 26.)
1. IfP, QandRbe the forces, wehare, by Lami’s Theorem,
P=Q=B.
(i)
1 ' sin150° sin150°“sin60°
hence P :Q: J2=1 : 1 :
2. IfP, Q andR be theforoes, wehave, by Lami’sTheorem,
P Q R
Bin120° sin150°“sin90°,*
hence P: Q : R=tJ‘d : 1 : 2.
3. Since 7Pisequaltqtheresultantof 6Pand 8P, wehave, if a
be therequiredangle,
(7P)2=(5P)a+(8P)a+2 5P . 8Pcosa,
whence cosa=-~> i*e, a=120°.
for4R.. DTrhaewsaidfeisguOrLe,asLNinaArntd.3N8O,woiftht1he2PtrfioarnPgl,e5OPLfoNraQr,eapnrdop1o3rP-
atinognlael OtoLN12,is5tahenrdef1o8rereaspreicgthitvealnyg;lea.nd, since (13)a-(12)a+5a, the
Again,tanLOAT=^=•4166667; thereforetheangleLON=22°87'.
Hencetheanglebetweenthedirectionsof the forces 5Pand 12P
=MOL=OLN=90°
betweenthedirectionsoftheforces 12Pand 13Ptheangle
=180°-22°&7'=157°23',
and, therefore, between the directions of the forces 18P and 5P the
angle«112°87'.