Mechanical j
Engineering
SolutionsManual
j
, Chapter1 j
Problemsj1-1jthroughj1-4jarejforjstudentjresearch.
1-5
(a)j Pointjvehicles
v
x
cars v 42.1vj−jv2
Qj=j =j =
hourj xj 0.324
Seekjstationaryjpointjmaximum
dQ 42.1j−j2v
=j 0j= ∴jv*j=j 21.05jmph
dvj 0.324
42.1(21.05)j −j 21.052
Q*j= =j1368jcars/h Ans.
0.324
(b) v
l x jl
2 2
µ ¶
j vj j 0.324 lj −1
Qj =j = +j
xj+jlj v(42.1)j−jv2j v
MaximizejQjwithjlj=j10/5280jmi
v Q
22.18 1221.431
22.19 1221.433
22.20 1221.435 ←
22.21 1221.435
22.22 1221.434
1368j−j1221
%jlossjofjthroughputj= =j 12% Ans.
1221
22.2j−j21.05j
(c)j %jincreasejinjspeedj =j5.5%
21.05
Modestjchangejinjoptimaljspeed Ans.
,2 SolutionsjManualj •j Instructor’sjSolutionjManualjtojAccompanyjMechanicaljEngineeringjDesign
1-6 Thisjandjthejfollowingjproblemjmayjbejthejstudent’sjfirstjexperiencejwithjajfigurejofjmerit.
• Formulatejfomjtojreflectjlargerjfigurejofjmeritjforjlargerjmerit.
• Usejajmaximizationjoptimizationjalgorithm.jWhenjonejgetsjintojcomputerjimplementa-
jtionjandjanswersjarejnotjknown,jminimizingjinstead jofjmaximizingjisjthejlargestjerrorjonejc
anjmake.
Σj
FVj =j F1jsinjθj −jWj =j0
Σj
FHj =j−F1jcosjθj −jF2j=j0
Fromjwhich
F1j =j W/sinjθ
F2j=j−Wjcosjθ/sinjθ
fomj= −$j=j −¢γj (volume)
.j
=j−¢γj(l1jA1j+jl2jA2)
F1 W l1
A1j =j j =j , l2 j = j
Sj Sjsinjθj cosjθ
¯jF2¯j j Wj cosjθ
A2j =j =j
¯j Sj ¯j Sjsinjθ
µj j l W l2jWjcosjθ ¶
2
fomj=j −¢γ +j
cosjθj Sjsinjθ Sjsinjθ
µ 2 ¶
−¢γjWl 2 j 1 j+jcos jθj
=
S cosjθj sinjθ
Setjleadingjconstantjtojunity
θj◦ fom
θj*j=j 54.736◦ Ans.
0 −∞ fom*j=j−2.828
20 − 5.86
−
30 −4.04 µ
Alternative: ¶
40 3.22 dj 1 j+jcos2jθj
=j 0
45 −3.00 dθ cosjθj sinjθ
50 2.87
−
Andjsolvejresultingjtran-
54.736 −2.828
scendentalj forjθj*.
60 —2.886
Checkjsecondjderivativejtojseejifjajmaximum,jminimum,jorjpointjofjinflectionjhasjbeenj found.j
Or,jevaluatejfomjonjeitherjsidejofjθj*.
, Chapterj1 3
1-7
(a) x1j+jx2j=j X1j+je1j+jX2j+je2
errorj=jej=j(x1j+jx2)j−j(jX1j+j X2)
=j e1j+je2 Ans.
(b) x1j −j x2j =j X1j+je1j −j(jX2j+je2)
ej =j (x1j−j x2)j−j(jX1j−j X2)j =j e1j−je2 Ans.
(c) x1x2j=j(jX1j+je1)(jX2j+je2)
ej =j x1x2j −j X1jX2j =j X1e2j+µj Xje2e1j+je1eje2j j ¶
=j X1e2j +j X2e1j =j X1jX2j +j Ans.
. 1j 2j
X1 X2
µ ¶
x1j X1j+je1j X1j 1 j+je1/jX1j
(d) =j =j
x2 X2j+je2 X2 1j+je2/jX2
µ ¶− 1 j µ ¶µ ¶j
je2jj . je2j je1jj je2jj . je1j je2j
1j+j =j1j−j and 1j+j 1j−j =j1j+j −j
X2 X 2 X1 X 2 Xj j 1 X 2
µ ¶
x1j X1j .j X1j je1jj je2jj
ej=j −j =j −j Ans.jx2
X2j X2 X1 X2
1-8 √
(a) x1j= 5j=j2.236j067j977j5
X1j =j 2.23 3-correctjdigits
√
x2j=j 6j=j2.449j487j742j78
X2j =j 2.44 3-correctjdigits
√ √
x1j+jx2j=j 5j+j 6j=j4.685j557j720j28
√
e1j =j x1j−j X1j =j 5j−j2.23j =j 0.006j067j977j5
√
e2j =j x2j−j X2j =j 6j−j2.44j =j0.009j489j742j78
√ √
ej=je1j+je2j=j 5j−j2.23j+j 6j−j2.44j=j0.015j557j720j28jSumj=j
x1j+jx2j=j X1j+jX2j+je
=j2.23j+j2.44j+j0.015j557j720j28
=j4.685j557j720j28j(Checks) Ans.
(b) X1j =j 2.24, X2j =j 2.45
√
e1j=j 5j−j2.24j=j−0.003j932j022j50
√
e2j=j 6j−j2.45j=j−0.000j510j257j22
ej=je1j+je2j=j−0.004j442j279j72jSumj
=j X1j+jX2j+je
=j 2.24j+j 2.45j+j (−0.004j 442j 279j 72)
=j4.685j557j720j28 Ans.
Engineering
SolutionsManual
j
, Chapter1 j
Problemsj1-1jthroughj1-4jarejforjstudentjresearch.
1-5
(a)j Pointjvehicles
v
x
cars v 42.1vj−jv2
Qj=j =j =
hourj xj 0.324
Seekjstationaryjpointjmaximum
dQ 42.1j−j2v
=j 0j= ∴jv*j=j 21.05jmph
dvj 0.324
42.1(21.05)j −j 21.052
Q*j= =j1368jcars/h Ans.
0.324
(b) v
l x jl
2 2
µ ¶
j vj j 0.324 lj −1
Qj =j = +j
xj+jlj v(42.1)j−jv2j v
MaximizejQjwithjlj=j10/5280jmi
v Q
22.18 1221.431
22.19 1221.433
22.20 1221.435 ←
22.21 1221.435
22.22 1221.434
1368j−j1221
%jlossjofjthroughputj= =j 12% Ans.
1221
22.2j−j21.05j
(c)j %jincreasejinjspeedj =j5.5%
21.05
Modestjchangejinjoptimaljspeed Ans.
,2 SolutionsjManualj •j Instructor’sjSolutionjManualjtojAccompanyjMechanicaljEngineeringjDesign
1-6 Thisjandjthejfollowingjproblemjmayjbejthejstudent’sjfirstjexperiencejwithjajfigurejofjmerit.
• Formulatejfomjtojreflectjlargerjfigurejofjmeritjforjlargerjmerit.
• Usejajmaximizationjoptimizationjalgorithm.jWhenjonejgetsjintojcomputerjimplementa-
jtionjandjanswersjarejnotjknown,jminimizingjinstead jofjmaximizingjisjthejlargestjerrorjonejc
anjmake.
Σj
FVj =j F1jsinjθj −jWj =j0
Σj
FHj =j−F1jcosjθj −jF2j=j0
Fromjwhich
F1j =j W/sinjθ
F2j=j−Wjcosjθ/sinjθ
fomj= −$j=j −¢γj (volume)
.j
=j−¢γj(l1jA1j+jl2jA2)
F1 W l1
A1j =j j =j , l2 j = j
Sj Sjsinjθj cosjθ
¯jF2¯j j Wj cosjθ
A2j =j =j
¯j Sj ¯j Sjsinjθ
µj j l W l2jWjcosjθ ¶
2
fomj=j −¢γ +j
cosjθj Sjsinjθ Sjsinjθ
µ 2 ¶
−¢γjWl 2 j 1 j+jcos jθj
=
S cosjθj sinjθ
Setjleadingjconstantjtojunity
θj◦ fom
θj*j=j 54.736◦ Ans.
0 −∞ fom*j=j−2.828
20 − 5.86
−
30 −4.04 µ
Alternative: ¶
40 3.22 dj 1 j+jcos2jθj
=j 0
45 −3.00 dθ cosjθj sinjθ
50 2.87
−
Andjsolvejresultingjtran-
54.736 −2.828
scendentalj forjθj*.
60 —2.886
Checkjsecondjderivativejtojseejifjajmaximum,jminimum,jorjpointjofjinflectionjhasjbeenj found.j
Or,jevaluatejfomjonjeitherjsidejofjθj*.
, Chapterj1 3
1-7
(a) x1j+jx2j=j X1j+je1j+jX2j+je2
errorj=jej=j(x1j+jx2)j−j(jX1j+j X2)
=j e1j+je2 Ans.
(b) x1j −j x2j =j X1j+je1j −j(jX2j+je2)
ej =j (x1j−j x2)j−j(jX1j−j X2)j =j e1j−je2 Ans.
(c) x1x2j=j(jX1j+je1)(jX2j+je2)
ej =j x1x2j −j X1jX2j =j X1e2j+µj Xje2e1j+je1eje2j j ¶
=j X1e2j +j X2e1j =j X1jX2j +j Ans.
. 1j 2j
X1 X2
µ ¶
x1j X1j+je1j X1j 1 j+je1/jX1j
(d) =j =j
x2 X2j+je2 X2 1j+je2/jX2
µ ¶− 1 j µ ¶µ ¶j
je2jj . je2j je1jj je2jj . je1j je2j
1j+j =j1j−j and 1j+j 1j−j =j1j+j −j
X2 X 2 X1 X 2 Xj j 1 X 2
µ ¶
x1j X1j .j X1j je1jj je2jj
ej=j −j =j −j Ans.jx2
X2j X2 X1 X2
1-8 √
(a) x1j= 5j=j2.236j067j977j5
X1j =j 2.23 3-correctjdigits
√
x2j=j 6j=j2.449j487j742j78
X2j =j 2.44 3-correctjdigits
√ √
x1j+jx2j=j 5j+j 6j=j4.685j557j720j28
√
e1j =j x1j−j X1j =j 5j−j2.23j =j 0.006j067j977j5
√
e2j =j x2j−j X2j =j 6j−j2.44j =j0.009j489j742j78
√ √
ej=je1j+je2j=j 5j−j2.23j+j 6j−j2.44j=j0.015j557j720j28jSumj=j
x1j+jx2j=j X1j+jX2j+je
=j2.23j+j2.44j+j0.015j557j720j28
=j4.685j557j720j28j(Checks) Ans.
(b) X1j =j 2.24, X2j =j 2.45
√
e1j=j 5j−j2.24j=j−0.003j932j022j50
√
e2j=j 6j−j2.45j=j−0.000j510j257j22
ej=je1j+je2j=j−0.004j442j279j72jSumj
=j X1j+jX2j+je
=j 2.24j+j 2.45j+j (−0.004j 442j 279j 72)
=j4.685j557j720j28 Ans.
