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Unformatted text preview: 278 Solutions Manual Problem 4.5ﬂ An atomic force microscope (AFM) is a state—ofthe—art device used to study
the mechanical and topological properties of surfaces on length scales as small
as the size of individual atoms. The device uses a ﬂexible cantilever beam AB
with a very sharp, stiff tip BC that is brought into contact with the surface to be studied. Due to contact forces at C, the cantilever beam deﬂects. If the tip
of the AFM is subjected to the forces shown. determine the resultant moment of
both forces about point A. Use both scalar and vector approaches. Solution For the scalar approach we use Eq. (4.1) on page l82. Noting that positive moment is counterclockwise, MA
is given by 10‘9N 10‘6m
~(l3nN)(12/tm) + (5nN)(l.5,um) = —l49nN;tm (l)
, nN ,um
= —1.49x10—‘3Nm. (2)
For the vector approach we use Eq. (4.2) on page 183. It follows that
FAC=(il2ie—l.5j)um (3)
ﬁ=(57+i3j)nN (4)
and MA is given by
= are x f = (5)
—l49l€nN,um (6)
—1.49x10“‘3I€Nm. (7) M Thh solutions manual. in tiny piint or electronic lomi. temains the propeir) ot i\v'ICGle\\rHIH [in It may be that] rind/oi possesxcd only by peiTnission AUgUSl 24. 2009
ol .‘vchrawHill. and must be \un'entlei'ed upon request of .\lc(jia\\rllil. .\n} tluplittinon or dixtiibution. either in print or CIEL‘II'OIHC l'oim. “liliout the
permission ol Mt‘Graanill. is prohibited. Statics 10 281 Problem 4.7l The ball of a trailer hitch is subjected to a force F. If failure occurs when the
moment of F about point A reaches l0,000in.lb, determine the largest value
F may have and specify the value of oz for which the moment about A is the largest. Note that the value of F you determine must not produce a moment F
about A that exceeds 10,000 in.lb for any possible value of oz. Solution The maximum moment for a given value of force F occurs when the moment arm
is at its maximum. Using the Pythagorean theorem in the ﬁgure to the right, the
maximum value of the moment arm d is given by d = ‘/(9in.)2 + (6 in.)2 =10.8in. (1) We then use Eq. (4.1) on page 182 to determine the maximum force F, given that the moment of this force
about point A may not exceed 10,000in.lb. Thus, MA 2 Fa’ = F(10.8in.) : 10.000in.1b => F : 9261b. (2) The angle 0: at which this maximum force acts is then found to be dcosﬁ = 9m. :> i3 = 33.6° (3) a: 180°—(l80°~f3—90°)= 123.6°. (4) ———————.—__—____—____—
This solutions mnnuttl. in tin) print or L‘IL‘LII'UIHL lonn. remains the [nopeny oi McGrtmrHill. Inc. It may be used and/oi possessed onI} h} DL‘l'llll\\)I1 August 24. 2009 oi McGimvllil]. and muxi be suricntleretl upon request ol‘ McGrtnvllil]. An) duplication oi distribution. eithcl In print or electronic iorm. thout the
permission of McGrmthll. is prohibited 288 Solutions Manual Problem 4.1M A ﬂat rectangular plate is subjected to the forces shown, where all forces are
parallel to the x or y axis. If F = 200N and P = 300 N, determine the
resultant momentof all forces about the (a) Z. axis. (b) a axis, which is parallel to the Z. axis. Solution Part (a) We use Eq. (4.1) on page 182 to ﬁnd F(300 mm) — F(500 mm) + P(200 mm) — P(6OO mm) (1)
——1.6X105Nmm= —l6ONm. (2) In writing the above expression, positive moment is taken to be in the positive :, direction, according to the
right—hand rule. Part (b) We use Eq. (4.1) again to ﬁnd : —F(300 mm) + F(200 mm) — P(500 mm) (3)
—~1.7X105Nmm=~170Nm. g (4) In writing the above expression, positive moment is taken to be in the positive a direction, according to the
righthand rule. ———————————*____—_ This wlutinm manual, Ill an} plinl 01' Electronic I'nnn. remains the pmpcny l)"l\/1C(llll\\’rlllll, Inc, [I may he U\CLl and/m possessed (ml) by permixsmn AUgUSI 24, 2009
ul tVlL’Gl'H“rHlll. and muxt be surrendered upon request of McGrmvllill. Any duplication in tlixmhulinn, euhcx in piinl or electronic hmn. \\ll10u( [he
pel mission or McGraqull, H pmhibued, 294 Solutions Manual
Problem 4.191 Repeat Prob. 4.18, using a scalar approach. Solution We begin by adding the two cable forces at point A to obtain [714, and adding the two cable forces at point B
to obtain FB, as follows .  4 —60j—80/€ _ 150j—80/2 A A  F 2F F =250N :230N = 221 —150 —318k N. 1
A Ac+ AD 100 170 ( l j ) () F“ ~ f + 1E —100N150fﬂ200k + lOON————150j—200k — (60A+60 ‘ 16012)N (2)
B — BD BE — 250 250 — I J  The components of FA and F3 are shown in the ﬁgure below. Part (3) Taking positive moment components to be in the positive coor—
dinate directions, we use Eq. (4.1) on page 182 to write (MA)x = —60N(120 mm) = 47200Nmm (3)
(MA)y = 60N(120 mm) = 7200Nmm (4)
(MA)z = 0 (5) Part (b) Taking positive moment components to be in the positive coor—
dinate directions, we use Eq. (4.1) on page 182 to write (Molt = —60N(200mm) + 150N(80 mm) = ONmm (6)
(Mon = 6ON(200 mm) + 221 N(80 mm) : 29,700Nmm (7)
(M0)z = 0 (8) —_——h_—____—_“— Thix mlutionx manual. in tiny print or electronic form. temains the plttpCI’l) ol \ltGrzm Hill, Inc It mu) he used and/or possexsetl only by pcnntninu August 24 2009
of McGraqutll. uml must be surrendered upon tequest of McGrtmHtll An} tlnpltmtmn tvr disttthuttun. eithct in print or electiuntc lot'm. without the
petmissiun m chGrzm rHlll. is ptohibitcd, 300 Solutions Manual Problem 4.241 Forces of 3 kN and 200 N are exerted at points B and C of the main rotor of a
helicopter, and force F is exerted at point D on the tail rotor. The 3 kN forces
are parallel to the z axis, the 200 N forces are perpendicular to the main rotor
and are parallel to the xy plane, F is parallel to the y axis, and or = 45°. (3) Determine the value of F so that the 3, component of the moment about
point 0 of all rotor forces is zero. (b) Using the value of F found in Part (a), determine the resultant moment of
all rotor forces about point 0. (c) If 01 is different than 45°, do your answers to Parts (3) and (b) change?
Explain. Solution Part (a) We use Eq. (4.1) on page 182 to evaluate the moment of the rotor forces about the z. axis, and as
instructed in the problem statement, this moment is required to be zero. Thus, we obtain (M0)z = —200N(3 m) 4 200N(3 m) + F(5m) = 0 2 F : 24ON. (1) Part (b) Using the geometry provided in the problem statement, with the aid of the sketch below to
determine x and y components of the main rotor forces, the following vector expressions may be written f3 = 200N (—sin 45° i — cos 45° j) + 3000N1€ (2)
EC = ZOON (sin 45° 1‘ + cos45° j) + 3000NI€ (3)
ft) = F (—j) (4)
Fog=3m(—cos45°i+sm45°j)+1.812m (5)
fog : 8m(cos45°f—sin45°j) + 1.812m (6)
PODz—sim+1.212m (7) Using Eq. (4.2) on page 183, the moment about point 0 of all rotor forces is M02703XﬁB+FOCXﬁC+FODXﬁD (8)
r j 12
= 3cos 45° —3 sin 45° 1.8 Nm ‘ —200 sin 45° —200cos 45° 3000 (9
A A ) i j k i j k + i3cos45° 3sin45o 1.8 Nml —5 0 1.2 Nm
200 sin 45° 200 cos 45° 3000 0 e240 0 = [i(»6109) — j(6618) + é(—600)] Nm + [i(6109) 2 j(—6618)+l§(—600)]Nm + [i(288)—J(0)+k(l200)]Nm. (10)
“WE——
This solutiom manual. In any print or electronic form. remains the property of .\1c(in\\>Hill. Inc. It may be used and/m p0\\6\88d only by nernnxxmn August 24 2009
of McGrawHill. und must be surrendered upon request ()1 MtGrauHill Any duplication or distribution. either in print m electronic turn]. \\[h\ltH the permission of McGrauHill. \ ptuhihuml Statics l e 301 M0 = (288?) Nm. Part (c) If 01 3E 45°, the answer to Part (a) does not change, since the moment produced by the main rotor
forces about the z axis is not a function ofoz. The answer to Part (b) does not change, for the following reason.
The main rotor forces produce moments about the x and y axes that are functions of a, but the moment arms for components of force at points B and C are always opposite (negative of one another) so that the net x
and y components are zero regardless of oz. which simpliﬁes to (11) ———————————__h This solutions manual. in an} punt or Elt‘L‘llUniC Ioi'ni, ienmns the pioperi) oi NICGJ'ZJHVHl”. lnt. Ii ma) be used nntl/or possessed onl} h} permission AllgllSl 24. 2009
ol McGrtmrllilL nntl must be surrendeietl upon tequesl ol .\ik(ilil\\riilii, :\n_\ duplication or disn'ibulion, eiihei in pIInl or electronic form. miliout the
pctniission ol McGImvl lill. l\ piohiblmd, 328 Solutions Manual Problem 4.4M The top View of a workpiece that ﬁts loosely in a ﬁxture for drilling is shown.
The drill bit has two edges that apply in—plane cutting forces F to the workpiece. (a) If F = 600 N, determine the forces Q between the workpiece and ﬁxture
so that the resultant couple momentis zero when 0: = 30°. (b) Does your answer for Q from Part (a) Change if or has different value? If
yes, then repeat Part (a) with a = 60°. Solution Part (a) The workpiece is subjected to two couples. Equation (4.12) on page 209 is used, with positive
moment being counterclockwise, to determine the resultant moment of the two couples as M : Q(150mm)—F(30mm). (1) As instructed in the problem statement, M = 0. Thus, we set Eq. (1) equal to zero. and using F = 600 N, we solve for Q to obtain
Q = 120N. (2)
The answer obtained in Part (a) is valid for any value of 01. Part (b) M Phi.» solurlonx manual. in an) print m electronic lonn. remains the propeny of McGruu l’llll. lnt ll ma) be met] and/or possessed only h) ]3€‘lll\\iull AUgUSl 24, 2009
ol Mt‘GrmvHill. and Illlhl be suncntlcred upon request ol‘ A1C<llklurHllL Any tluplicnnon m lelIIhLIlllllt. enlm in prim (n electinnic lmm. \xillmur [he
pernnssion ul McGluu rHlll. is prohibited Statics 10 329 Problem 4.473 Three tugboats are used to turn a barge in a narrow channel. To avoid producing HSOﬁaHOH
any net translation of the barge the forces applied should be couples. The 3012140011, tugboat at point A applies a 400 lb force. ,2 {:
B (a) Determine F3 and FC so that only couples are applied. " "
I’D/SS1,
(b) Using your answers to Part (3), determine the resultant couple moment that is produced. (0) Resolve the forces at A and B into X and 1' components, and identify the
pairs of forces that constitute couples. Solution Part (a) In order to have loading due to couple forces only, the resultant of all forces must be zero. Hence,
FRX=ZFX=O: (4001b)sin30°—F3sin45°=0. (1)
FR, 2 Z Fy = 0; (400 lb) cos 30° — FB cos 45° _ FC 2 0. (2) Solving these equations provides FB : 282.81b. (3)
FC 2 146.4lb. (4) Part (b) With the values given in Eqs. (3) and (4), the system of forces are couples, and the moment
produced by these, taking positive moment to be counterclockwise, is MR : Z M3 = “(400 lb) cos 30°(50 ft) — (4001b) sin 30°(44 ft) — FC (40 ft), (5) MR 2 —31.980 ftlb. (6) Note that the same value of MR would be obtained regardless of the summation point used in Eq. (5), because
the forces are couples. which provides Part (c) This sketch below illustrates two force systems that are equivalent. The original force system is
shown on the left, with forces resolved into .\~ and y directions. The force system on the right shows more
explicitly the three couples that are applied to the barge. 200 lb 346.4lb 146.41b T 146.4 lb
' 146.4 lb l l 200 lb y Statics 10 333 Problem 4.51 l A structure built in at point 0 supports 70 and 85N couples and a tip mo—
ment. Determine the resultant couple moment, using both scalar and vector
approaches. Solution From the geometry provided in the problem statement, we write vector expressions for the forces applied at
points B and D as ﬁB=(7ON)%(~6i+3j—ZI€). (1)
ED 2 (85 N) %(—8i— 12)“ + 912). (2) For the scalar approach, we evaluate the moments of each of the x, y, and 1 components of the 7ON and
85 N forces, taking positive moments to be in the positive coordinate directions according to the right—hand
rule. Thus, using Eq. (4.12) on page 209, we obtain _. 3 6 A
9 A A 8 A A
+85N ﬁ 2ml+0j+85N ﬁ ka—ZOOINm, (3) which provides M = (—50t+120j+8012)N~m. (4) For the vector approach, the following position vectors are needed fAB =—2km, 7CD=2fm (5) We use Eq. (4.11) on page 209 to write M=7ABXﬁB+;CDXﬁD+A7IE (6)
70N~m ’ 1 k 85Nm ’ 1 k A
= 7 0 0 —2 + 17 0 2 O —2001Nm (7)
*6 3 —2 —8 —l2 9
70 Nm A _ A 85  A A A A
= 7 [i(6)—j(—I2)+k(0)]+ [1(18)—j(0)+k(16)]—2001Nm. (8)
which provides
M = (—50i+120j+80/€)Nm. (9)
—————————_———__——
This solutions manual. in tiny pnnt or electronic tm'm. remains the pmpcn) 0il\1c(jlt\\r'el‘llll. Inc [I ma} he used tmd/tu posu‘xscd onl) h) pCl'IHi\\tt\I1 August 24 2009
nt .N’lL'Gl'llurHlll. and must he \utiendered upon request of McGttmlhll An) duplu‘ntmn or distribution. either in print or eluttinntt hum. \\tthnu the permission til Mcﬁttm 7H1“. is pinlnhned 338 Solutions iManual Problem 4.561 Forces F and T exerted by air on a rotating airplane propeller are shown. Forces
F lie in the .ry plane and are normal to the axis of each propeller blade, and thrust forces T act in the z direction. Show that the forces F can be represented
as a couple or system of couples. Solution The ﬁgures shown at the right are views y
looking down the z axis toward the pro i
peller; forces T are not shown and these
are irrelevant to the question being asked.
We resolve the forces F into horizontal
and vertical components, as shown in the
second ﬁgure at the right. We then ob
serve that the vertical force F cos 30° in
the positive y direction and the vertical
force F cos 30° in the negative y direction
constitute a couple. The sum of the two
horizontal forces F sin 30° in the negative
x direction and the force F in the positive
x direction constitute another couple. View of the propeller
looking down the z axis ————§__——_“_ This solutions manual. in any print or electronic tonn. remains the prime") of McGiianill. l1\_, It may be used and/or possessed oiin h} permission August 24 2009
oi McGi’an Hill .ind must be surrendered upon request Di lVlL‘Gl‘anHlll .\ny duplication oi distribution. either in prim oi ClL‘CII'UIHt‘ Iorin. without the
permission oi MtGitmrllill. is prohibited Statics II? 339 Problem 4.571 If the structure is subjected to couple forcesAapplied at points A and B and the ‘
force applied at A is F 2 (8? + 10] — 40 k) lb, determine the moment of the z \ a
couple about line a. Line a has direction angles 6X = 72°, 8y 2 36°, and
c
0 A\
F— y QZ = 60°.
A (16, 36, 10) in.
x B (—8, 24, 18) in. Solution Using Eq. (4.] l) on page 209, the moment M of the couple is is given by szgAxlEAz(247+l2j—81€)in.x(8i+l()j~401€)lb (1)
i j 12 = 24 12 ~8 111.113 (2)
8 10 ~40 = (—4001“+ 896] +14412)1n.1b. (3) The component of M in the direction of line a is then given by Ma = A7! ~51 = (—400; + 896] 4 14412)1n..1b(cos72°1+ cos 36° j + cos60°12) (4)
(5) M This soluiions manual. in any prinl or electronic ton“, remains 1h: plopcn} ol .\lc(l111\\'rllil. Inc [1 may be used and/or possessed only by p61 mission August 24. 2009
ol Mchleill. and must be sun'cndered upon requexl ol McGlzanill A11) dupliczllion or rlislnbution. either in print or elemonlc form. without the
permission ol McGrmvHill. is prohlhuetl. ...
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